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1 then am is an element other than s of (ai, ... (an, _1)P. Since the latter is a p-product, because of the induction assumption, then
ate,=ai...aM-1 where it is prime top and every ik (k > 2) equals 0 or is prime to p. From
a=ai ... am-, Proposition 4 then follows.
mPtlmam+l...ann
FINITE ABELIAN GROUPS
372
PROPOSITION 5. If in' two p -products the first factors are different and all their factors occur among the factors of one and the same Hajds product, then these p-products have no common factor. For the purpose of the proof we assume that the assertion is false. After
cancelling some of the last factors of the two given p-products we reach a point where only the last factors of both p-products are equal. Let this common factor be (cc)p. Then the p-products obtained are of the form P(a)p and Q(a)p, respectively, where P, Q are some p-products. It follows that ap (A s) is an element of P as well as of Q. But the factors of PQ are different and are factors of one and the same Hajbs product. This contradiction proves Proposition 5. LEMMA. I f al, ..., a" (n >_ 1) are elements of an Abelian p -group with the property that for G. = {ocl, . . ., a,,) and for all the Gk = {a1,, ..., a;k)
(1 I we assume
the assertion for smaller values of o(al) ... o(a,J. It should be noted that, by (92.12), o(al), ..., o(an) >--_ p.
First we consider the case, where > always holds in (92.12). Then O(Gk) Z pk+1
(92.14)
for all the Gk with 1 aac (= ma)
(a E G)
(93.5)
certainly are characters of G. If in this way all the different characters of G occur, it means that, conversely, c0 may be considered as the char-
icter group of G. In fact, this case always happens, as is implied by THEOREM 219. Every finite Abelian group d is isomorphic to its character
group G, and furthermore by (93.5) cPi may be regarded as the character group of G. In the proof we shall actually construct the character group G of 4t. For this purpose we need a prime-power basis
al, ..., a,
(93.6)
of d1. The invariants of dt are then the orders Pi = o(al), . . ., P, = o(ar);
(93.7)
m = [PI, ..., P,]*.
(93.8)
furthermore, we have
By,8," we denote the cyclic subgroup of ,g of order m, which consists of those elements of 8 whose order is a divisor of m. We select from $.
378
FINITE ABELIAN GROUPS
arbitrary elements j,
., 3r with
o($I) = F1,. .., o0r) = Pr
(93.9)
which shall remain fixed throughout the following. We now write the elements a of 4 in their basis representation
a=a...ar
(ik=0,...,Pk- 1; k=
(93.10)
and prove that all the characters of 4, i.e. the elements a of the character group G, are given by a
as = $i'' ... $:'r
(jk = 0, ..., Pk - 1; k = 1, ..., r).
(93.11)
On the one hand, by the homomorphism (93.4), every character a of C4 satisfies o(aak) I o(ak); hence, by (93.7) and (93.9), the existence of certain equations (jk = O, ..., Pk - 1; k = 1, ..., r) aak = $k follows. This shows, by (93.2) and (93.10), that a is in fact given by (93.11).
On the other hand because of (93.9), it is evident that only distinct characters, i.e., homomorphisms a of 4 into 8, are given by (93.11). This proves the statement relative to (93.11). Now let us take the elements ak (k = 1, . . ., r) of G, which are obtained
from (93.11) by putting jk = 1 and j, = 0 (10 k; 1 < l 5 r). By (93.3) and (93.11) it follows that
a = al' ... aJrr
(jk = 0, ..., PA: - 1;
k = 1, ..., r)
(93.12)
are all the distinct elements of G, and furthermore that
o(ai) = PI, ..., o(ar) = Pr
(93.13)
follows from (93.9). Hence we see that ai, . . ., ar is a prime power basis of the character group G and its invariants agree with those of (4. Consequently the first assertion of Theorem 219 has been proved.
Since according to this O(G) = 0(d4), it will suffice to prove with respect to the second assertion of the theorem that distinct mappings (of G into 8) are provided by (93.5) for different a. Because of (93.2), it is sufficient to show that as can be the unity element of $ for all the a (E G) only if a is the unity element of c.0. As this is a trivial consequence of (93.10) and (93.11), Theorem 219 has been proved.
This theorem means, among other things, that all the characters of At and G are simultaneously given by (93.4) and (93.5) respectively. Since (93.4), (93.5) arise from each other simply by interchanging a and a, we have the important duality principle for finite Abelian groups, which is as
CHARACTER GROUP
379
follows: Every proposition, which follows from the definition of the character
group of a finite Abelian group, retains its validity after interchanging the roles of these two groups. In future, co and G will have the same meaning as in Theorem 219, while rand V will designate the lattice of all the subgroups of CO and G, respectively. For a subgroup V of 4 we denote by the set of those elements
a of G, for which
as = 1
(for all a E X) .
(93.14)
From (93.3) it follows that these a constitute a group, so that A' is a subgroup of G (i.e., A' is the group of those characters, which map onto 1). Accordingly
' ->
'
(,° E ?)
(93.15)
is a mapping of the lattice ?' into the lattice V. Similarly let the dual mapping
(of V into ?) be denoted by H -> H'
(93.16)
(H E V),
in which the group H' (c c0) of those elements a of 4, for which
as = 1
(for all a E H)
(93.17)
is assigned to the subgroup H of G (i.e., H' is the group of those elements of (4, which are mapped onto 1 by the characters in H). Since characters were first
considered by Dirichlet, we call (93.15) and (93.16) the first and second Dirichlet mappings of ? into V and of V into r, respectively. Furthermore "' and H' are said to be the Dirichlet images of and H, respectively. THEOREM 220. The first Dirichlet mapping maps the lattice ?' of a finite Abelian group 4 one-to-one onto the lattice V of the character group G of 4t. The dual proposition holds for the second Dirichlet mapping where the roles of ?, V are interchanged. These two mappings are inverse to each other Finally, the first (second) semilattice of C.4 and the second (first) semilattice of G are mapped isomorphically one onto the other. The following two (dual) results follow immediately from the definitions:
H c K H' K'
,
(93.18)
where,
and H, K denote subgroups of 4t and G, respectively. Similarly from the definitions follow the (dual) properties
l°" Q X, H"
H
(93.19)
380
FINITE ABELIAN GROUPS
for every subgroup X and H of c4 and G, respectively, where we understand that A"' = (a°')' and H" Then we prove the (dual) properties
c4/,Y :
G/H ., H',
(93.20)
for the mappings (93.15), (93.16). It will suffice to prove the first of these.
We consider a character A (with respect to 8) of the factor group cP/'P and define a mapping a --> as of ci (into 8) by
as = A(a')
(a E (P)
.
(93.21)
Then for a, b E c
aab= Therefore we have a E G. In particular, for all a E 2f°, as = AX = 1, so that a even E X'. Conversely, if a E :V', then as = 1 (a E t'). Hence, because of (93.2), it follows that for two elements a, b (E d) with aX = bX we have as = ab. Thus if we put A(a ) = ma, we have defined a (one-to-one) mapping a2P --> A(aX) of cP/X (into 8). Here, because of (93.2), A(a 2P ba") = A(ab
') = aab = as ab = A(a2f°) A(bX )
.
Hence A is a character of c4/.T. Consequently we have shown that (93.21) gives exactly all the characters
a E a', if A runs through the characters of d /X. Since different a evidently correspond to different A the character group of c0/V is mapped one-
to-one onto 7' by A --* a. Finally, if B, 9 is a second pair of characters assigned to each other, then
afla = as #a = A(a.P) B(aT) = AB(aMj . This completes the proof of (93.20). This easily leads to the proof of Theorem 220. From (93.20) it follows that
O(P)O(Yt') = 0(c4),
O(Y')O(Jt") = O(G).
CHARACTER GROUP
381
Because 0(4) = O(G) it follows that O(X") = 0(,7). This means that (93.19) is refined to
X"=', H"=H.
(93.22)
Accordingly (93.18) may be refined to
ar c 7 a 2L°' DH c K a H' D K'.
(93.23)
Since the lattices r, V are finite the first assertion of the theorem follows
from (93.22). The second (dual) assertion is then also true. The third follows from (93.22) and the last from (93.23). Consequently we have proved Theorem 220. NoTE. In our definition of the characters of finite Abelian groups, we can take the group of the field of complex numbers for the group 8. Further valuable properties of characters will arise from this specialization, so that usually only this case is referred to. (Cf. § 155, Example 9.) EXAMPLE 1. The first and second semilattices r1, rY of all subgroups of a finite
Abelian group 4 are isomorphic. Consider two subgroups 7, .7 of 4 with . ° c X. For their images 7, .7' in the character group G of 4 we have 7' z .7' by Theorem 220 or (93.23). On the other hand, take an isomorphism a -- q'a of 4 with G according to Theorem 219. Then q '.5Z'' are subgroups of 4, and evidently 97-'. °' op-',5I'"'. Hence we have the isomorphism
V1 - A (Jr -
(93.24)
EXAMPLE 2. From Example I it follows that in a finite Abelian group of order n the number of subgroups of order d is exactly the same as that of the subgroups of
order d- In (d I n) (91.181 is a special case of this). EXAMPLE 3. For an element a of a finite Abelian group 4 and for an element at of its character group G aa" = a"a
(93.25)
because of (93.2) and (93.3). As an application we prove that the Dirichlet image of 4" is equal le, where a denotes the unity element of G and -Je the n1° radical of e. Firstly 4" consists of all the a" (a E 4). The Dirichlet image of 4" consists of those a (E G), for which the equation ma' = 1,
i.e.,
a"a = 1,
is satisfied by all the a (E 4). This is so if, and only if, a" = e, i.e., a E . IE. This proves the assertion.
§ 94. The MSbius-Delsarte Inversion Formula Let a finite or infinite set C5 of finite Abelian groups be given such that, if a group is an element of C5, all its subgroups belong to S. Take an arbitrary module M. Let f and F be functions with the domain of definition C and the
FINITE ABELIAN GROUPS
382
range M. The problem raised and solved in a special case (see below) by MOBius, and in general by DELSARTE (1948), is as follows : What is the condition for these two functions f, F to satisfy the system of equations E f(H) = F(G) (G E C5) . (94.1) HcG
("H c G" here means that H runs through all the subgroups of G.) Of course, we are concerned with the case where the function F is given and the function f is unknown and so may be called the solution of the problem. It is almost trivial that exactly one solution always exists; the problem is to obtain the formula for solving the equation system (94.1). This can be done using the Delsarte function µ(G), which we define for
any finite Abelian group G as follows: If G is elementary of order p'° (p a prime number), then µ(G)
_
(-1)kp(2)
(k
(94.2)
0) .
If G is a non-elementary p-group, then p(G) = 0. Finally, if G is arbitrary and P,, ..., P, are its primary components, then u(G) = p(PI)
... p(P,)
(94.3)
.
THEOREM 221 (DELSARTE'S theorem). The system of equations (94.1) is equivalent to
f(G) _ Y u(G /H) F(H)
(G E C)
.
(94.4)
HcG
Further for every finite Abelian group G
_ HC-G µ(H
1 for O(G) = 1 0 for O(G) > 1
(94.5) .
Before proving this, it should be noted that (94.4) is the formula for the solution of (94.1). We call (94.4) the Delsarte inversion formula. We see that (94.5) itself constitutes that special case of (94.1), where S denotes the set of all finite Abelian groups, where isomorphic groups are regarded as
equal, and we have M = 7+. Furthermore F(G) denotes that function which equals I for the unity group and vanishes for other G. We begin the proof with the verification of (94.5). The left-hand side of (94.5) will be denoted by Y_ G. If
G=P,®...®P,
383
MSBIUS-DELSARTE INVERSION FORMULA
is the direct product decomposition of G into its primary components PI, ..., Pr, then we obtain all the different subgroups of G in the form
H=Q1®...®Qr, where Q1 has to run through all the subgroups of P1 (i = 1, ..., r). Since then, according to (94.3), we have µ(H) = µ(QI) ... µ(Q,), >G = Y- P,
.. Y- P,
follows.
Accordingly it will suffice to prove (94.5) for the case where G is a pgroup. Evidently, from the definition of the function µ, it follows that I G = Y- G,,,
where Go denotes the subgroup of those elements of G whose order is at mostp. Thus it will suffice to prove (94.5) for the case where G is elementary of order p" (n >-- 0). Of this the case n = 0 is trivial. For n >_ 1 the assertion becomes by (94.2)
(n = 1, 2, ...)
k J (- l )k p(k) = 0 k=O
(94.6)
,
L
I denotes the number of subgroups of order pk of an ele-
where
k mentary Abelian group of order p". Let us generalize 1
1 as in (91.18).
I
k
Then we may write the left-hand side of (94.6) in the form pk I
n- 1 k
k
1)k p(_)
+E
Ink
-1
- 1, (- 1)k
p(:)
,
(94.7)
where( we have to sum over all the k (E J). The first sum in (94.7), because 11,
k + 2I = Ik
is equal to
2
In - 11(-
k-1
I)'-' p(2)
Hence we see that (94.7) vanishes. Hence (94.6) and (94.5) have been proved.
We now see that, by (94.5),
µ(G/H) _ H9G
1
for O(G) = 1,
0 for O(G) > 1
.
(94.8)
FINITE ABELIAN GROUPS
384
For, if we apply (94.5) to the character group of G instead of G and apply the formula in (93.201), then we obtain (94.8).
The first assertion of Theorem 221 still remains to be proved. For this, we substitute, on the one hand (94.1) into (94.4), and, on the other, (94.4) into (94.1). This gives
f (G) = E E µ(G/H) f(K)
(G E (S) ,
E Y µ(H/K) F(K) = F(G)
(G E Cam),
HcG KcH
(94.9)
or
(94.10)
HcG KcH
respectively. It will suffice to prove that (94.9), (94.10) are identical, i.e., they are satisfied for all the functions f and F, respectively. These assertions come from the following: Kc
Gµ(
K SHSG
G
l
(1 for K=G, = to for KcG,
µ(H/K) =
94.11
(
)
1 for K= G, j0 for K c G,
where K is a subgroup of G and H has to run through those subgroups of G,
for which K c H S G. Now (94.112) is true, because this formula is obtained if we apply (94.5) to G/K instead of G. Since, according to
the second isomorphy theorem (Theorem 126') G/H
(G/K) / (H/K)
,
(94.111) follows from (94.8), by applying it to G/K instead of G. Consequently we have proved Theorem 221. We shall now apply this theorem to the special case where C5 is the set of all finite cyclic groups. Since these groups are essentially given by their
orders, so for this case we may introduce the notations u(n), f(n), F(n) instead of µ(G), f(G), F(G), where n is the order of the (cyclic) group G (n = 1, 2, . . .). We call µ(n) the Mobius function, which has, according to the above general definition, the following meaning: µ(n)
_
-
( {
1)k
0
if n is the product of (k >_ 0) different prime numbers, if n has at least one multiple prime factor. (94.12)
The system of equations (94.1) for this case is
E f(d) = F(n)
(n = 1, 2, ...) .
dIn
Again from Theorem 221 we obtain the following
(94.13)
MdBIUS-DELSARTE INVERSION FORMULA
385
THEOREM 222 (MSBIus's theorem). The system of equations (94.13) is equivalent to
f(n) = > u(d-'n) F(d)
(n = 1, 2,...) ;
(94.14)
din
furthermore
µ(d) din
1 for n = 1
= 0forn>1.
(94.15)
This special case (94.14) of (94.4) is called the Mobius inversion formula. EXAMPLE. The formula
v(n) _ E d1(nd-1)
(94.16)
din
for Euler's function follows from (94.15) and (84.22). EXERCISE. Formula (94.8), and consequently Theorem 221, may be proved without using the first main theorem for finite Abelian groups and the character concept.
§ 95. Zeta Functions for Finite Abelian Groups We shall define here certain functions, which allow the lattice of subgroups of a finite Abelian group to be examined more closely.
In this paragraph we denote by [S] the number of the elements of an arbitrary finite set S. (Consequently, if G is a finite group we could write [G] for O(G), nevertheless we retain for this case the notation O(G).) We denote an arbitrary non-negative integer by n and the set by 1.
(For n = 0 we take t = O.) Let z (= 1,2 ....) denote a positive integral variable. Let a finite Abelian group G be given with unity element e and arbitrary
(not necessarily distinct) subgroups A1, . . ., A. of it. We denote by Ag
(C c %) the product of the subgroups Ai with i E C5. (If e = 0 or 0). Prove that if we take all the pairs of non-negative integers m, d with d I m and the ring defined
by the equations ma = 0, aQ = da, then we obtain all the non-isomorphic cyclic rings.
CHAPTER VI
OPERATOR MODULES Operator modules are of great importance for the whole of algebra. The theory of operator modules is also called linear algebra and includes a great part of ring theory. We have already met operator modules. Although
this chapter is exclusively devoted to them, only a small selection of the important problems concerning them can be discussed here. § 97. Operator Modules and Vector Spaces
Let an s-module M be given. This means that M is a module and the is a ring, which we apply - unless otherwise indicated operator domain as a left operator domain; this means that for a (EJ2), a (E M) the operator product as (E M) is defined and for arbitrary a, b (E-I), a, fi (E M)
a(a + P) = as + a#,
(a + b)a = as + ba,
aba = a(ba).
(97.1)
Mainly we shall consider only a unitary $-module M (cf. Theorem 144). This means that JP has a unity element e which is an identical operator, i.e., em = a (a E M). For the present, we take the former more general case into consideration. As submodules of M only the admissible submodules M' will be considered, for which therefore
ME M'
(aEt.,aEM')
is satisfied, i.e., which are s-modules. Likewise, a homomorphism of M will always mean an R-homomorphism a - a', for which therefore (aa)' = am'
(a E J2, a E M)
is satisfied.
Any elements al, ..., a of M are called linearly independent (with respect 0 for each i and an equation such as
to or over tR), if,
alai + ... +
an E JP) 395
(97.2)
OPERATOR MODULES
396
can hold only if
slot, = ... =
(97.3)
0.
(This amounts to an extension of the definition of linear independence in modules without operators, given in § 45.) If, in particular, JP is a skew field .7 and the module M is unitary, then we can replace this definition by a simpler one. For, from an equation such as as = 0
(a E Y, 0 0)
it follows that a-laa = 0, i.e., a = 0. Accordingly, in this special case the linear independence of the elements al, ..., a means that from the equation (97.2) we obtain (97.3')
Here it need not be assumed from the very first that al, from 0, for if, e.g., al = 0, then (97.2) is fulfilled for
..., a are
distinct
a100,
a the definition of linear independence to arbitrary subsets M
of an .'-module M, where we call M linearly independent, if all finite subsets of M are linearly independent. Then linear independence is a property of finite character. The.'-module {a} generated by the element a is called a cyclic J-module or a cyclic module. This consists of all the
is + as
(i E 7, a E .),
(97.4)
which need not all be different. In the unitary case the first term may be omitted in (97.4).
Let A (9 M) be a complex of M. The elements a (ER) with aA = 0 then constitute, because of (97.12,3), a left ideal a of JP, which we call the
annihilator left ideal of A in 2. The elements of a are briefly called the annihilators of A (in JP). If, in addition, A is an admissible submodule of M then a is also an ideal, as is obvious. In this case, as well as for a commutative J2, a may be called the annihilator ideal of A. The definition of the annihilator left ideal can also be applied to the case where A is a complex of an arbitrary ring R, so that we consider A as a complex of the R-module R+. Therefore the annihilator left ideal a of A will then consist of those elements a of R, for which aA = 0. If, in addition, A is an admissible submodule, i.e., a left ideal of R, then a is
OPERATOR MODULES AND VECTOR SPACES
397
an ideal. In the special case A = R, a consists, according to our former definition (§ 20), of all the left annihilators of R. THEOREM 227. If {a) is a unitary cyclic s-module and a the annihilator left
ideal of a in JP, then we have the J9-isomorphy +/a+.
{a)
(97.5)
COROLLARY. For a skew field.7 every unitary cyclic 7-module other than 0 is isomorphic with Y. For the proof, consider the mapping a --> ax
'-homomorphism of 39+ onto {a). Its kernel consists of the a with as = 0, therefore it is equal to a. Hence, and from the bomomorphy theorem (Theorem 60), Theorem 227 follows. The corollary is self evident. NOTE. Conversely, if a is a left ideal of A, then the factor module in (97.5) is a unitary cyclic .39-module, which is generated by the residue class e (mod a). Accordingly, the determination of all the unitary cyclic ,R-modules is essentially identical with the determination of the left ideals of '. More precisely, we are interested in the determination of all the s-left submodules of ,51+ (these being the left ideals of JP) and of the subsequent formation of the related factor modules, among which the nonisomorphic ones have still to be determined. All this implies the homomorphy problem of the .51-left module J'+. Theorem 227 is the special case n = 1 of the following generated by THEOREM 228. For a unitary 39-module M = {al, ..., finitely many elements and the l'$-vector space V with a basis Q1, ..., Sl of. 39 onto {a}. This is, by (97.12 3), an
V ^' M (a1JQ1 +
... + a,yQn -+ alal + . .. + aaan),
(al, ..., an E e51). (97.6)
Consequently,
M
V/V',
(97.7)
where V' is a proper submodule of V, the kernel of the homomorphism (97.6).
For, since the a1Q1 + ... + an.Q are all the different elements of V, so (97.6) follows from (97.1). Hence the theorem has been proved. Since, conversely, the right-hand side of (97.7) is a finitely generated -submodule V' of V, just as unitary A-module for an arbitrary above we see that the determination of the finitely generated unitary Jpmodules is reduced to the homomorphy problem ofd-vector spaces. On the other hand, Theorem 228 retains its validity for arbitrary (instead of finitely generated) . 9-modules.
OPERATOR MODULES
398
THEOREM 229. Over a skew field .7 every unitary module is an .5"-vector space.
For, let M be a unitary .-module. According to the lemma of TEicHMULLER-TUKEY (Theorem 17), M contains a maximal linearly independent
subset M = . Every element a of M may then be written in the form
a = a1a1 + a2a2 + ...
(a1, a.,,
... E,71
(97.8)
where only finitely many coefficients are different from 0. If a lies in M, then it is obvious. Otherwise, the union set M U a is not linearly independent,
whence the existence of an equation of the form
cm+ c1a1+... +
(c,c1,...,CnE-7)
with c :A 0 follows. After multiplying by c-1 the required equation (97.8) is obtained. This representation (97.8) of the elements of M is, because of the linear independence of M, also unique. Hence and from (97.1) it follows that M is the -vector space with the basis al, a2, .... Consequently the theorem is proved. THEOREM 230. All the submodules of an n-dimensional vector space over a principal left ideal ring R with unity element and without zero divisors are also J2-vector spaces of dimension at most n. Let Vn denote such a vector space with the basis al, . . ., ac.. Similarly
(for n > 1) will denote the ,-vector space with the basis al, ..., a submodule of V. For n = 0, Vo = V' = 0, therefore the theorem is now trivial. For n >-- I we assume the theorem for n - 1. The elements of V' are uniquely written in the form
a = alas + ... + anan
(a1, ..., an E A).
(97.9)
If always an = 0, then V' c V. -1; consequently, on account of the induction assumption, the theorem is then true. If an 96 0 for at least one a, all the possible an in (97.9) evidently constitute a left ideal of A different from 0. This is a principal left ideal Ra for some suitable a (E 2, 96 0). Then V' contains an element a, for which an = a holds in (97.9). By (97.9) the
ca + fi
(c E A ,B E v. _1(1 V')
(97.10)
are then all the different elements of V'. But now V _1 n V' constitutes, according to the induction assumption, an (m - 1)-dimensional vector then it follows space with m < n. If a basis of it is denoted by from (97.10) that the ca + b1N1 + ... + bm-iNra-1
(c, b1, ..., bn_t E.t` )
OPERATOR MODULES AND VECTOR SPACES
399
are all the different elements of W. Consequently V' is the /?-vector space with the basis a, F'1, ..., An -1- Thus the theorem is proved. EXAMPLE 1. The previous theorem is false for arbitrary integral domains !k. If !k = 9[xl and V1 is the Q -vector space with the basis al, then the elements
(2.f (x) + x 9(x)) al
(1(x), 9(x) E "k)
constitute an !k-submodule of V1, which is not an !k-vector space. EXAMPLE 2. If a is an element of an 0-module (i.e., of a module without operators),
then the annihilator ideal of a in J is just the principal ideal (n), where n = o+(a) is the additive order of a. Accordingly the notion of the annihilator left ideal is to be regarded as a generalization of the (additive) order of an element. Therefore, for an element a of a commutative ring R (with regard to the R-module R+) the annihilator ideal of a in R is sometimes also called the order ideal of a ; this consists of those elements
e of R, for which ea = 0. EXAMPLE 3. The submodules of a finitely generated unitary module M over a left principal ideal ring !k are likewise finitely generated. Then M, by Theorem 228, is the homomorphic image of a finite-dimensional ek-vector space V, so that, by Theorem 58, every submodule M1 of M is the homomorphic image of a submodule V1 of V. Now V1, according to Theorem 230, is also a finite dimensional !k-vector space and so finitely generated, whence the same follows for its homomorphic image M,.
§ 98. Determinant Divisors and Elementary Divisors
Here we consider matrices (of an arbitrary type) over a commutative ring
with unity element 1, which will later be assumed to be Euclidean.
The notions to be introduced will then be applied to JP-modules. We consider a matrix A of type m x n over and put p = min (m, n).
The determinant of an arbitrary square submatrix of A is called a subdeterminant of A. The order of such a subdeterminant is then always one of the numbers 0, . . ., p. The maximum r of the orders of all non-vanishing subdeterminants of A is called the rank of A. Accordingly
0 m, the
previous conclusion may be repeated. Hence Theorem 232 is proved. The first part of the proof might be replaced by a simple direct proof.
We now suppose that A is also a commutative Euclidean ring. In this case we want to get more precise determination of the submodules of an .-vector space than in Theorem 230. Let an n-dimensional,A-vector space V be given with a submodule U, which, by Theorem 230, is an m-dimensional s-vector space (m < n). They then have bases wt, . . ., w and o,, ..., Q., respectively. These are connected by certain equations n
o; = Y=ta wi
(ate E *)
(98.2)
with uniquely determined coefficients a,,. Their matrix of type m x n is denoted by
A = (a,;).
(98.3)
(Later it will be evident that A is necessarily of rank m.) Now, the choice of the bases col, . . ., w,, and et, ..., em may be made in different ways. if we carry out a basis transformation on each of them with the matrices N and M, respectively, then the matrix A, with respect to the new bases. according to Theorem 161, becomes the matrix
MAN".
(98.4)
Here, according to Theorem 159, M, N may be any invertible matrices over ,2 of types m2 and n2, respectively. Thus, in order to make the survey of the relation of U to V easier we want to reduce (98.4) to the simplest form possible by a suitable choice of M and N. Therefore we now discuss.
this problem. We allow somewhat more generality since we admit an arbitrary matrix A (over R). The answer is implied in the following THEOREM 233. For every matrix A of type m x n (m _ 2).
(98.10)
If a residue bij is other than zero, then we add - qij times the first row of A
to the ith row, then the first column to the ja' column. By this the element aij in A is turned into bij. Similarly as above we can replace all by this bij. But since ail, we can in this way decrease ail more and more, until finally in (98.10) all the bij vanish completely, i.e., in (98.9) all l aij
(i, j >_ 2)
.
(98.11)
If Al is a zero matrix, nothing remains to be done. In the other case AI may be carried over to
A1I`( ant
404
OPERATOR MODULES
where a22 I a;, so that thus,
(i, ] ? 3).
an I a221 au
It should be noted that, by (98.9), the elementary transformations of A, are equivalent to those of A. Continuing this process a diagonal matrix finally arises from A where all I a 2 2 1
This means that we have carried out our programme. We have only to prove the last assertion of Theorem 233. This results from (98.7), since the determinant divisors d1, . . ., are, apart from unit factors, uniquely determined. Consequently we have proved Theorem 233.
From this, and from what has been said above we obtain: THEOREM 234. If V is an n-dimensional vector space over a commutative Euclidean ring JP and U is a submodule of V, which, according to Theorem 230, is an m-dimensional vector space over V (m 1. Evidently, two primepower modules are isomorphic if, and only if, their annihilators are equal. We now consider the general case in (99.7) in order to determine the sub-
modules of (a). The generating elements of such a submodule are of the form cla, c2a, ... (cl, c2, ... ER). This submodule is then a cyclic module {da}, where d = (ci, c2, . . .). On the other hand, as = 0 and therefore d I a. We have thus obtained the proposition that the submodules of a cyclic J *-module {a} are again cyclic, and of the form {da}
(d E JP, d I a),
(99.8)
where (a) is the annihilator ideal of {a}. Further if d 0 the annihilator ideal of {da} is evidently equal to (d-' a). In the following we distinguish between the cases a = 0, a 0. The case a = 0 is very easy. For, d may now be any element of JP in (99.8). If we have d :A 0, then the annihilator ideal of (da) is equal to 0. like that of {a}. Because of (99.7), {a} and also its submodules other than 0
are now all isomorphic with 92+. Since, furthermore, two submodules {dla}, {d2a} (d,, d2 # 0) have an intersection other than 0, (a) is indecomposable.
For the case a # 0 we proceed as follows. If
a = be
(b, c E'; (b, c) = 1),
(99.9)
then we have the direct decomposition {a} _ {ba} ® (ca).
(99.10)
For, the equation
xb+yc= 1
(x,yE.
)
is solvable, so it follows that at = xba + yca, so that firstly we have
{a} _ {ba) + {ca}. Further, the intersection of the modules {ba}, {ca} consists of the za with z E JP and b, c I z. Hence by (99.9), a I z and so za = 0. Thus we have proved (99.10). By repeated application of (99.10) it follows that {a} may be decomposed into the direct sum of prime-power modules and also the
product of the annihilator ideals of the direct summands is equal to the annihilator ideal of {a.}. In addition to this we still have the obvious consequence from (99.8) that prime-power modules are indecomposable.
MAIN THEOREM FOR ABELIAN GROUPS
409
Hence, from Theorem 235 and the theorem (Theorem 146) we obtain the following THEOREM 236 (theorem of uniqueness for finitely generated Abelian groups). Every finitely generated unitary module M over a commutative Euclidean ring,`? is a direct sum of cyclic.-submodules, which are partly isomorphic with J1 , partly prime power modules, and, apart from order and isomorphy, uniquely defined.
We also have the following trivial SUPPLEMENT. If Mo and MP denote the sum of those direct summands of M,
which are isomorphic with .-,9+ or are annihilated by an ideal (p") (p a prime
element of -2, k = 1, 2.... ), respectively, then M has the direct decomposition
M=MoeIMP,
(99.11)
P
where p has to run through finitely many different prime elements. The direct summands MP are uniquely defined; furthermore, because
M0 ., M/IMP,
(99.12)
P
Mo is, apart from isomorphy, uniquely defined. EXAMPLE. Since .1 is Euclidean, all that has been said above also refers to the finitely
generated modules without operators. In particular, the additive finite Abelian groups are included here, so that some of the results in the §§ 90, 91 may also be obtained by particularizing from the above. If P1, . . ., P, are the invariants of a finite Abelian group, then (PI), ..., (P,) are exactly the annihilator ideals of those prime-
power groups, into which this group is decomposed by Theorem 236 or Theorem 214. Accordingly, in the general case the annihilator ideals of the prime-power modules in Theorem 236 and the number of the direct summands isomorphic with (R+ are called the invariants of this module. These again constitute a complete independent system of invariants.
§ 100. Linear Dependence over Skew Fields
In this paragraph we define the notion of "linear dependence" and set
forth in detail the fundamental facts referring to it. The concept of "linear independence" introduced previously will now appear to be a derived one. We follow VAN DER WAERDEN'S presentation, in such a
manner that the results may also be applied to certain concepts to be introduced later, e.g., to that of "algebraic dependence". Let a skew field 3 and a unitary module M over.7, which, according to Theorem 229, is then a vector space over .7, be given. The whole paragraph refers to this. Small Roman and Greek letters (perhaps with indices) always denote elements of Y and M, respectively.
410
OPERATOR MODULES
An element o is said to be linearly dependent on the elements wt, ..., co,., (with respect to .7), if an equation such as (100.1)
holds. In particular, this linear dependence will mean that o = 0 for n = 0. We prove that the following three fundamental properties hold: FUNDAMENTAL PROPERTY I. Every wi (i = 1, . . ., n) is linearly dependent on w1, ..., co.. FUNDAMENTAL PROPERTY II. If o is linearly dependent on co,, ..., wn, but not linearly dependent on col, . . ., co,,,, _then co,, is linearly dependent on Lo, w1, ..-,(on -t-
FUNDAMENTAL PROPERTY III. If a is linearly dependent on 01, ..., Lon, . ., m) is linearly dependent on co,, ..., co,,, then a
and every p; (i = 1, .
is linearly dependent on w), . . ., con. Of these, I and III need no proof. 11 is proved as follows: By hypothesis, (100.1) holds with an # 0. Thus
wn = ant (O - a1w, - . . - a,, .
-1w. - ,).
Therefore the assertion is true. The proofs of Theorems 237 to 242 do not rely on the above definition
of linear dependence, but only on the fundamental properties I to III. by which the more general applicability of these theorems is guaranteed. From I and III immediately follows THEOREM 237. If a is linearly dependent on ..., om and every ei (i = 1, ..., m) occurs among the w,, ..., to,,, then a is linearly dependent On w1, ..., w,. It is worthwhile to note the special case of this theorem, when 01, agree apart from order with the wt, . . ., w;,. For this case the theorem
means that in the proposition "a is linearly dependent on cot, ..., (A)n" the succession of the elements wt, ..., co. is immaterial. The elements wl, ..., w,, are said to be linearly independent, if none of them is linearly dependent on the others. In particular the empty system is said to be linearly independent. From Theorem 237 it follows that every partial system of a linearly independent system is also linearly independent. This definition is not new, but agrees in content with the special case for
Q = .7" of our former definition [(97.2), (97.3)], as it results from the note made there [(97.3')]. THEOREM 238. If w1, .... are linearly independent, but w1, ..., (') are not linearly independent, then w is lin?arty dependent on w1, ..., wn_t.
LINEAR DEPENDENCE OVER SKEW FIELDS
411
Otherwise an w; (i # n) would be linearly dependent on the other w1, ..., w,,. We may suppose that con -1 is linearly dependent on w1, ..., (On-21 CO.- On the other hand, wn_I is not linearly dependent on wl, ..., wn_2, so that wn, by II, is linearly dependent on 0 ) 1. . .., wn_I. Consequently Theorem 238 is proved.
THEOREM 239. Among arbitrary w1, ..., wn there are always linearly independent elements o1i ..., em (m ? 0), such that all the elements are linearly dependent on them. Let us choose from among the w1, . . ., con a maximal number of linearly independent elements CI, ..., Cm. Every w; different from these is linearly
dependent on them by Theorem 238, and, according to 1, also every e . This proves Theorem 239. Two systems of elements 01, ..., qm and w1, ..., w are called linearly equivalent, if every t), is linearly dependent on the w1, ..., con and every w; is linearly dependent on the 91' , Pn,.
By this we have defined an equivalence relation, since the symmetry follows immediately from the definition, and the reflexive and transitive laws follow from I and III. If an element is linearly dependent on one of two linearly equivalent systems, then it is also linearly dependent on the other, from III. From Theorem 239 it also follows that every system of elements w1, ..., con is linearly equivalent to a linearly independent partial system of these elements. THEOREM 240 (STEINITZ'S exchange theorem). If OI, ..., Pm are linearly independent and every Qi is linearly dependent on w1, ..., wn, then m < n. and one can suitably order the w1, ..., co,, such that the two systems
w1, ..., co,,; QI... , Cm , wm+I , ..., co,
(100.2)
are linearly equivalent. We say that the system (100.22) originates from the system (100.21) by
the exchange of the elements w1, ..., w,n (for the elements e1, . . ., on,). The theorem is trivial for in = 0. If m z 1 we assume its truth for m - 1. Since e1, ..., CC.-1 are linearly independent, because of the induction assumption, m - 1 < n, and after a suitable rearrangement of the elements (o1, ..., co,, the two systems 0) I, ..., (0n; 01, .... gm - I
,
Loon, ..., wn
(100.3)
are linearly equivalent. Since Ln, is linearly dependent on (100.31), it is also linearly dependent on (100.32). Since, on the other hand, nom is not linearly dependent on eI, ..., 2m_1, it follows that m < n. Further, by
OPERATOR MODULES
412
suitably ordering the wm, . . ., co,,, we can say that for a suita ble t (m < t < n)
em is linearly dependent on 91, -
, Um -1 ,
wm.... 1011,
(100.4)
but is no longer so, if any co, (i = m, ..., t) in (100.4) is eliminated. From 11 it follows that wm is linearly dependent on
01,..., nr
wmfl,-..,IO,.
and so, by Theorem 237 on 601...
wm+1, ... w .
(100.5)
But since, as stated above, 9m is linearly dependent on (100.32), it follows that (100.32), (100.5) are linearly equivalent, as are also (100.31), (100.5). This implies Theorem 240. THEOREM 241. If L01, ..., Am and co], ..., co,, are linearly independent and linearly equivalent, then m = n. This follows from the fact that, according to Theorem 240, m:!5; n and n _< m.
We say that a subset M of the .%--module M is of finite rank (over 3), if in M there are finitely many elements, on which every element of M is linearly dependent. Then there are in M, according to Theorem 239, linearly independent elements w1, . . ., co,, such that every element of M is linearly depend-
ent on them. In addition, by Theorem 241, n is uniquely defined. This n is said to be the (linear) rank of the set M (over or with respect to .-".) Since in M, by Theorem 240, there are then at most n linearly independent elements, we may define the rank of M as the maximal number of linearly independent elements of M. THEOREM 242. If a set M is of finite rank, then so is every subset M' of M. Moreover, the rank of M' is at most equal to that of M. In M'there may be at most as many linearly independent elements as in M. Therefore the theorem is true.
§ 101. Vector Spaces over Skew Fields Let . " be a skew field. As is known from Theorem 229 every unitary.? module is an .7"-vector space. We now wish to study .f"-vector spaces of finite rank in fuller detail. We have defined the rank of such a vector space V in two ways. According to the definition of the previous paragraph, it
means the maximal number of linearly independent elements of V. According to a former definition (¢ 64) it means the number of elements of an arbitrary
VECTOR SPACES OVER SKEW FIELDS
413
basis of V. This second notion of rank refers to vector spaces over any ring with unity element and in general it has no invariant meaning; its invariance
has been proved, by Theorem 232, only for commutative fundamental rings. We now prove the following THEOREM 243. In a vector space V over a skew field" the elements w],. . .. to,, constitute a basis if, and only if, n is the maximal number of linearly independent elements in V and col, . . ., co. are linearly independent. Consequently for finite-dimensional vector spaces over skew fields the rank defined in § 64 is identical with that defined in § 100, and so is invariant.
The assertion "if" can be proved in the same way as Theorem 229, or even somewhat more easily. In order to verify the assertion "only if", we suppose that wl, . . ., w is a basis of V. Evidently o , . . ., co,, are then linearly independent; furthermore every element of V is linearly dependent on them, whence it follows by Theorem.240 that n is the maximal number of linearly independent elements in V. Hence the theorem has been proved. The (admissible) submodules of a vector space we briefly call its subspaces provided that they are also vector spaces over the same fundamental ring (this is always valid for fields). THEOREM 244. If V is a finite-dimensional vector space over a skew field.? and U is a subspace of V, then every basis of U may be completed to one of V.
For the proof we denote bases of U and V by ol, ..., a,, and w 1 , . . ., w,,, respectively. Then Theorem 240 is applicable, so that m suitable elements of the w1, . . ., co,, may be replaced by the ol, ..., om in such a manner that again there arises a basis of V. Consequently Theorem 244 is proved. By applying the results on.7-vector spaces we obtain the following THEOREM 245. In the full matrix ring .-` over a skew field .rte' every matrix
is either invertible or a zero divisor. Consider a matrix
A=(aid)(E n)
(aiiE. )-
We make use of an n-dimensional.7-vector space V with the basis wl, .... n0,,. The necessary and sufficient condition that the
also constitute a basis of V may be given in two ways. On the one hand, this
condition consists, according to Theorem 159, in the invertibility of A, on the other hand, in the linear independence of ol, ..., o,,. This independence means, furthermore, that the equation Y_ cioi = i=1
J
i cia, cot = 0
i=1;=1
(c, E -t)
(101.1)
OPERATOR MODULES
414
can be satisfied only with c1= ... = cn = 0. Since the w1, ..., con constitute a basis, equation (101.1) is equivalent to the system of equations n
I=1
ciaij=0
(j
I... ,n).
This is representable in the form (c1, ..., Cn)A = 0
.
(101.2)
where the left-hand side is to be interpreted as a product of matrices. But, since (101.2) can be satisfied with a vector (cl, . . ., cn) other than 0 if, and only if, XA = 0 has a solution X other than 0 in ., we have obtained the following: the invertibility of A is equivalent to the fact that A is not a right zero divisor (in .5 ). Since invertibility is a self-dual property, Theorem 245 has been proved. EXAMPLE 1. In an 'k-algebra of finite rank n, and so also in the full matrix ring Q. of rank n2, these ranks have an invariant meaning, if 'k is a commutative ring with unity element or skew field. EXAMPLE 2. From Theorem 245 it follows that if r,Q is an integral domain, then in the full matrix ring Q. every left zero divisor is a right zero divisor, and conversely. EXERCISE. Theorem 244 also holds for infinite-dimensional . '-vector spaces. Prove this using the lemma of TEICHMULLER-TUKEY (Theorem 17). PROBLEM. Does Example 2 hold for subrings rk of skew fields?
§ 102. Systems of Linear Equations over Skew Fields Let us consider a system of linear equations over a skew field Y n
E ajtj=bt j=1
(i= 1,...,m),
(102.1)
where the "coefficients" aij and the "constants" bi are arbitrarily given elements of.7 and the tj denote the unknowns, which are likewise to be taken in Y. The number of equations and unknowns is designated by m and n.
respectively. In order to determine the solutions we use the linear forms n
Li = E aijxj j=1
(i=
(102.2)
from the polynomial ring _ [x1, . . ., xn]. In this all the linear forms, i.e., the homogeneous linear polynomials, constitute an."-vector space of rank n. The linear forms (102.2) have rank r, which we also call
415
LINEAR EQUATIONS OVER SKEW FIELDS
the rank of the system of the linear equations (102.1). Accordingly r is the maximal number of linearly independentelements among the L1, ..., Of course
r < min (m, n) .
(102.3)
We prove the following THEOREM 246.
The system of linear equations (102.1) of arbitrary
rank r over a skew field 3 is solvable if, and only if,
k,Ll+...+k»,L,,,=0=k1b, +... +kn,b,,,-0
(k i E .7). (102.4)
When the equations ([02.1) are solvable, these equations and the unknowns
tl, ... , t may be numbered so that LI,
,Lr,.Yr+l,.
.
,x
(102.5)
are linearly independent, whence, with uniquely determined c,.. d,j (E 7), we have n
r
xi = I cl Lj + Y d,jx; j=r+l
j=1
(1 = 1,
..., r).
(102.6)
Then all the solutions t1...., t,, of (102.1) are given by the formula n
r
t, = E cjbj + Y d,jtj j=1
i=r+1
(i = 1, ..., r)
(102.7)
. . ., t we have to substitute arbitrary elements of 2-. The condition of solvability (102.4) may be expressed by saying that
where for tr+1
(102.1) is solvable if, and only if, every linear relation of the form (102.41) remains valid after the substitution L; -* b, (i = 1, . . ., m). The equations
(102.7) may be called the solution formulae for the system of linear equations (102.1). (Of course these need not be uniquely determined, since the L1, . . ., Lr may generally be chosen in several ways.) For the proof we assume first of all that (102.1) is solvable. We consider a linear relation (102.41). This remains true if the indeterminates x1, . . ., x are replaced by arbitrary elements of t We perform the substitution xj -+ tj
(j = 1, .
.
.,
n), where t1,
. .
., t,,
is a solution of (102.1). Then (102.41),
because of (102.2), turns exactly into (102.42), wherefore (102.4) is, in fact, a necessary condition for the solvability of (102.1).
In the following proof of the other assertions of the theorem we may assume that the condition (102.4) is satisfied. We number the linear forms
OPERATOR MODULES
416
(102.2) such that L1, ..., Lr are linearly independent. First of all we show that if the first r equations n
(1=1,....r)
Yaljtj=b,
(102.8)
j=t of the system (102.1) are satisfied, the other m - r equations (102.1) are automatically fulfilled.
We consider an Lp (p = r + 1, ..., m). Then r
LP=>11L1
(102.9)
l=t
with suitable coefficients 11, ..., 1, in 7". Because of (102.4), r
l;h1,
by
i.e., according to (102.8), d
r
by = Y Y l,aljtj .
(102.10)
1=1j=1
On the other hand, it follows from (102.2) and (102.9) that n
r
E avjxi-
j=1
apl=
E Ilauxi 1=1j=1
(j= 1,...,n).
hall
l=j Hence and from (102.10) follows the validity of thep`h equation of the system (102.1), i.e. the validity of our assertion. We apply STEtrnTZ'S exchange theorem (Theorem 240) to the linearly independent Ll, ..., Lr and to the likewise linearly independent x1, ..., xn.
It follows that for a suitable order of the latter (which is practically equivalent to the renumbering of the unknowns t1, . . ., 1,,) the linear forms (102.5) are linearly independent, as was asserted. It still remains to be proved that (102.7) exactly yields all the solutions of (102.1). It will suffice if we show that the equations (102.8) follow from (102.7). To do this we substitute (102.6) in (102.2): r
Ll =
f
I aj
(t ejkLk + k =rt1 n
r
n
djkxk
aljxj
+
(i = 1..... !') .
j=r+t These equations remain true because of the linear independence of the elements (102.5) when they are replaced in order by j=1
k= L
b1i
, ,
., br,
tr+1, ..., to
.
Thus we obtain (102.8), which is what was required to be shown.
417
LINEAR EQUATIONS OVER SKEW FIELDS
Secondly, we show that (102.7) is satisfied for every solution of (102.1), and even for every solution of (102.8). For this purpose we substitute (102.2)
in (102.6). By the substitution x, --+ ti (i = 1, ..., n) we obtain (102.7) because of (102.8). This completes the proof of Theorem 246.
It is both theoretically and practically important that the rank r of the system of equations (102.1) as well as the solution formulae (102.7) may be determined by the following process, which is called successive elimination. We choose an x;, which occurs (with a coefficient other than 0) in one of the equations (102.2). We solve this equation for x? and substitute the expression so obtained for x; in the other equations (102.2). (This means that in the
basis x1...., .r the element x; is replaced, in the sense of Theorem 240, by a suitable L,.) We proceed similarly with the system of m - I equations obtained in this way and continue this process until finally the x1...., x no longer occur in the remaining equations, but at most the LI, ..., L,,,. After renumbering the equations (102.1) and the unknowns t; (i.e., the equations (102.2) and the indeterminates x;) the equations, with the help of which the above elimination took place, will be of the form
(i = I,
xi=f,(L1,.. ,L,,
,r),
(102.11)
where the right-hand sides are linear expressions in all the elements (with coefficients from ") and the coefficient of L, in f,. is different from 0 (i = 1, ..., r). From (102.11) it is evident that the systems
are linearly equivalent, whence it follows that LI, . . ., L, are linearly independent. If in the last m - r equations of the system (102.2) the xj, . . ., x, are successively replaced by the right-hand sides of (102.11), then all the xI. ..., x,, must disappear, since otherwise the process could be continued.
Consequently equations are obtained of the form
L,=f(LI,....L,)
(i=r+ 1, . . ., m) .
(102.12)
In conformity with this the L1, ..., L. are linearly dependent on LI...., L,.. Therefore r is necessarily the rank of (102.2), i.e. of (102.1). Every linear dependence between L1, . . ., L,,, is then a consequence of (102.12), otherwise L1, ..., L, would not be linearly independent. Consequently the condition of solvability (102.4) is expressed by the equations
b,=f,(b1.....b,)
(i=r + 1,...,m).
Finally we see that the equations (102.11), after the substitution
x;--+t;, L,-*b,
(j=
i= 1....,r)
OPERATOR MODULES
418
become the solution formulae for (102.1). ((102.6) may be obtained by replac-
ing the indeterminate x1 for i = r, r - 1, ..., 2 in the first i - I equations of the system (102.11) by the right-hand side of the 1`h equation.) Since the process of elimination above is based exclusively on the coeffi-
cients aij, b1 in (102.1), the following extension of Theorem 246 holds. THEOREM 247. The coefficients cij, d1j in the solution formulae (102.7) for (102.1) lie in the sub-skew-field of 3 generated by the coefficients aij, b;. Thus if (102.1) is solvable in Y, then it is solvable already in this subskew-field.
Theorem 246 is much simplified for a homogeneous system of linear equations.
(i = 1, ..., m) .
a;jij = 0
(102.13)
Now the condition of solvability (102.4) is trivially satisfied, wherefore (102.13) is always solvable. Indeed we see immediately that t1 = ... = t == 0 is a solution, known as the trivial solution of the homogeneous system of linear equations. We see moreover that on the right-hand side of (102.7) the first sum is zero, so that this right-hand side is now homogeneous in the
tr+l...., t.-
For a solution 11, ..., t of (102.13) we call (tl_, . . ., of this system. According to (102.7) all the solution vectors are: rn
(ti, ..., t,,) = (
dijtj, j=r+1
a solution vector
rr
'
.,
j=r+l
clrjtj ir+1 ,
, tn)1
(tr+1... ,t,,E, ).
if we replace one of the tr+1, ..., t here by the unity element I of.7 and the others by 0, then we obtain the special solution vectors r+i
at-(dl,r+i,
1-1
1 ...,0)
(i=1,.. (102.14)
All the solution vectors are obtained from this in the form (102.15) als1 + ... + a,, rsn- r , where the scalar factors s1, ..., are arbitrary elements of.i Accord-
ingly all the solution vectors constitute an (n - r)-dimensional Y-rightvector space with the basis al, . . ., Therefore the solution vectors (102.14) are called the fundamental solutions of the homogeneous system of the linear equations (102.13). In conformity with this we have the following theorem and corollary.
LINEAR EQUATIONS OVER SKEW FIELDS
419
THEOREM 248. A homogeneous system of linear equations with n unknowns
of rank r over a skew field has exactly n - r fundamental solutions. Consequently the trivial solution is the only one if, and only if, r = /I. COROLLARY. A homogeneous system of linear equations over a field with an equal number of equations and unknowns has at least one non-trivial solution
if, and only if, the determinant of the system is 0.
We now consider a matrix A of type m x n over a skew field 3. We shall always consider its rows and columns as elements of an n-dimen-
sional .7-left-vector space and an m-dimensional .7-right-vector space, respectively. Then both the row and column vectors of A have a (linear) rank r' (< m) and r" (- n), respectively, called the row rank or the column rank of this matrix. THEOREM 249 (rank theorem for matrices over skew fields). The row rank and the column rank of a matrix A over a skew field.7 are equal, and, if." is commutative, then they are equal to the rank of A. Because of this theorem we may talk of the rank of the matrix A also
in the case of a skew field .7 and understand by it the row rank (or the column rank) of A. In order to prove the theorem, we denote the row and column rank of A by r' and r", respectively. If A is diagonal, then the assertion r' = r" is trivial.
On the other hand, by the help of elementary transformations A may be changed into a diagonal matrix. Thus to prove r' = r" in general, it suffices to show that r' and r" are invariant for example with respect to elementary row transformations. For the interchange of two rows, this assertion is trivial. For the proof of the remaining assertion, we denote the row vectors of A by aI, .., am
(102.16)
If, e.g., ca.. is added to at (c E .7), then the new row vectors are
aI + ea.,, a2,...,a,,,.
(102.17)
The systems (102.16), (102.17), conceived as systems of elements of a left vector space, are linearly equivalent, from which the asserted invariance for r' follows. Furthermore the change from (102.16) to (102.17) for the column vectors of A, which we denote by bt, . . ., b,,, means that in every bI the second component, multiplied on the left by c, is added to the first one. Every linear dependence is then evidently still true, whence the invariance of r" follows. This proves the first part of the theorem.
OPERATOR MODULES
420
Now let .7 be commutative and denote the rank by r, i.e.. the maximum of the orders of the non-vanishing subdeterminants of A. It is sufficient to prove that r = r'. Again this is trivial if A is a diagonal matrix. So it will suffice to show that r is invariant with respect to elementary transformations. On account of the duality principle for determinants and the invertibility
of the above transformations, it is sufficient to show that r will not be less, if we carry out on A, e.g., the row transformation considered in (102.17). (For row interchanges even the invariance of r is evident.) Let d run through those subdeterminants of order r of A, in which the first row of A participates, and the second row does not participate; moreover let d'run through
the other subdeterminants of order r of A. In our row transformation. according to Theorems 166, 167, every d is changed into a d + cd', while all the d' (partly by Theorem 168) remain invariant. Thus if all the d' vanish,
then all the d remain invariant. But since, by hypothesis, not all the d and d' vanish, so the same follows for the subdeterminants of order r after the row transformation. Consequently Theorem 249 is proved. This theorem permits the replacement of the conditions of solvability (102.4) for the system of linear equations by one condition only. This is the content of the following THEOREM 250. For the system of linear equations (102.1) over the skew field,t"let A denote the matrix of coefficients a;j and A' that matrix which
arises from A by adding a (last) column, consisting of the constants bl, . . ., b,,,. Then the necessary and sufficient condition for the solvability of (102.1) is that A and A' are of equal rank. If a,, . . ., a,,, b denote the column vectors of A', then the system of equations (102.1) may be condensed into the single equation
alts+...+at,;=b. The solvability of this equation is equivalent to A and A' having the same column rank. Hence, and from Theorem 249, Theorem 250 follows. THEOREM 251. A square matrix A over a field Y is invertible if, and only if.
its determinant is not equal to 0. By Theorem 245 the invertibility of A is equivalent to the solvability
of the equation
AX= U
(X, U E Y ; U the unit matrix).
This means that in Y" all the systems of linear equations (with the same number of equations and unknowns) with the matrix of coefficients A are solvable. By Theorem 250, this is true if, and only if, the rank of A is the greatest possible, i.e., equal to n. Hence Theorem 251 follows.
LINEAR EQUATIONS OVER SKEW FIELDS
421
Another method of considering the above topics is given by EGERVARY (1955), where he takes practical applications into consideration as far as possible. An extension of Theorem 246 to the case of infinitely many unknowns and equations has been accomplished by GACSALYI (1951-52) and SZELE (1951-52b). A gener-
alization of Theorem 246 for semisimple rings (see § 108) instead of skew fields is given by KERTFSZ (1955-56). EXAMPLE 1. If a solution t, = c, (j = 1, ..., n) of (102.1) is known, then the question
of the determination of all the solutions may be reduced to the homogeneous system (102.13). Because
aic,=b,
(i = 1,
M),
f=1
(102.1) is representable in the form
ault; - c,) = 0
(i = 1, .... m).
This system of equations is homogeneous in the unknowns t; - c (j = 1, ..., n). If, moreover, we introduce the notation
c = (cy...., then, by making use of (102.15), it follows that the solution formulae (102.7) appear in the form
Thus, the solutions of the system of linear equations consist of a special solution of the general system of equations and of the general solution of the special (viz., homogeneous) system of equations. EXAMPLE 2. It is essential in Theorem 249 that in the definition of the row and
column rank of a matrix, we have regarded its rows and columns as elements of a left and right vector space, respectively. In order to prove this we consider the matrix bb
I ca
'
where a, b, c ( : A 0) are elements of a skew field. Because
c(a, b) = (ca, cb) the row rank is now equal to 1. The column rank must be likewise equal to I. and we indeed find that (a, ca) a- 'b = (b, cb).
But if we considered the column vectors as elements of a left vector space, too, then
the new "column rank" of the matrix would, in general, no longer be 1, since the equation
q(a, ca) = (b, cb)
need not have a solution q in the given skew field. For, it would have to be q = ba-
then ba 'ca = cb, i.e., ba 'c = cba ', and this does not always hold. EXAMPLE 3. For the solution of the system of equations (102.1) for a commutative fundamental field JK - after determining the rank r of the system - we can make use
of Cramer's rule. For this purpose we renumber the equations and unknowns in
OPERATOR MODULES
422
such a manner that the determinant i ail 1r of order r, formed from the ai, (i, j 1, ..., r). is not 0. If we then write the first r equations of (102.1) as n
r
i=t
aiiti
b, - F, atiri
(r = 1, ..., r),
1=r--l
then Cramer's rule yields the solution formulae of (102.1). EXAMPLE 4. If A is a matrix of type in x n over a skew field .7-, then there are (square) invertible matrices M, N of type m' and n', respectively, over .F-, for which
MAN is a diagonal matrix all of whose elements are 1 or 0 and such that in the diagonal the elements 1 precede the elements 0. This matrix is called the normal form of the matrix A (over the skew field .F). It is uniquely determined, and the number of diagonal elements equal to I is the rank of A. The normal form of a matrix over a field was defined differently on p. 402, but the difference is immaterial.
§ 103. Kronecker's Rank Theorem if one of the matrices At, A.,, say At, is a submatrix of the other, then we
call A2 a hypermatrix of A,. THEOREM 252 (KRONECKER'S rank theorem). 1j' a matrix A over a skew field .7 contains a square submatrix B of type r2 and of rank r, which has the property that (in A) all the square hypermatrices of B of type (r + 1)2 are of rank r, then r is the rank of A. For the proof we may suppose that B lies in the first r rows and columns
of A. Let A be of type m x n. We assume that the theorem is false, i.e. the rank of A is greater than r and prove that B (in A) has two hypermatrices
of types (r + 1) x is and m x (r + 1), which are of rank r + 1. This proves the theorem, since by applying this fact twice, the existence of a (square) hypermatrix of B (in A) of type (r + 1)'2 and of rank r + I follows, in contradiction to the supposition. It is sufficient to prove the first of the assertions, as the second may be similarly proved. In conformity with the supposition A certainly contains r + 1 linearly independent row vectors. But since the first r rows are linearly independent, it follows from STEINITZ's exchange theorem (Theorem 240) that the first r rows and a further row of A are linearly independent. These rows then constitute a submatrix of A which is of type (r + I) x n and has
rank r + 1. Consequently the above theorem is proved. § 104. Schur's Lemma We mention by way of introduction that a module M may be considered as an 3'-module, where 8' is the full endomorphism ring of M or a subring of this ring. If 8' is a skew field, which contains the identical endomorphism of M, then M is unitary over 8'. and thus, according to Theorem 229, an
'-vector space.
SCHUR'S LEMMA
423
The simple modules other than 0 (without operators) are those of prime order p. If M is such a module, then it is immediately obvious that its full endomorphism ring is a field r " with 0(Y) = p, for which so that M is now an Y-vector space. However, this is not an isolated fact, since for every simple operator module over an entirely arbitrary ring,
which need not even contain a unity element, we have the following: THEOREM 253 (SCHUR'S lemma). The R-endomorphism ring .7 of a simple
,R-module V (0 0) is a skew field,V and itself is therefore anY vector space. The identical endomorphism of V belongs to Y and is the unity element
of it. Let k be an element of Y other than 0. It will suffice to show that k is an automorphism of V, for then the inverse mapping k-1 likewise belongs
to Sr. Since k is an 9-endomorphism of V, so
k(ax) = a(ka)
(104.1)
for all a Et and a E V. So, if ka = 0, then k(am) = 0. According to this the a with ka. = 0 constitute an R-submodule of V, which is, because k:0 0, different from V. But since V is a simple .52-module, this submodule
is equal to 0. On the other hand it follows from (104.1) that the image kV of V under k is an. *-submodule of V other than 0, i.e., kV = V. This, together with what has gone before, means that k is an automorphism of V. So the theorem is proved.
§ 105. The Density Theorem of Chevalley-Jacobson
To begin with we give the following definition: for a semiordered set S (or for its elements) the minimal condition is satisfied (or C5 is a semiordered set with minimal condition), if every descending chain a1 > a2 > ...
(a,, a..... E C5)
(105.1)
is finite. So, e.g., a ring with minimal condition for the set of left ideals means that every descending chain of its left ideals is finite. The minimal condition admits an equivalent formulation: for the semiordered set 6 the minimal condition is satisfied if, and only if, every nonempty subset of C5 contains at least one minimal element. (Hence the name "minimal condition".) If namely C5' is a subset of G without minimal elements, then if we pro-
ceed from an arbitrary element a, (E s'), an infinite chain of the form (105.1) may be constituted from the elements of G'. Conversely, if e con-
tains an infinite chain (105.1), then there is no minimal element in the subset as
(a E .')
(105.2)
of'V is dense (with respect to V as a rector space over .3). If, furthermore, the minimal condition is satisfied for the left ideals of J', then V is finitedimensional (over Y), so that S' is then the full .7-endormorphism ring of V. For the proof we need the following PROPOSITION. If'wi, ... , co. (n >_ 1) are linearly independent elements of V, then there is an element a of -.99 with
awl =...=awi_1=0,
(105.3)
For n = I we have only to prove that 2wl # 0. With this end in view we suppose that Jiwi = 0. Then the set of a (E V) with Pa = 0 is an ,
-submodule other than 0 of V, hence -5Ia = V. From this it follows that
.V = 0 which contradicts the supposition that PV # 0. For n > 1 we assume the assertion for n - 1 instead of n. We denote by J9P' the set of elements a of. -*with
awl = ... =
(105.4) z = 0. Evidently 2' is a left ideal of }', so is an - -submodule of V. On the other hand, according to the induction assumption, 0, so that 1
i = V. We have only to show that there is an a (E P') with We suppose this to be false. Then for the elements a of .1' awn- ,
= 0 > a(o = 0.
(105.5)
aw
0.
(105.6)
DENSITY THEOREM OF CHEVALLEY-JACOBSON
425
In order to derive a contradiction from this, we show that (a E,.*')
awn_ 1 -* awn
(105.7)
defines anR-endomorphism of V. Because of (105.5), (105.7) is a (possibly many-valued) mapping of V into itself. Furthermore, this mapping is unique, since if awn_1 = a'wn_1 for two
elements a, a' of 2', then (a - a') w.,_1 = 0. By (105.6), (a - a')wn = 0 and hence awn = a'wn. Since 2' is a left ideal of .-.*, for arbitrary a, b (r 2') and c (E 19) the elements a + b, ca belong to *', whence it follows that the elements
awn-1 + bwn-1 (= (a +
e.awn_1( = cawn-1)
are mapped by (105.7) onto ((a + b)wn =) awn + bw,,,
(ca(On =) c awn,
respectively. Accordingly (105.7) is in fact an '-endomorphism of V. Consequently, there is a k (EY), for which (105.7) agrees with the mapping
cc -' ka
(aE V).
This implies that acon = k(awn_1)
for all a (Es'). But since k is an R-endomorphism of V, the right-hand side is equal to a(kwn-1), thus
a(w - kwn-1) = 0 for all a E JP'. Hence it follows, by (105.4) and the induction assumption. that wl, ... wn_n, wn - kwn-1
must be linearly dependent. Then w ...., co,, are also linearly dependent. This contradiction proves the proposition. Now, in order to prove the first assertion of the theorem, let us also denote ?" be the left ideal
by w1, . . ., wn linearly independent elements of V. Let
of V consisting of the a with
Because of the proposition
c.
and
"w,, are an .. V-submodule ( : A 0)
OPERATOR MODULES
426
of V and so equal to V. Consequently, there is, for every en (E V), an an (E R) with anwl = ... = anwn-1 = 0, anwn = en
Apply this for 1, ..., n instead of n. It follows that for a = al + ... + a,, the equations awl = 21, ... aw,, = en
bold, where el,. .., en are arbitrarily given elements of V. This implies that the ring 9 of the endomorphisms (105.2) of V is dense.
Finally, in order to prove the second assertion of the theorem, we suppose that V is of infinite dimension over Y. Then we take countably infinitely many linearly independent elements w1, w2, ... of V and denote by Jan the left ideal of consisting of those elements an for which From the proposition it follows that,55'1 D 1*2 .... But this is in contradiction to the supposition that for the left ideals of . the minimal condition is satisfied. Consequently the theorem is proved.
§ 106. The Structure Theorems of Wedderburn-Artin Before considering important applications of the results obtained above, we need the following theorem, important in itself. THEOREM 255. The full matrix ring f n of rank n2 over a skew field .7 is simple. Further the ring (.7 )° opposed to Y,, is isomorphic with the full matrix ring (.f°)n over the skew field9" ° opposed to Y.
We consider an ideal a other than 0 of .Y,,. We have to prove that a = Y,,. To this end, we take an arbitrary element, other than 0, n+
a = L aijE,i i.j=1
of a, where the Eij denote the matrix units of Y,,, and at least one coefficient, say au, is different from 0. Because a,j' E,,c'-Ejs = E,.,
(r, s = 1, ..., n),
a contains all the E,,s, consequently all the elements of Y . So the first part of the theorem is proved. The second part of the theorem is easy to prove, namely,
(T,)°
(-170). (A -A),
where A' denotes the transpose of the matrix A (E F,,).
427
STRUCTURE THEOREMS OF WEDDERBURN- ARTIN
As a generalization of the concept of nilpotent elements, a left ideal (right ideal or ideal) a of an arbitary ring is called nilpotent, if there is a natural number n with a" = 0. A ring with minimal condition for left ideal is called semisimple, when it has no nilpotent left ideals different from 0. This nomenclature is justified by the fact that every semisimple ring is a direct sum of finitely many simple rings. We prove much more than this assertion. Namely the semisimple rings constitute a fully discovered structure class for which first of all the following extremely important theorem holds. THEOREM 256 (first structure theorem of WEDDERBURN-ARTIN). Every semisimple ring ( 0) is a direct sum of finitely many full matrix rings over skew fields. Let R (# 0) be a semisimple ring and V a minimal left ideal of 'R. Then V is a simple R-leftmodule. Also R V 0, since if R V = 0 then, a fortiori, VZ = 0, while V contains no nilpotent left ideals different from 0. Therefore we can apply the density theorem Of CEV P ALLEY - JACOBSON (Theorem
254). If we take into consideration the second parts of Theorems 158 and 255, then it follows that V is a vector space of finite dimension n over a skew
field 7, and the ring, which consists of the distinct .7-endomorphisms
a -* as
(a E -*)
(106.1)
of V, is isomorphic with the full matrix ring of rank n' over the skew field 3° opposed to.7. But this endomorphism ring is obviously a homo-
morphic image of S'z, because it is isomorphic with .9/a, where a is a left ideal (and so an ideal) of .
which annihilates V. Consequently ._`P/a
(70)..
(106.2)
We show that 5z' has the direct decomposition
=V ®a.
(106.3)
Next, it is evident that VV is an ideal of _. For the intersection
fl a, b2=bb
1=0.
=0.
Since S'z contains no nilpotent left ideal different from 0, it follows that b = 0. Therefore we have only to show that
-92 =V + a. We suppose this to be false. Since R/a, according to (106.2) and the first part
of Theorem 255, is simple, it follows that idthe eal (VV + a)/a of *la
OPERATOR MODULES
428
is equal to 0. This means that VR S a. Because b = 0, then V'* = 0, consequently V2 = 0. This contradiction proves (106.3).
Now for the first term of the right-hand side of (106.3), we have, by (106.2), the isomorphy
Vt
(3°)" .
(106.4)
But because of (106.3) every left ideal of a is also one ofd. Accordingly, together with JP, a (c 2) is a semisimple ring, with which (instead of '2) we can repeat the above process. But because of the minimal condition this must be broken up into finitely many steps. Hence, and from (106.3) and (106.4), follows the truth of Theorem 256. THEOREM 256' (second structure theorem Of WEDDERBURN-ARTIN). Every simple ring ( 0) with minimal condition for left ideals is isomorphic
with a full matrix ring over a skew field, or is a zero ring of prime order. Let 2 denote a ring, for which the conditions are satisfied. If it has no nilpotent left ideal other than 0, then it is semisimple and simple, therefore on account of Theorem 256 the theorem is true. In the other case let a denote a nilpotent left ideal of V different from 0.
If a
_ W, then a"2 = * (n = 1, 2, ...). This is impossible as a is nil-
c A. But since the left-hand side is an ideal of the right-hand side and this is simple, it follows that a 2 = 0. Thus the ideal consisting of the left annihilators of R is different from 0 and consequently equal to V. Thus 22 = 0. Since in conformity with this J` is a zero ring, the module '+ is itself simple, thus of prime order. Therefore the theorem potent. Accordingly a
is proved.
Theorems 256 and 256' with their converses, to be considered later, make a complete characterization of simple rings with minimal condition for left ideals or of semisimple rings possible.
THEOREM 256" (converse of Theorem 256'). The full matrix ring .3" of rank n2 over a skew field .7 (which is simple, according to Theorem 255), is also
semisimple. Also the .9,, left module .7 is the direct sum (106.5)
of minimal left ideals 11, ...,1", where l; consists of those matrices A (E which have only elements 0 outside the j`h column.
THEOREM 256" (converse of Theorem 256). Every direct sum of finitely many full matrix rings over skew fields is semisimple.
For the proof we need the following THEOREM 257. In every group with composition series for the normal divisors
the minimal condition is satisfied.
STRUCTURE THEORLMS OF WEDDERBURN- ARTIN
429
This theorem also holds for operator groups, and the corresponding theorem holds for rings with operators. If n is the length of the composition series of the group to be considered, then, from the theorem of JORDAN-HOLDER (Theorem 136), it follows that
every normal divisor of it has a composition series of at most length n, so Theorem 257 is proved. We now prove Theorem 256". It is obvious that 11, . . ., 1 are left ideals of .7; and that (106.5) holds for them. To prove the assertion that 11, . . ., 1 are minimal, it is sufficient to show it for 11. We have recourse to the matrix units E,1 of .7,,,. Then 11 consists of the matrices n
A=
i=1
(al...., an E .`) .
a.E,1
Of these we take an A different from 0, where e.g., ai is different from 0. It will suffice if we show that t-5rn A = 11. Because
Cra, 1Er,A=CrErl
(Cr E
;r
1,....n),
contains all the C,Erc, and so also the I?
Y c Erl
r=1
whence the assertion .7.A = 11 follows. Moreover we have to show that no left ideal I (# 0) of , , is nilpotent. For that purpose we take an element n
A= of 1. Let, e.g.
(air E 9r)
ai1E,1
ars=c'#0.
The matrix
B = ES,A = : ar,Esj j=1
lies in 1. Furthermore n
CCn
B- _ E ar;arjEs;Esj = L arsarjEsj = cB . 1,j=1
i=1
Since, according to this, B is not nilpotent, then neither is I. Now (106.5) means that -9',+, considered as an .5z-left module, is completely
reducible, since the minimal left ideals of .fin are the simple 5 -left submodules of 7,+,. Consequently it follows from Theorems 137 and 257
OPERATOR MODULES
430
that in .7. the minimal condition for left ideals is satisfied. This completes the proof of Theorem 256". In order to prove Theorem 256", we consider a ring
R=R1®...®R,,,, which is the direct sum of full matrix rings R1, . . ., R,,, over skew fields. By Theorem 256" R is the direct sum of finitely many minimal left ideals. So it
follows, as above, that the minimal condition for left ideals holds in R. Consider further a nilpotent left ideal I of R. The R; component of I is then a nilpotent left ideal of R, (i = 1, ..., m), which, by Theorem 256", equals 0, whence I = 0. Consequently Theorem 256'" has been proved. More generally, we can obta inimportant information concerning the socalled Artin rings, i.e. rings with minimal condition for left ideals, from the first structure theorem of WEDDERBURN-ARTIN (Theorem 256). The follow-
ing theorem is a necessary preliminary. THEOREM 258. The sum of two nilpotent left ideals of a ring is again nilpotent;
further, every nilpotent left ideal is contained in a nilpotent ideal. For the proof, let a denote a left ideal of a ring R with as = 0 for a natural
number a. Let Pk(a) = ... OC1 ... OC2 ...... OCk ...
((C1,
..., Sc, E a)
be an arbitrary product of elements from R with at least k factors from a. Since Pk(a) = (... OCl)
(... y.,) ... (... OCk) ...
and the partial products, included within parentheses, lie in a, it follows from as = 0 that all the PP(a) vanish. From this observation it is easy to derive the theorem. For, if b is a further ideal of R with b° = 0, then every element of (a + b)at8-1 is a sum of products Pa(a) or Pb(b), and thus equal to 0. Therefore a + b is nilpotent. Moreover, every element of (aR)a is a .Pa(a), thus the left ideal aR is nilpotent, consequently the ideal a + aR (? a) is also nilpotent. So the theorem is proved. A left ideal consisting entirely of nilpotent elements is called a nil left ideal.
Every nilpotent left ideal is then, afortiori, a nil left ideal. The following definition is of great importance : the sum of all the nil-
potent left ideals of a ring is called the radical of the ring. A ring with radical 0 is called radical ,free. THEOREM 259. The radical of a ring is always a nil ideal, that of an Artin ring is a nilpotent ideal. Let R denote a ring and it its radical. From the second part of Theorem
258 it follows at once that n is the sum of all the nilpotent ideals of R,
STRUCTURE THEOREMS OF WEDDERBURN-ARTIN
431
consequently it is itself an ideal of. Furthermore, every element of it lies in a sum of finitely many nilpotent left ideals of R, and so, because of the
first part of Theorem 258, is nilpotent. This proves the first part of Theorem 259.
Then let R be Artinian. We even prove that every nil left ideal a of R is nilpotent. Since a a2 ? ... is a descending chain of left ideals, there is a natural number a with aa = as+I (= as+z We have to show that a° = 0. With this end in view we suppose that the left ideal b = a° is not 0. Then b2 = b 96 0. From among the left ideals m of R, contained in b, with bm # 0, we take a minimal one. This m contains an element ,u with bu 0. But since by (C m) is a left ideal of R and bb,u = b,u 0, it follows that by = nt. Accordingly there is a (3 (E b) with fla = p. So ,u = pit = l32µ = = ..., i.e., µ = 0, since (3 is nilpotent. This contradiction proves Theorem 259. THEOREM 260. If the radical it of a ring R is nilpotent, then the factor ring R/n is radical free. According to the Supplement of Theorem 128', every left ideal of R/n may be taken as I/n for some left ideal f (;? n) of R. We suppose that I/n is nilpotent, i.e., for some natural number I (I/n)' = 0
.
We have to prove that I = it. The product of 1 arbitrary residue classes mod it of I is 0, so the product of 1 arbitrary elements of I lies in it. This implies that I' c it. Therefore from the nilpotence of it, that of I follows. Since accordingly 19 it, we have I = it as required. For this paragraph cf. SZELE (1954). EXAMPLE. Theorem 130' may be obtained from Theorems 256' and 256" as follows:
Let R (96 0) be a ring without proper left ideals. If R is a zero ring, then O(R) must be a prime number. Otherwise, it follows that R is a full matrix ring F. of rank n$ over a skew field F and necessarily n = 1, so that R = F. PROBLEM. Give for a skew field F a complete system of non-isomorphic F-subrings of the full matrix ring FA.
CHAPTER VII
COMMUTATIVE POLYNOMIAL RINGS Some fundamental facts on commutative polynomials of one indeterminate will be discussed here. Polynomials of several indeterminates will be examined partly by the use of results obtained here and partly by studying more closely the so-called symmetric polynomials among them. The fundamental domain will mostly be an arbitrary ring with unity element, but some-
times an integral domain or field, in connection with special problems. § 107. McCoy's Theorem
Polynomial rings with zero divisors have, so far, been given far less attention than those without zero divisors. The following theorem is very important. THEOREM 261 (McCoy's theorem). If the polynomial f(x) (# 0) is a zero divisor in the polynomial ring R[x], then there is an element c (# 0) of J with cf(x) = 0.
For the proof we take f(x) in the form f(x) = aox" + ... + a,,
(n >_ 0; at E .T; ao # 0) ,
(107.1)
and denote by g(x) (A 0) a polynomial in fi[x] with
fix) g(x) = 0 .
(107.2)
If g(x) is constant, then the theorem is proved. From now on we assume that
g(x) is of degree at least 1. It will suffice to prove that in R[x] there is a polynomial h(x) (# 0) of smaller degree than g(x) with the property that f (x) h(x) = 0 , since from repeated application of this the theorem will follow. Let us consider the polynomials aog(x), .
. .
432
a"g(x)
(107.3)
McCOY's THEOREM
433
If these all vanish and c (# 0) denotes the leading coefficient of g(x), then
aoc = ... = ac = 0. Because of (107.1) we then have f(x)c = 0, therefore the theorem is now true. Otherwise there is an r (0 5 r S n) with
aog(x) _ ... = ar-1 g(x) = 0,
(107.4)
a g(x) s 0 .
(107.5)
Now (107.2) may be written, by (107.1) and (107.4), as:
(arx"-' +
... + a,Jg(x) = 0.
Hence it follows that the polynomial
h(x) = arg(x)
(107.6)
is of smaller degree than g(x). By (107.5), h(x) # 0. Finally from (107.2) and (107.6) we obtain (107.3). Hence the theorem is proved. Cf. McCoy (1948). EXAMPLE. Theorem 261 is false for non-commutative polynomial rings 12[x]. Consider in this respect the polynomial 1
f (x) = -1
-1, x- (1 l)
in 9$[x]. Although for
g(x) =
1 -1
11 1,1
l) x + (l -1)
the equation g(x)Ax) = 0 holds, we have (e d
for every element (a dl ( c
f(x) # 0
0) of 9=, as is obvious.
§ 108. Differential Quotient As is well known, in function theory the differential coefficient of a function is defined as a limit. The following definition of the differential quotient of polynomials is not based on the notion of a limit. First of all let R be an arbitrary ring with unity element, for which we shall later also postulate commutativity. We consider an arbitrary poly-
nomial f(x) in R[x]. Form by the help of a further indeterminate y the polynomial f(x + y) (in the polynomial ringR [x, y]) and write it in the form
Ax + y) =.fo(x) + Yfi(x) + Ysf2(x) + ... ,
(108.1)
COMMUTATIVE POLYNOMIAL RINO3
434
where the coefficients fi(x) are uniquely defined polynomials in '*[x]. If we put y = 0, then fo(x) = f(x). Hence, and from (108.1), we have the congruence
f(x + y) = f(x) + yf1(x) (mod y2) with respect to the ideal (y2) of R[x, y]. The polynomial f1(x), which we call the differential quotient of f(x) and denote by f'(x), is uniquely defined
by f(x). Therefore its definition is contained in the congruence J (x + y) = f(x) + yf'(x) (mod y2) .
(108.2)
We denote a further polynomial from R[x] by g(x). Then g(x + y) = g(x) + yg'(x) (mod y2) . Hence, and from (108.2), after addition and multiplication we obtain
f(x + y) + g(x + y) =.f(x) + g(x) + y(f' (x) + g'(x)) (mod y2) , J (x + y) g(x + y) = f(x) g(x) + y (f'(x) g(x) + f(x) g'(x)) (mod y2) . Comparison with definition (108.2) yields the formulae for the differential quotient of the sum and product of two polynomials:
(f(x) + g(x))' = f'(x) + g'(x) ,
(108.3)
(Ax) g(x))' = f'(x) g(x) + f(x) g'(x) .
(108.4)
Of these (108.3) means that the mapping f (x)
f' (x) is an endomorphism
of the module A[x]+. Formula (108.4) is known as Leibniz's rule. Both formulae (108.3), (108.4) may be immediately generalized to k
ll'
i=1
k
(108.5)
i=1
Ma1 ... ak I, _ E a1 ... ai-1 ai ai+1 ... Mk ,
i=1
/
i=1
(108.6)
where al,..., ak are arbitrary polynomials of Jp[x] (k >_ 1). We now wish to determine f'(x) explicitly. From (108.2) we obtain
c'=0, x = 1
(cE, ).
Hence, and from (108.6), first (xk)' = kxk-1
(k >_ 1) ,
(108.7)
435
DIFFERENTIAL QUOTIENT
then, by (108.4), (cxk)' = kcxk'1 (c E 2). So, because of (108.5), the explicit formula required for the differential quotient of an arbitrary polynomial of i[x] reads: (ao + a1x +
... +
ax")' = a1 + 2a2x +
... + nax"-1.
(108.8)
We see that (108.8) agrees formally with the rule from function theory for the differentiation of polynomials. However, formula (108.8) means much more, as it refers to an arbitrary polynomial ring rk [x], whereas in function theory usually only the field of real or complex numbers is admitted for the fundamental ring rk. Of course we could have defined the differential quotient immediately by (108.8)
but the idea of the original definition (108.2) is clearer, and the application of (108.2) is often more convenient than that of (108.8).
We gather from (108.8) that the differential quotient of a polynomial
f(x) = ao +
... +
(108.9)
vanishes if, and only if,
o+(a,)Ii
(i=1,...,n).
(108.10)
Accordingly we may even have f'(x) = 0, when f(x) is non-constant. The polynomials f(x) for which f(x) = 0 are called polynomials with vanishing differential quotient. These are thus characterized by condition (108.10). It is in the case of a zero-divisor-free fundamental ring .2 that the question of the vanishing of f'(x) is of the greatest importance. By Theorem 38,
the characteristic of R is then equal to 0 or to a prime number p. In the first case (108.10) implies that a1 = ... = a" = 0, i.e., that f(x) is a constant. Since, in the second case, o+ (a;) = p for a1 0 0, (108.10) now implies that in the sequence ao, al, ... at most the terms a0, ap, a2p, .. .
are different from 0, i.e., f(x), according to (108.9), is of the form
ap + apxp + a2p x2p + ... ,
i.e., a polynomial in x". This result, for both cases, is contained in the following
THEOREM 262. In a zero-divisor free polynomial ring '[x] of characteristic
p (= 0 or a prime) all the polynomials with vanishing differential quotient are the f(xp) (f(x) E 2[x]). From (108.8) we also see that for a non-constant f(x) the differential
quotient.f'(x) is of smaller degree than f(x). If f(x) is of degree n with the leading coefficient c, then f'(x) is of degree n - 1 if and only if, nc
0.
COMMUTATIVE POLYNOMIAL RINGS
436
We define the ia` differential quotient (or the differential quotient of order 1) f te(x) of f (x) by
(i = 0, 1, ...) ,
(f 0(x))'
where it is understood that f) = f. In particular f(l) = f'. For f 2), f« we also write f",f"'. From function theory we borrow the following notations: ddxr)
=
dd-x'f(x) =.f"(x)
and in particular df(x) dx
=
d .l(x) =f'(x)
dx
Since a polynomial f (x1, ..., x,,) (E [xi, ..., x"]) may be considered as a polynomial in xk over 9[xl, ..., xk-v xk+v ..., we can correspondingly form the differential coefficient
dk f(xI,...,xa, which is called the (partial) differential quotient of f (x1, ...,
with
respect to xk, and to avoid possible misunderstanding, we denote it by a
axk J XII .
.
., x") .
We also write frr(xj, ..., for this or, sometimes, fk(xj, . . ., x.). 263 (EuLui 's theorem). For a homogeneous polynomial f (x1, .,
of degree r in an arbitrary polynomial ring tR [xI, . . ., x"] k
i=1
a
y
xr - (xl, ..., xa) = rf (x1, ..., x,,) ax,
It suffices to prove the theorem for a one-termed polynomial of the form
(k1+... +k"=r). Since we now have xI
x f (x1, ..., x,,) = krf(xI, ..., x,,) , f
we obtain the theorem by summing for i = 1, ..., n.
DIFFERENTIAL QUOTIENT
437
If in the polynomial P'(x) the indeterminate x is replaced by c (ER) or by g(x) (E 51 [x]), then we denote the result of this replacement by
f (')(c) or by f tn(g(x)). A similar notation is applied in connection with several indeterminates.
From now on let us suppose that * is commutative. If f(x), g(x) are arbitrary polynomials in,* [x], then we obtain by repeated application of (108.2) :
f(g(x + y)) m f(g(x) + yg'(x)) = f(g(x)) + yg'(x)f'(g(x)) (mod y2 . Again, because of (108.2), this implies that (108.11)
(f (g(x))' =f'(g(x)) g'(x) ,
where the differential quotient of the iterated polynomial f(g(x)) is denoted on the left. Especially we have
(f (x + c))' =Ax + c)
(108.12)
(c E JT) .
We prove the more general rule d
f(gI(x),
, gn(x)) = E fi(gl(x)
, gn(x))
,
t=1
dg,{x) dx
,
(108.13)
where
a
(i = 1, ..., n).
f (xl, ..., x,. ) = ax, f ( x t , ..., x n ) Write
F(x) = .1(gl(x), ..., gn(x)) then according to (108.2)
F(x + y) ° .1(gl(x) + ygi(x), ..., gn(x) + ygn(x)) (mod y2) .
(108.14)
But from (108.2) we obtain the congruence
f(x1 + ti, ..., xn + tn)
f(x1, x2 + t2, ..., X. + t,.) + tlfl(x1, x2 + t2i ..., X. + t .)
(mod ti)
in the polynomial ring S[x1, ..., xn, t1i ..., tn]. If a denotes the ideal generated by all the tits, then we obtain from the above congruence by continued application of (108.2): n
.1(x1 + 1 1,--.
E tif, (xl, ..., x,) (mod a) . .Xn + tn) =.1(x1, ..., xn) + =1
438
COMMUTATIVE POLYNOMIAL RINGS
After that we carry out the substitutions xr = 9r(x), ti = yg (x)
Since every element of a becomes an element of the ideal (y' and because of (108.14) we obtain
F(x + y) = F(x) + y
r=1
9i (x) ff (91(x), ..., 9"(x)) (mod ys).
This, together with (108.2), proves (108.13). THEOREM 264 (TAYLOR'S theorem). For every polynomial f(x) of degree n
over a field .7 of characteristic 0 or p > n
AX + c) = E =o
i!
x'
(c E .7)
(108.15)
.
[In (108.15) the denominator i! is to be regarded as an element of J r, namely as i t times the unity element.] By hypothesis
J(x+ c)=bo+... +bx"
(bo,...,b"E. ').
After differentiating i times, we obtain for x = 0
f(O(c)=i!b,
(1=0,...,n); -
then bi = (fl)-'f() (c). Thus the theorem is proved. EXERCISE. In an arbitrary ring R following BOURBAKI (1939) we understand by
differentiation any endomorphism a -- a' of R+ with the additional property that (a3)' = a'/3 + afl' ((X, fl E R). This yields a generalization of differentiation of polynomials. Rules (108.5) and (108.6) also hold here. If, moreover, R is an integral domain, then a - a' may be extended to give a differentiation in the quotient field F of R as follows a
a l'
T
a'p - a#' #2
(a, P E R ;
0)
In the ring J the zero endomorphism of J+ is the only differentiation.
§ 109. Field of Rational Functions We denote a field by -9'. Since the polynomial ring . '[x1, ..., x"] is an integral domain, it has a quotient field, which we call the field of rational functions (or rational function field) of the indeterminates x1, ..., x" over and denote by ,7(x1, ..., x"). Its elements
r(x1, ..., x") =
.(x1, . . ., x,)
9(x1, ..., xa
(9(x1, ..., x') # 0)
(109.1)
FIELD OF RATIONAL FUNCTIONS
439
are called the rational functions of the indeterminates x1, . . ., x over Y, where numerator and denominator are elements of -'X[x1,..., xn]. If we consider the polynomial ring .7 [x1, x2, ... ] with infinitely many indetermi-
nates, then we obtain in the same way a field of rational functions of infinitely many indeterminates, which we correspondingly denote by '(x1, x2, . . .). Each of its elements contains only finitely many indeterminate,, so it is likewise a rational function. The term "rational function", borrowed from function theory, is, admittedly, not very suitable as it has nothing to do with functions as such. On the contrary, (109.1), after replacing the x1, ..., xn by variable elements t1, ..., t,, of f, becomes a proper function .
g(t1,
.., Q
where, of course, only the t1, . . ., t such that g(t1i . . ., t,,) # 0 are admitted. Since, according to GAUSS'S theorem (Theorem 207), -9'[x1, ..., xn] is a ring with prime decomposition, we may suppose in (109.1) that the numerator and the denominator have no common divisor. Then (109.1) is called
the reduced form of the rational function or a reduced rational function. In this the numerator and the denominator are, apart from a common
constant factor c (E"
0), uniquely defined. We occasionally refer to them
as the numerator or denominator, respectively, of this rational function. The maximum of their degrees is said to be the degree of the rational function. In particular, a rational function a1x1 + ... - anx,, + a b1x1 + ... + bnxn + b
of the first degree is called a linear rational function and, if a = b = 0, a homogeneously linear rational function.
In (109.1) numerator and denominator may also be considered as polynomials in xn over .52'[x1, ..., xn_1]], whence it follows that
-7'(X1, ..., xn) _ (.7[X1..... xn-1J)(xn)
According to this Y(x1, ..., xn)
(_7(X11
..., xn-))(Xn) ``
Since this also holds trivially with Q instead of S, it follows that
J,(x1, ..., x,.) _ (7(x1, ..., Xn-1))(Xn) Because of this recurrence formula it will be sufficient to restrict ourselves,
in everything concerning fundamental questions, to the consideration of .7(x). 15/a R.- A.
440
COMMUTATIVE POLYNOMIAL RINGS
§ 110. The Multiple Divisors of Polynomials
First we consider an arbitrary commutative ring R with unity element. Of two elements a, j9 (E R) we call a a k ple divisor or a k ple factor of f
(k=0, 1,...), if
ak
Y
(110.1)
ak+1
and we use for this the notation (110.2)
akIIP
If k >_ 2, then we say that a is a multiple divisor or multiple factor of P. If, moreover, R is zero-divisor-free, and so an integral domain, then (110.2) evidently means that there is an element y of R with # = aky , a,f'y
If a is a unit, or P = 0, then (110.2) is not satisfied by any k. In what follows we are concerned only with the case where R is an integral domain with prime decomposition. Apart from the exceptional cases just mentioned, there is exactly one k for each pair a, f (E R) for which (110.2) holds. The case where a is a prime element is important. In this respect we note the following: if e?CT
... 79r
is the prime-power decomposition of f, where N is a unit and
v1, .. y 7cr
are the distinct prime factors of j9, then
r) . We now consider a commutative polynomial ringR[x]. We know from the corollary of Theorem 204 that an element c (E ') is a zero of a poly-
nomial f(x) (E 2[x]) if, and only if, x - c I f(x). More generally, if (x - c)k I I f(x)
(110.3)
we call c a k ple zero of the polynomial f (x) or a k ple root of the equation f(x) = 0. If k Z 2 we use the terms multiple zero or multiple root. THEOREM 265. If, for two elements f (x), g(x) of a commutative polynomial
ring 2[x] and an integer k (>_ 1), (g(x))k I fix)
,
(110.4)
then (g(x))k-I If(X)
(110.5)
441
MULTTPLE DIVISORS OF POLYNOMIALS
According to the given supposition f(x) _ (g(x))kh(x)
(110.6)
for some h(x) E .R[x]. Hence
f'(x) = kg'(x)(g(x))k-I h(x) + (g(x))kh'(x),
(110.7)
and, consequently, the theorem is valid. It is important that under certain circumstances this theorem should hold
with "! l" instead of "I". The following theorem brings out this point. THEOREM 266. Let -W be an integral domain of characteristic p (= 0 or a prime number) with prime decomposition. If, for an irreducible polynomial g(x) and for a further polynomial f (x) in R [x], (g(x))k I I f(x)
,
k ? 1 , p , k , g'(x)
0
,
(110.8)
then
(g(x))k -' ! ! f'(x)
(110.9)
According to GAUSS'S theorem (Theorem 207) R[x] is also an integral domain with prime decomposition. Thus if we assume that f(x) has the form (110.6), where now, because of (110.8), Ax) 'f' h(x)
then (110.7) reveals the absurdity of (g(x))k f'(x). Hence, and from Theorem 265, Theorem 266 follows. From this follows immediately
TimOREM 267. If ii (and so also fi[x]) is an integral domain with prime decomposition and j (x) a product of irreducible linear polynomials in _ [x], then (f(x), f'(x)) = 1 is a necessary and sufficient condition for f (x) to have no multiple divisors. § 111. Symmetric Polynomials We take an arbitrary ring with unity element. A polynomial f (x1, ..., xA) is called symmetric if it is invariant with respect to all permutaover
tions of the indeterminates x1, . . ., With the help of a further indeterminate z we form the polynomial
2-...+(-1)"s. , (111.1)
COMMUTATIVE POLYNOMIAL RINGS
442
where s1, .
.
., s belong to * [x1,
. .
It is evident that by this the
., xn].
s1i ..., s are uniquely defined and symmetric. We call s; the ith elementary symmetric polynomial in the indeterminates xl, .... xn (i = 1, . . ., n). Since the left-hand side of (111.1) is homogeneous in the indeterminates z, x1, . . ., xn, the right-hand side shows that 1 is at the same time the degree of s,. By (111.1), the explicit definition of the elementary symmetric polynomials is as follows:
S1=x1+x2+...+xn, S2=XIX2+X1X3+...+XJxn+...+Xn-lXn, S3 = X1X2X3 + X1X2X4 + ... + Xn-2Xn-1Xn,
TtnoRa1268 (main theorem for symmetric polynomials). Iff(x1i ..., x,,) is a symmetric polynomial over a ring R with unity element, then there is exactly one polynomial F(x1i ..., xn) over *such that -
f(x1, ..., x,,) = F(s1, . . ., s,) ,
(111.3)
where si is the ith elementary symmetric polynomial of x1, ..., xn (i = 1, ..., n). Furthermore the coefficients of F(x1i ..., x,,) belong to the submodule of $ generated by the coefficients of f(x1, ..., xn). The first part of the proof, in which we verify the existence of F, provides at the same time a process for determining the coefficients of F. We order f(xl, . . ., x,) lexicographically and denote by ax,k,
...
X,k,n
(aE
_
), #0)
(111.4)
the lexicographically last term of it, which we call its principal term. Since . ., x,.) is symmetric, it contains all the terms resulting from (111.4) by the permutation of the x1, . . ., x,,, whence
fxl, .
k1 ?
> kn.
We then consider gSki-k'S2'-k..
S,k,-1_knsnn
By (111.2) the principal term of (111.5) is obviously axi`_k' (x1 x2)k'-k'... (X1
...
xn-1)ka-I-kn (x1
... xn)kn
(111.6)
SYMMETRIC POLYNOMIALS
443
Since this is equal to (111.4), it follows that all the terms of
f1(x1, ..., x,.) = (x1, ..., x") -
asi'-k2
f
... sn"
(111.7)
in the lexicographical sequence precede the term (111.4). Now f1(x1, ..., is also symmetric in x1, ..., x", wherefore the process may be repeated with it instead of f(x1i ..., x"). The principal terms, occurring at each step, constitute, according to what was said above, a lexicographically decreasing sequence and they are all of the form
bxl'...xt
(b ER)
with
11>...z1", 115k1. Since there are only finitely many possibilities for these exponent systems 11, ...,1., it follows that the process terminates after finitely many steps. If we write (111.7) in the form
f(x1, ..., x,,) = fl(X1...., x,,)
asi'-k' ... Jrs"
we see that it leads to the determination of a required polynomial F(x1,..., x,). It is evident that the last assertion of the theorem is satisfied by this. We have still to prove the assertion of the theorem regarding uniqueness. Evidently it will suffice to show for a polynomial G(x1, . . ., over.2 other than zero that G(s1i . . ., s") is also different from zero. With this end in view we consider an arbitrary term of G(x1, ..., x,,.), which we write in the form axi'-k' ... --k" xn" (a E R; k1 >_ ... z k") . (111.8) If we replace x1, ..., x by s1,.. ., s", then according to (111.2) we obtain a polynomial in the x1, . . ., x" whose lexicographically last term is evidently (111.6), i.e., (111.4). If moreover (111.8) is the principal term of G(x1, ...,
x"), then the term (111.4) of G(s1, ..., s") is not eliminated. This completes the proof of Theorem 268. ExAMPLE. We have (cf. § 115)
xi+xq=s1-2s2, xi+x$=sj-3s,s2 (s,=x,+x2; s2=x,x1) § 112. The Resultant of Two Polynomials To decide whether two polynomials have a non-constant common divisor is a fundamental problem. We shall deal here with the simplest case, where
there are two polynomials f(x), g(x) over a field. The question can be answered by means of the Euclidean algorithm, which immediately gives the
444
COMMUTATIVE POLYNOMIAL RINGS
greatest common divisor; on the other hand, the above question of existence is, as we shall see, much easier to answer. First of all we take an arbitrary commutative ring R with unity element and in it form a so-called Sylvester determinant
ao...am rows
R =
ao
bo ..b
.. am
(112.1) rows
bo
b
of order m + n (m, n > 0), where the a,, b, denote elements of R and with only zero elements in the empty spaces. By virtue of the Theorem 270 below, R is called the resultant of the polynomials
f(x)=aox'"+...+am g(x)=box" +... +b
(m > 0),
(n > 0),
(112.2) (112.3)
so that we also use the notations R = R(f(x), g(x)) = R(f, g) .
(112.4)
We often consider R to be a polynomial in (,R =) _7 [ao, ..., am, bo, ..., The product as bn of the diagonal elements of (112.1) is called the leading
(principal) term of the resultant R. It should be noted that R(f, g) is not uniquely defined by the polynomials f, g, since, in general, m, n, denote the formal degrees of f and g. Thereby, if necessary, we have to use the notation R,,(f, g) for the resultant. One could ensure uniqueness by demanding that ao 96 0, bo 96 0, but this would detract from the general validity. From now on, m, n will denote fixed chosen numbers, so that we may keep to the shorter notation (112.4).
In the resultant R(f, g) the order of succession of the polynomials f(x), g(x) makes a difference, but by (112.1)
R(g,f) = (-1)m"R(f, g).
(112.5)
THEOREM 269. The resultant R of the polynomials f(x), g(x) given in (112.2) and (112.3) over a field 9 is equal to 0 if, and only if, these polynomials have a non-constant common divisor, or when ao = bo = 0.
445
RESULTANT OF TWO POLYNOMIALS
If ao = bo = 0 then, by (112.1), R = 0. Thus the theorem is true for this case. Therefore in the following we suppose that a0, bo do not both vanish. We have to prove that the greatest common divisor d(x) _ (f(x), g(x))
(112.6)
is not constant if, and only if, R - 0. First of all we show that d(x) is not constant if, and only if, the equation (112.7)
f(x) g1(x) -I- g(x)f1(x) = 0
can be satisfied by two polynomials fl(x), g1(x), which do not both vanish and are of smaller degree than m or n.
For that purpose we first assume that d(x) is not a constant. Then, because of (112.6)
g(x) _ -d(x) 91(x) J(x) = d(x)f1(x), for two polynomials fi(x), g1(x), which possess all the required properties [including also (112.7)]. Conversely, we now suppose that the required polynomials f1(x), g1(x) exist. It has then to be proved that d(x) is not constant. We may suppose that ao # 0, whence it follows that f(x) is not constant. If g(x) = 0, then the
statement is self evident, wherefore we have only to consider the case g(x) # 0. Since f1(x) = g1(x) = 0 is not true, it follows from (112.7) that neither f1(x) nor g1(x) is equal to 0. It will suffice for the following that f1(x) s 0. Since according to (112.7) we have .f(x) I g(x)f1(x)
and f1(x) has a smaller degree than f (x), it follows that g(x) is divisible by at least one prime factor of f (x). This means, by (112.6), that d(x) is not a constant. By this means we have proved the statement concerning (112.7). We now take
f1(x) = uox-1 + ... + um-1 91(x) = vpxn-1 + ... + vn-1
(u, E .}) ,
(v. E 9)
.
Equation (112.7) is then identical to +bou0 + blu0 + b0u1 + b2uo + blur + b0u2
a0vo
a1v0 + aovl a2v0 + a1v1 + aov2
amen-2 + am-lvn-1 amen-1
+ bnum-2 + bn-lum-1 + bnum-1
=0, =0, = 0,
= 0, =0,
446
COMMUTATIVE POLYNOMIAL RINGS
where the number of equations and that of the unknowns is m + n. By what was stated above d(x) is not constant if, and only if, this homogeneous
system of equations is not trivially solvable. For this, on account of the corollary to Theorem 248, it is necessary and sufficient that the determinant of this system of equations should be equal to 0. Since this determinant is exactly the transpose of the Sylvester determinant (112.1), and so is equal to this, Theorem 269 is proved. THEOREM 270. For the resultant R of two polynomials f(x), g(x) we have
f(x) F(x) + g(x)G(x) = R
(112.8)
with two suitable proper polynomials F(x), G(x) whose degrees are smaller than the given formal degrees of g(x) andf(x), and whose coefficients lie in the ring generated by the coefficients off(x) and g(x). For the proof we add, for j = 1, . . ., m + n - 1, the j`h column of the determinant (112.1), multiplied by xm+n-', to the last column. This leaves
the value of this determinant unchanged but its last column, because of (112.2) and (112.3), will consist of the elements xn - I.f l x ) , ..., I ( x ) , X, - Ig(x), ..., g(x)
Thus if it is expanded with respect to the last column, the theorem follows. In general, by the weight of a one-term polynomial cxp
... xr' ... ... zl r zQ°
(c # 0)
in .xo...., x ..., zo, ..., zf we mean the sum
it+2i2+...+rir+... +11+21.,+...+tlj. A polynomial f(xo, ..., x,., ..., zo, . . ., zr), is called isobaric if all its terms are of the same weight, which is then called the weight of the polynomial. It should be noted that, if st, . . ., sn are the elementary symmetric polyis a homogeneous symmetric polynomials in x1, . . ., x and f(xt, . . ., nomial of degree k, and so, according to Theorem 268, equal to a polynomial F(st, ..., sn), then it is isobaric with respect to s1, . . ., sn and of weight k. THEOREM 271. The resultant R given in (112.1) is a homogeneous isobaric polynomial of the ao, ..., am, b0, . . ., bn of degree m + n and of weight mn. With respect to ao, ..., am and with respect to bo, . . ., b", R is likewise homogeneous of degree n and m, respectively.
The assertion on the homogeneity and the degrees are trivial. We have
only to prove that every term of R is of weight mn. For that purpose
RESULTANT OF TWO POLYNOMIALS
447
we first consider an arbitrary determinant
D=IC1,I of order m + n, and note at once that under the replacement C
(1 < i < n) ,
= { a,_1
(n+1 0) ,
(112.10)
g(x) = bo(x - z2) ... (x - zn)
(n > 0),
(112.11)
f(x) = ao(x - yl) .
.
we have the three product representations m n
R=aobo fi l(y, -z1),
(112.12)
R = ao 11 g(Yk) ,
(112.13)
k=11=1
k=1
R = (-1)'nn bi fl f (z,)
.
(112.14)
/=1
First we prove (112.12). Here, we may assume that ao, bo, yl, Ym, z1, ..., zn are indeterminates over.5ro, so that after suitable substitutioila the general validity of (112.12) will follow.
448
COMMUTATIVE POLYNOMIAL RINGS
We write the given polynomials (112.10), (112.11) in the form (112.2), (112.3), where then
ar = (-1)'aosr
(1 = 1, ..., m) ,
(112.15)
br = (-1)'botr
(i = 1, ..., n),
(112.16)
while the elementary symmetric polynomials of yl, ..., y, and are designated by s1, ..., sm and t1, . . ., tn. The divisibility
z1, ..., z, (112.17)
ao bo I R
follows from (112.1), (112.15) and (112.16).
We show further that
Yk - zlI R
(k= 1,...,m; 1 = 1,...,n).
(112.18)
If yk is replaced by z1, then by (112.10) and (112.11) f(x), g(x) have the
common divisor (x - yk =) x - z1. Hence, because of Theorem 269 it follows that R vanishes after this substitution. This implies, by the corollary to Theorem 204, the divisibility (112.18). Since, by Gauss's theorem (Theorem 207) 9ro [ao, bo, Yl, Ym, Z1, Zn] is now a ring with prime decomposition, it follows from (112.17) and (112.18) that R is divisible by the right-hand side of (112.12). But it follows from Theorem 271, having regard to (112.15) and. (112.16),
that the polynomial R is homogeneous of degree n with respect to ao and also homogeneous of degree m with respect to bo. Furthermore it is of weight mn with respect to sl, ..., s,, t1i . . ., t,,. This means that R is homogeneous
of degree mn with respect to y1i ..., ym, z1, ..., zn. From the divisibility above m
,
k=1r 1 for some integer q. In order to determine this we substitute y1 = ... = ym = 0. By (112.15) a1= ... = am = 0, thus our equation, on account of (112.1), becomes then m
n
aob. = gaobo
(- zr)I r= I
The right-hand side is equal to (-1)gaobo t'", and so, according to (112.16), equal to gaob'. Hence it follows that q = 1, implying the truth of (112.12). From (112.12) because of (112.11) we obtain equation (112.13) and also,
by (112.10), equation (112.14). Consequently Theorem 272 is proved.
RESULTANT OF TWO POLYNOMIALS
If JV [x1, .
.
.,
449
xn], Y[x1, ..., xn] are two polynomial rings, in which 5°
is an overring of '9, then we say that 9'[x1, ..., xn] is obtained from .51+[x1, ..., xn] by an extension of the fundamental ring. A non-constant of a commutative polynomial ring .Ji[x1, ..., xn] polynomial f(xl, . . ., is called absolutely irreducible, if there is no commutative polynomial ring So[x1i ..., xn] arising from *[x1, . . ., xn] by an extension of the fundamental ring with the property that f(x1, ..., x,,) splits into a product of two non-constant polynomials in ..?[x1..... xn]. THEoREM 273. The resultant (112.1) is an absolutely irreducible polynomial
of ao, ..., a., bo, ..., b,,. We take a commutative polynomial ring .9'[a0..... am, b0..... arising from .51[a0, ..., am, lo..... bn] by an extension of the fundamental ring, and consider in it a decomposition of (112.1) into two factors F, G:
R=FG.
(112.19)
We subtitute (112.15) and (112.16), where s1, ..., sm and t1, ..., t again denote the elementary symmetric polynomials of the indeterminates Ym and z1, ..., z,,. Then R becomes the right-hand side of (112.12), Y1, while F and _G become symmetric with respect to y1, ..., ym as well as to z1, ..., z,,. From (112.19) it follows that YI - z1 is a divisor of F or of G. We may suppose that y1 - z1 I F. Because of the above symmetry F must then be divisible by all the factors Yk - zr in (112.12), which we denote irrespective of order by 11, ..., l,,, and show that F = 11... 1mnFo
f o r some polynomial F 0 from R = Y[ao, bo, y1.
(112.20) Ym, z1.
In addition to this we assume that
F= 11... 1,F1, for some r such that 1 5 r < mn and for some F1 from R. It is sufficient to prove that this is also valid for r + 1 instead of r. Let 1,+1 = Yk - z1. On account of the residue theorem for polynomials (Theorem 204), F* = 0 follows from 1,+1 F, i.e., If ... 1; Fl = 0 where "*" denotes the substitution yk -+ z1. But since the li, ...,1*, according to McCoy's theorem
(Theorem 261), are not zero divisors, it follows that F; = 0. Again by Theorem 204, it is implied that 1,+1 1 F1. This proves (112.20). Since now, according to (112.12), R = 11... lmnaobo and the 1, are not zero divisors, it follows from (112.19) and (112.20) that aobo = F0G. On the other hand, because of (112.1) ao, ho (before the substitution), are not divisors of R. Thus, by (112.19), neither are they divisors of G. Hence it follows immediately that G is a constant. This proves Theorem 273.
450
COMMUTATIVE POLYNOMIAL RINGS
EXAMPLE 1. The resultant of the polynomials ax + al, box + bl is ao a, bo b,
EXAMPLE 2. The resultant of the polynomials aox$ -,- alx + a2, box2 + blx + b2 is ao al a2 0 0 ao al as bo bl b2 0
0 bob, b2
ao a, a20 ao as
bo bl b2 0
= - 0 ao al a2 =
I
bo b2
(-
ao a,
al a.
bo bI
b, bs
0 be b, b2
EXAMPLE 3. Theorem 269 is false if, instead of the field .F', we take as a basis the integral domain 9[ J- 5], as in this the polynomials (cf. § 88, Example 2)
3x2+4x+3,
3x+2+V=5-
have no non-constant common divisor, although their resultant is 0. EXAMPLE 4. The proposition in Theorem 269 that the resultant R is equal to 0, if f(x), g(x) have a non-constant common divisor, also follows immediately from Theorem 270, even for an arbitrary integral domain instead of the field 31' . EXAMPLE 5. For the resultant R(f, g) of the polynomials given by (112.2) and (112.3) use the more precise notation R,,,,,(f, g). Then it follows from (112.1) that Rm.n+.t(f,g) = au Rmn(f, g)
(k = 1, 2, ...).
EXAMPLE 6. From the product formula (112.14) it follows that Rm+n. p(fg, h) = Rmp(f, h) R,(g, h)
(112.21)
for three polynomials f(x), g(x), h(x) of formal degrees m, n, p (> 0) primarily for the case where they may be decomposed into a product of linear factors, but also generally, by Theorem 299, (to be discussed later). We call (112.21) the multiplication theorem for resultants. EXAMPLE 7. The polynomial x2 + y3 is absolutely irreducible, but x2 + y2 is not
absolutely irreducible, since in the field of complex numbers x2 + y2 = (x + yI)
(x - A.
§ 113. The Discriminant of a Polynomial Let a polynomial
f(x) = aox" + ... + aA
(n > 0)
(113.1)
be given and suppose for the time being that its coefficients ao, ..., a,, are indeterminates (over Y). Its differential quotient is
f(x) = naox"-I + ...+ a"_I.
(113.2)
In the Sylvester determinant (112.1) defining the resultant R(f, f'), the elements of the first column are divisible by ao, i.e.,
in .7[ao,
. .
.,
1 R(f, f') is a polynomial ao
a"]. For certain reasons to be clarified later, we assign the
451
DISCRIMINANT OF A POLYNOMIAL
factor (-1)(2) to this polynomial; we call it then the discriminant of the polynomial J (x) and denote it by
D = D(f(x)) = D(f) _ (- G) 1 R(ff').
(113.3)
ao
We immediately extend this definition to the case where f(x) lies in an arbitrary commutative polynomial ring R[x], so that then 40, ..., a, are to be replaced by the coefficients of f(x). Definition (113.3) fails if n = 1,
but in this case we write D(f) = 1. In order to transform (113.3) we perform the permutation
n=
1
3... n-1 nn+1 n+2... 2n-1
2
24
6
2n
2
2I
5
3
1
n -
on the rows of the determinant D. Since sgn n = (- 1)(2 it follows from 113.3) (after dividing by ap) that I
(n - l) a1... a1...
n
an-1 an-1 an
... an-1
nao (n - 1) a1
D=
a0
... an-1 an
a1
...
nap
2n - 1 rows
an-1
To explain this formula we compute the discriminant of apx2 + alx + a2: 2
a1
01
2ap
0
= a2 - 4apa2
a1
(113.4)
a1 a2
and that of a0x3 + alx2 + a2x + a3: 13
2a1
a2
0
0
1
a1
a2
as
0
0
3a0
2a1
a2
0
0
ap
a1
a2
a3
0
0
3a0
2a1
a2
= ai a2 - 4ai a3 _4a04 + + l8a0a1 a2 a3 - 27ap a3 . (113.5)
Both discriminants are lexicographically ordered. If necessary, we use the more precise notations
Dn(f) or D(ap xn + for D(f).
... + an)
COMMUTATIVE POLYNOMIAL RINGS
452
We now prove the formula
D(Ox" + alx"-1 + ... + a") = a,D(a1x"
+ ... + a") .
(113.6)
The left-hand side is obtained from the determinant (113.3') by replacing ao by 0. If we apply Laplace's expansion theorem to the first two columns, we then obtain (113.6). THEOREM 274. The discriminant D of the polynomial f(x) (& 0) in (113.1)
over a field,r "is equal to 0 if, and only if, f(x) has a non-constant multiple factor or an irreducible factor with vanishing differential quotient, or when ao = a1 = 0. If the formal degree n of f(x) is equal to 1, then D = 1, so that the theorem
is then trivially true. It is also true if a = a1 = 0, since then, according to (113.3'), D = 0. We now assume that n 2 and ao, a1 do not both vanish. If ao
0, then according to (113.3), D is equal to 0 if, and only if,
R(f,,f') = 0. On account of Theorem 269, this means that f(x), f(x) have a common irreducible divisor g(x). Furthermore this is the case if, and only if, g(x) is a multiple factor of f(x) or is a factor of Ax) and g'(x) = 0 [see (110.7)]. Thus the theorem is true for ao 0 0. Finally, let ad = 0, then a1 0 0. This may be reduced to one of the cases already discussed, since
now fl x) = a1x"-' + ... + a,, and, according to (113.6), D = 0 means that D(a1x"-1 + ... + a") = 0. Consequently Theorem 274 is proved. THEOREM 275. For the discriminant D of a polynomial f(x) of formal degree
n (> 2) we have
f(x) F(x) + f'(x)G(x) = D
(113.7)
with two proper polynomials F(x), G(x) whose degrees are at most n - 2 and n - 1, respectively, and whose coefficients lie in the ring generated by the coefficients off (x). Add the first column of the determinant (113.3'), multiplied by
aox2"-2,
and, for i = 2,..., 2n - 2, the 1th column, multiplied by one. Since this then consists of the elements
'x"-1f'(x), x"-2f (x),
x"-2./
'(x),
x"-3./
x2"-1-I, to the last
(x), ...,./ '(x),
the theorem follows from (113.3'). THEOREM 276. The discriminant of the polynomialf(x) in (113.1) is a homogeneous isobaric polynomial of the ao, . . ., a,, of degree 2n - 2 and of weight
n(n - 1). This follows at once from (113.3) and Theorem 271. THEOREM 277. The discriminant D of the polynomial
f(x)=ao(x-x1)...(x-x")
(n>_ 1,ao#0)
(113.8)
453
THE DISCRIMINANT OF A POLYNOMIAL
has the following two product representations D = ao"-2 D=(_
fl
(Xk
15 n are equal to 0. EXAMPLE 2. We call
1 xl ... xi-
V.=I -' I =
(115.8)
11X....4' the Vandermonde determinant of the elements x1, . . ., x,,, which may belong to an arbitrary commutative ring with unity element. We prove that
V. = [I
(Xk - x,). (115.9) lei0; r+s=n), then bo, co * 0 (mod a) follows from (118.1), (118.2) and (118.5). Then from (118.6),
g(x) = box' (mod a), h(x) - coxs (mod a), thus in particular br , cs m 0 (mod a), consequently
brc, - 0 (mod a2)
.
But, because of (118.1), (118.5) and (118.7), a" = b,cs, so that we obtain a contradiction with (118.4). This proves the theorem. Hence, and from GAUSS'S theorem (Theorem 207), we obtain as a special case the following THEOREM 283 (EISENSTEIN'S theorem). If ` is an integral domain with
prime decomposition and the conditions
p,f' ao; p I aI,
, an-1; p
(118.8)
II an
are satisfied, for the coefficients of the polynomial f(x) in (118.1) and for a prime element p of JP, then f(x) is irreducible over the quotient field of EXAMPLE 1. The so-called ptu cyclotomic polynomial (cf. § 135)
FP(x) = z _ is irreducible over
1
xP_= +
... + I
(p prime)
"o, since for
F'P(x + 1) =
(x + Ix )P _ 1
conditions (118.8) are satisfied.
16 R.-A
= xP-' +
=
xP_1 + ( p ) x- +
... + (p
p
1
464
COMMUTATIVE POLYNOMIAL RINGS
EXAMPLE 2. If a is an integer containing a simple prime factor, then because of Theorem 283 x" + a is irreducible over . a. On the other hand:
x4+4=(x2+2x+2)(x2-2x+ 2). (Cf. Theorem 428.)
§ 119. Hilbert's Basis Theorem THEOREM 284 (H1LBERT's basis theorem). If every ideal of a ring J" with unity element is finitely generated, then this is also true for the polynomial ring
J'[x]. (The term "basis theorem" dates back to the time when the generating systems of an ideal were called its bases.) For the proof, let us consider an ideal SIX of R[x]. Let 1 (n z 0) denote the set of those elements of K whose degree does not exceed n. Furthermore let a and a denote the set of leading coefficients of the polynomials belonging to W and ul", respectively.
We prove that both a and a are ideals of V. With this end in view we denote two elements of a by a, b and an element of R by c. There are two elements of 21 of the form
f(x)=ax'+..., g(x)=bx' + ...
(119.1)
We can suppose that r < s. Then
xs-'f(x) - g(x) = (a - b)e + ...; also belongs to 2l, whence it follows that a - b belongs to a, thus a constitutes a module. Since moreover
cf(x) = cax' + ...,
f(x)c = acx' + .. .
also belong to 21, it follows that ca, ac E a. Thus a is in fact an ideal. The proof of the same assertion regarding a" is parallel, with the only difference that only the cases r, s < n are admitted in (119.1).
We assign an ideal 3 of fi[x] to the ideal a in the following manner. By hypothesis
a = (a,, ..., a")
(119.2)
for a finite number of a,, ..., a, (E 9'). Moreover 21 contains elements of the form
f1(x) = a, x"' + ...; ...;f"(x) = ax"" + ..., for which we put
58 = (fi(x), ..., f"(x))
HILBERT'S BASIS THEOREM
465
Further, we write
n = max (n1, ..., and denote certain ideals of R [x] assigned in a corresponding manner to the ideals ao, . . ., ai_1 by F80,..., respectively. All the 18, It (i = 0, .. .
..., n - 1) are generated by a finite number of elements of W. We shall prove that K = (^', 8n-1, ...1580), whence the theorem will follow. It is sufficient to show that every element of the left-hand side also belongs to the right-hand side. For this purpose we consider an arbitrary element
F(x) = ax" + .. . of K and show that there is an element F.-I(x) of ^n-1 with F(x) = F.-I(x) (mod 18) . (Here we understand that S2C_1 = 0.) It will suffice to consider the case N >-- n. Since a E a, then by (119.2) a is the sum of a finite number of terms of the form kat, cat, at c', dat d' (k E 7; c, c', d, d' E (119.3) where, of course, k, c, c', d, d' change from term to term. We see that the polynomials
kx'-1f(x), cx"-nf(x), xN-nf(x)c'
,
dx-°`fi(x)d'
have as leading coefficients just the elements (119.3) and are all of degree N. So if they are all subtracted from F(x), then we obtain a polynomial of
degree at most (N - 1). After a finite number of repetitions we obtain the required polynomial Fn-1(x). In the case n z 1 one proves similarly that Fn-1(x) = Fn-2(x) (mod Zn-1) ,
where Fn_2(x) is a suitable element of `1n_2. We then have
F(x) = Fn-2(x) (mod (8, After sufficient repetitions we finally obtain
F(x) = 0 (mod (Z,n_1, ..., 58o)) The theorem is now proved.
466
COMMUTATIVE POLYNOMIAL RINGS
§ 120.* Szekeres's Theorem To obtain all the different ideals of a ring is an important problem which has only been solved for a few cases. The solution for a polynomial ring %R[x] over a (commutative) principal idealring -Rwith prime decomposition will be given here.
In general an ideal (# 0) of a commutative ring R is called primitive, if it may not be written as the product of a principal ideal different from 1 and a further ideal of R. It is obvious that in the polynomial ring R[x] every ideal different from 0 may be uniquely decomposed into the product of a principal ideal and a primitive ideal. Therefore it is sufficient to give the primitive ideals of fi[x]. With this end in view we introduce a norming of the elements in the fundamental ring R by singling out of each class of associated elements a representative as normed element (§ 79). Since 'R is a ring with prime decomposition, we arrange the norming so that the normed elements constitute a free commutative semigroup with the same unity element as 9P. It is also necessary to give to every normed element a (# 0) of a system of representatives RR(a) of the residue classes mod a, so that all the 91(a) contain the element 0 [which implies that every class 0 (mod a) is represented by 0]. In particular then l1(a), for the unity element a = 1, consists only of the element 0. THEOREM 285 (SZEKERES'S theorem). Let a system of representatives 1(a), containing the element 0, of'the residue classes mod a be given to every normed element a (# 0) in a principal ideal ring li with prime decomposition. Then if we take an arbitrary, but finite, number of the normed elements other than 0 ai, ..., a m (E JI) (m ? 0; am # 1) (120.1) and, to every ak, further k elements
rik , ..., rkk (E R (ak)) then from
go(x) =
al
(k = 1, ..., m),
(120.2)
k
... am , akgk (x) = xgk-I(x) +
i=I
rikgi-I(x)
(k = 1, ..., m)
(120.3)
(after cancelling by ak) we obtain further polynomials go(x), ..., gm(x) in _V[x] and if a = (go(x) , ..., 9m(x))
then a gives all the primitive ideals of 92 [x], each exactly once.
(120.4)
SZEKERES'S THEOREM
467
Let us call (120.4) the Szekeres normal form for the primitive ideals of ,R[x]. Before beginning the proof, it should be noted that, by our theorem, the elements (120.1) and (120.2) constitute a complete system of invariants of the primitive ideals of.$[x]. The elements (120.1) depend only on t-*, a and on the norming in * [the ideals (a), . . ., (am) thus depend only on and a] but the elements (120.2) also depend on the choice of the systems of representatives 91(a) (a E A 0 0). We begin the proof by verifying that go(x), ..., gm(x) are in fact polynomials in R[x]. We show from (120.3) that for the polynomials defined by
f0(x) = 1, fk(x) = xfk-1 (x) +
/
i=1
rikai ... ak_lf-1(x)
(k= 1, .., m)
(120.5)
/(k = 0, ..., m).
9k(X) = at +1 ... am fk(x)
(120.6)
This is obvious because the equations (120.5) are obtained from the equations
(120.3) by substituting (120.6) and cancelling a1... am or ak ... am. Then we show that by (120.4) only primitive ideals of R[x] are given. By (120.5), fk(x) is evidently a principal polynomial of degree k. By (120.6), gk(x) is a polynomial of degree k. Since gm(x) = fm(x) is a principal polynomial of degree m and, according to (120.31), go(x) is a constant other
than 0, it follows that for m > 0 we obtain by (120.4) primitive ideals of JP[x]. The same holds for m = 0, since then the polynomial sequence is reduced to go(x) = 1, so that (120.4) now yields the unity go(x), ..., ideal.
From now on, let a denote an arbitrary primitive ideal of _ [x]. It still remains to be proved that a is given exactly once by (120.1), (120.2), (120.3), and (120.4). As a preliminary, we define certain invariants of a. Let Mk be the 'R-module
consisting of those elements of a whose degree is at most k. Evidently Mo c M1 c
..
(120.7)
The leading coefficients of the elements of Mk constitute an ideal of,2, which, since 2 is a principal ideal ring, we take in the form (ck). Moreover it may be supposed that the co, c1, ... are normed elements of so that these become invariant. Evidently
(co) c (c) c ...,
(120.8)
i.e., Ck I Ck-1
(k = 1, 2, . . .).
(120.9)
468
COMMUTATIVE POLYNOMIAL RINGS
We obtain a further invariant in the following manner: since each ideal of may be generated by one element, so, according to HILBERT'S basis theorem (Theorem 284) a is finitely generated. Consequently there is a minimal k, for which a is generated by the elements of Mk. We denote this k by m(a)
(120.10)
.
Now, we temporarily disregard postulate (120.2) [thus we provisionally admit in (120.3) arbitrary elements rik of 92] and show that for the ideal a the conditions (120.1), (120.3) and (120.4) can be satisfied. For this purpose we prove that
M0 00,
(120.11)
i.e., that a contains constant elements other than 0. If this is not correct, then we suppose that
h(x)=ax'+...
in a is a non-constant element of minimal degree. Every element f(x) of a other than 0 is then of degree n (>_ 1). For this a"_'+l f(x) = q(Y) h(x) + r(x)
with two polynomials q(x), r(x) in R [x], of which the latter is of smaller degree than h(x). Since evidently r(x) E a, it follows that r(x) = 0. Hence h(x) I
an-,+ If(x) which, by GAUSS'S theorem (Theorem 207), gives ho(x) I f(x) ,
where ho(x) is a primitive polynomial of degree 1. We have now obtained
the proposition that the elements of a have the common divisor h0(x), while, however, a is a primitive ideal. This contradiction proves (120.11). Now (120.11) implies that co 0 0, whence by (120.9)
co,c1,... 00.
(120.12)
Since, furthermore, Mk contains polynomials with the leading coefficients ck, it also contains those of degree k. We introduce for (120.10) the notation m = m(a)
(120.13)
and choose, in one way or another, from among each of the M0, a polynomial
gk(x) = ckxk + ...
(k = 0, ..., m).
..., M. (120.14)
Since then, for every element f(x) of Mk (k > 0), there is an a (E R), for
SZEKERES"S THEOREM
469
which f(.x) - agk(x) lies in Mk_ 1, it follows by induction that the go(x), . . ., gk(x) constitute an 2-basis of the module Mk:
(k = 0, ..., m) .
Mk = { 90(x), ..., A(x)}
(120.15)
The validity of (120.4) follows from the special case k = m because of the definition of m = m(a). It still remains to be shown that go(x), ..., gm(x) can be defined by (120.1) to (120.3).
To do this, we define the elements a1, . . ., am (E,59) on the basis of (120.9) by akck = Ck-1
Then, by (120.12), a1,
. .
(k = 1, . . ., m).
(120.16)
., am are not 0 and are normed so that
Ck = ak+1 ... amCm
(k= l...., m)
(120.17)
follows from (120.16). By (120.14) and (120.16)
akgk(x) - xgk-1(x) E Mk-1
(k = 1, ..., m) .
(120.18)
Hence, and from (120.15), the validity of (120.32) for the elements r,k of follows. From (120.14) and (120.17)
9o(x) = a1 ... a,,, c,,, .
(120.19)
By a similar proof to that of (120.5) and (120.6), we obtain the divisibilities ak+1
amC, 1 9k(x)
(k = 0, ..., m)
from (120.19) and (120.32). Since then we have cm I go(x), ..., 9m(x), it follows that cm = 1 from (120.4). By this (120.19) becomes (120.31). Taking all these statements together it has now been proved that all the equations (120.3) are valid. It still remains to be shown that the condition am 1 in (120.1) is also
satisfied. (This assertion concerns only the case m > 0, since for m = 0 the system (120.1) is empty.) If am = 1 (thus m > 0), then it would follow from (120.18) for k = m that, because of (120.15), gm(x) is contained in the ideal (9o(x), ..., 9m-1(x))
But then because of (120.4) this ideal would be equal to a. Because of (120.13) this contradicts the definition of m(a). Consequently the assertion is proved. Conversely, we consider in the primitive ideal a of ,A[x] all the systems go(x), ..., gm(x) (for any m), for which the conditions (120.1), (120.3)
COMMUTATIVE POLYNOMIAL RINGS
470
and (120.4) [consequently also (120.5) and (120.6)] are satisfied. We have
already seen that there is at least one such system. It still remains to be proved that among these systems there is exactly one, for which conditions (120.4), too, are satisfied. For this purpose we show that every element of a is uniquely representable
in the form .f(x)9m(x) + bm-19m-1(x) + ... + bogo(x)
(120.20)
(f(x) E fi[x]; bo, ..., b,,,_1 E 39) If we write (120.32) in the form k+1
xgk(x) = ak+19k+1(x) - , ri,k+19,-1(x) i=1
(k = 0, ..., m - 1)
,
we can transform the elements xigk(x) of a into an expression of the form (120.20) by repeated application. But since, according to (120.4), every element of a is a sum of terms of the form axlgk(x), the same follows for all the elements of a. The uniqueness of this representation (120.20) is evident, since the element 0 is only representable by f(x) = b,"_1 =
=b0=0.
Hence the validity of (120.15) necessarily follows. We show also that (120.13) necessarily holds (so that the number of the
9k(X) is invariant). By (120.4) and (120.15) m(a) 5 m. But, according to (120.6), we have am 19o(x), . , 9n,-1(x) and, by (120.1), am 1, so the ideal (go(x), ..., g,,,-1(x)) is not primitive. Therefore, because of (120.15), it follows that m(a) > m - 1. This proves the validity of (120.13). From (120.5), (120.6) and (120.15) it follows that (ck) = (ak+,... am) and so, since all the ck, ak are normed, ck = ak+1 ... am
(k = 0, ..., m) .
Since the ck are invariants of a, the uniqueness of the system (120.1) follows. It still remains to be shown that the additional postulates (120.2) can also be satisfied, thus uniquely defining the go(x), . . ., g,,,(x) (and system (120.2)). This holds for g0(x) because of (120.31). We suppose that for some k (= 1, . . ., m) we have already chosen the go (x), . . ., gk_1(x) so that the coefficients ril (i = 1, . . ., 1; 1 = 1, ..., k - 1) in (120.32) satisfy conditions (120.2). Because of (120.15) the
9k(x) = 9k(x) + dk-19k-1(x) + ... + dogo(x) (d0, ..., dk-1 E are all the polynomials which may replace gk(x). (It should have been taken into consideration here that, by (120.5) and (120.6), gk(x) has the leading
SZEKERES'S THEOREM
471
coefficient ak+l ... am and this is an invariant of a.) From (120.32) we get k
akgk(x) = xgk-1(x) + Y_ (rik + akdi-)gi-l(x) i=1
This shows that we can determine the do, ..., dk_l in exactly one way so that for the new coefficients
(i = 1, ..., k)
r k = rik + akdi-l
conditions (120.2) are satisfied. Hence by induction the proof of the theorem is complete. As regards the case /k=J of Theorem 285 cf. SZEKERES (1952). EXAMPLE 1. The above theorem may, in particular, be applied to the case rk = 9.
For the normed elements we then take the non-negative integers and as a residue system W(a) (a > 0), the numbers 0, . . ., a - 1. In this way we obtain all the different ideals of 9'[x]. EXAMPLE 2. In the polynomial ring
[x, y] over an arbitrary field J%', all the different ideals may be obtained by means of Theorem 285, since .59x] is a principal ideal ring with prime decomposition.
§ 121. Kronecker-Hensel Theorem
Szekeres's theorem was (for the case
2=,7) formerly obtained by
KRONECKER - HENSEL (1901) in the following less simple but somewhat more explicit form: THEOREM 286 (KRONECKER-HENSEL theorem). Let JI, 91(a) mean the same as in Theorem 285. If we take natural numbers
n1 0 (i.e., p is a prime number). Evidently
0,e,...,(p - 1)e
(124.2)
constitute a subring of F,,. We now show that this subring is a field, and so is F,, itself. The assertion already follows because the ring is finite and zero-
divisor-free. A direct proof is made possible by the fact that, because o+ (e) = p, the equation
(p, a; a,bEJ) is equivalent to the congruence ax = b (mod p) which has a solution. Now it is also evident that the isomorphism F, ,:; .gyp (as -* a (mod p)) holds. This proves : THEOREM 289. To every possible characteristic there is, apart from isomorphism, exactly one prime field.
In conformity with this theorem we may henceforth briefly speak of the prime field of characteristic p. If F is an arbitrary skew field, then its unity element, and so also the prime field generated by it, is contained in all the sub-skew-fields of F. Therefore we may speak of the prime field of a skew field (field), which is the only
prime field contained in it and is evidently also definable as the intersection of all its sub-skew-fields.
Thus, after imbedding, every skew field will become an extension of some _57 ,, (p > 0), but we do not always carry out this embedding. In accord-
ance with this, Fp will often designate an arbitrary prime field of characteristic p. THEOREM 290. A prime field has no automorphism except the identical one.
Since in every automorphism of a field its unity element is mapped into itself and a prime field is generated by the unity element, the theorem is true.
§ 125. Relative Fields
We have already defined the notion of the relative field F I
sue' (§ 62,
Example 13). This means that F is a field and .f a subfield which acts as an operator domain of F, where the operation is defined as the multiplication valid in F. We also say that F is a relative field over.? (or with respect to .7). We often omit the attributive "relative" and speak briefly of the field F I.5r. Jr itself is called the fundamental field of F 1 .7.
RELATIVE FIELDS
479
For a field F I Y we usually denote the elements of F and Y by small Greek and Roman letters, respectively, as is also usual in connection with operator structures, but we denote the common unity element of these fields by 1. Since the operator product as is a special case of the product in F, we have the rules:
1a=a,
(a+b)a=as+ba, aba=a(ba), (125.1)
aa# = (am) P = a(a9).
(125.2)
Of these (125.1) means that 7+ is a unitary .i-module, i.e., an -7-vector space. Further (125.1) and (125.2) together mean that F is hypercomplex over Y. Thus in a relative field the usual kinds of operations are involved. Having clarified this we wish to establish the further consequences arising from the field F being treated as a relative field F Y. Namely we have to examine the meaning of (admissible) subfields and Sr-isomorphisms of F 1'.7. (As we are considering fields we need to consider only isomorphisms from among the homomorphisms; however, everything regarding these also refers to automorphisms and meromorphisms.) A subfield G of F I.5X means a subfield of F for which
aaEG
(125.3)
In particular, for a = 1, (125.3) means that all the a belong to G, i.e., that .9"is a subfield of G. Conversely, if this is true, then (125.3) is satisfied. Accordingly the subfields of F 1.9r are those subfields of F which are in turn
overfields of .7:
r9G9; F.
(125.4)
The subfields G with this property are sometimes called between-fields (namely between .9 and F). Of course we understand every subfield G of F 1.7 to be a relative field G I r. An.7-isomorphism of two fields F 1.5', F 1.7 (according to our general convention for operator structures) will be denoted as F I .-7
F I Jr (a -+ a') .
This means that the usual isomorphism F additional condition (aa)' = aoc'
(125.5)
F' (a -p. a') is subject to the
(a E .7, a E F)
.
(125.6)
480
THEORY OF FIELDS
But since (ax)' = a'a', (125.6) means that
a' = a
(a E 7) .
(125.7)
According to this (125.5) implies an isomorphism F Zt: F' (a - a') in' which the elements of the fundamental field . " are fixed elements. We also call the ,7-isomorphism (125.5) a relative isomorphism of the fields F, F' (over.}') or an isomorphism of F 1,91- with F I X. In conformity with this a relative automorphism of the field F over its subfield Y - briefly an automorphism of F 1-9- - means an automorphism of F in which the elements of 7 are fixed. A relative meromorphism of F over . " or a meromorphism of F 1.7 is correspondingly explained as an isomorphism of F I Jr with an (admissible) subfield.
Moreover, in a field F 1.7 the fundamental field Jr will play a further important role inasmuch as, in general, we consider the notions connected with
F in relation to J1. (Hence the term "relative field".) This principle is called the principle of relative fields. We shall pay special attention here to some of its important applications. Two extension fields F, F of _7 are called equivalent (over _91') when there exists a relative isomorphism (125.5). An element at of a field F 1.7 is called algebraic or transcendental, respectively, according as to whether there exists or not a non-constant polynomial f(x) over -9' with the zero at (cf. § 74). We also say that at is an algebraic or
transcendental element, respectively, of F over (or with respect to) Sr. Further, we say that F 1.7is an algebraic field (or F is algebraic over,7), when all the elements of F are algebraic over.'. Otherwise F 1 Y is called a transcendental field (or F itself is called transcendental over."); this means t hat F has at least one transcendental element over .7-. If at is an algebraic element of F 1 ,7, there is a principal polynomial f(x) over.rte" of minimal degree with the zero at. This minimal degree is called the degree of the element at (over .7) (cf. § 74). We now show that this polynomial f(x) is uniquely defined, therefore we call it the minimal polynomial of the element at (over .F). If f(x), g(x) were two distinct principal polynomials, so that they were both of the same degree, then f ( x ) - g(x) ( : A 0) would be a polynomial of smaller degree with the zero at. This is impossible, thus the assertion is true. The following theorem is also valid. THEOREM 291. The minimal polynomial f(x) of an algebraic element at of a field F 1 .7 is irreducible over .7 and is also definable as the only irreducible principal polynomial in F [x] with the zero at. Furthermore at is the zero of a polynomial F(x) over." if, and only if, f(x) I F(x). Of the three assertions of this theorem the second is a consequence of the other two; consequently it is sufficient to prove these.
RELATIVE FIELDS
481
We consider a decomposition f(x) = g(x) h(x) such that g(x), h(x) E 9' [x]. Then 0 = floc) = g(oc)h(a) .
Accordingly either g(a) = 0 or h(a) = 0. Because of the minimal property of f(x) it follows that one of the factors g(x), h(x) is of the same degree as f(x). Consequently f(x) is in fact irreducible. Furthermore we carry out on F(x), f(x) the Euclidean division:
F(x) = q(x)j(x) + r(x)
.
Then F(a) = r(oc). But as r(x) is of smaller degree than f(x), the equation
r(a) = 0 implies r(x) = 0, i.e., f(x) I F(x). This completes the proof of the theorem. In order to distinguish the relative fields we call a field without an operator domain an absolute field. If Jr is an arbitrary field, then its prime field is
- as noted above - contained in all the subfields of F. Therefore, and because of Theorem 290, "absolute field" and "relative field over a prime field" are identical concepts. Correspondingly, an element a of a field F is called algebraic or transcendental, according as a is algebraic or transcendental over the prime field of F. If necessary a is called an absolute algebraic or absolute transcendental element. Similarly, the degree, or more precisely the absolute degree, of an algebraic element from a field .7' means the degree
of this element over the prime field of F. Likewise an absolute algebraic field (briefly, an algebraic field) or an absolute transcendental field (briefly, a transcendental field) means an algebraic or transcendental field over its prime field. These examples may be sufficient to explain the meaning of
the adjective "absolute" in other similar cases, some of which will occur later.
Finally, it should be noted that in the course of our considerations, if a fundamental field 9r is already established, then every extension field F of .7 is usually automatically regarded as a relative field over `" without calling attention to it by the use of the notation F I -'/- or in any other way. the indeterminates x1, ...,x EXAMPLE 1. In the rational function field F(xl, ... , are transcendental over F. EXAMPLE 2. The complex field over .moo is algebraic, and its elements are all of degree I or 2.
§ 126. Field Extensions If G is an overfield of the field F and S an arbitrary subset of G, then we say (in the general terminology of § 35) that the subfield { F, S) of G, generated by the union set F U S, arises from F by the adjunction of S (or of the
THEORY OF FIELDS
482
elements of S) to F and use the notation F(S) for this field {F, S}. The elements of S are then called the adjoined elements. When S = , we write F(al, a2, ...), instead of F (S). We also write F(S1, S2, ...) = F(S1 U S2 U .. .) ,
where S1, S2 ,... are arbitrary subsets of G. Since every element of F(S) may be written as an expression with a finite number of elements of F and S, it is obvious that F(S) is the union of all the F(S1), F(S2), ..., where Si, S2, ... denote all the finite subsets of S. For this reason many questions may be reduced to the case where only a finite number of elements are adjoined at one time. The rule F(S, T) = (F(S))(T) is trivial; for this we write, with simpler notation on the right-hand side, F(S, T) = F(S) (T) (S, T S G). By repeated application we obtain F(aj,
. .
., an) = F(a) ... (an) ,
where al, . . ., a,, denote the adjoined elements. Accordingly the adjunction of a finite number of elements may be reduced to the successive adjunction
of each of these elements. § 127. Simple Field Extensions The extension fields F(i) arising from a field F by the adjunction of only one element t9 are called simple extension fields of F. We also say that these arise from F by a simple adjunction. The element 0 itself is called a primitive
element of F(t). The study of simple field extensions is the key to the whole theory of fields. We now give some important definitions as an introduction to their study. We say that a skew field G is of finite degree over its sub-skew-field F, if the
F-module G+ is of finite rank. The rank of this module is then called the degree of the skew field G over F and is denoted by
[G : F].
(127.1)
Further, every basis of the F-vector space G+ is called a basis of the skew field G over F or an F-basis of G. In the commutative case, we speak of the degree or the basis of G I F.
SIMPLE FIELD EXTENSIONS
483
THEOREM 292. Let F, G, H be three skew fields such that F S G S H. H is of finite degree over F if, and only if, H is of finite degree over G and G s of finite degree over F, and then
[H : F] =[H : G] [G : F] . If furthermore aL, ..., a; is an F-basis of G and A1,
(127.2) . .
., A. a G-basis of H,
then
(i=
a;Ai
(127.3)
is an F-basis of H. If H is of finite degree over F, then, according to Theorem 242, G is likewise of finite degree over F. Further, it is evident that H is of finite degree over G. Accordingly, we have only to prove the last assertion of the theorem, since the other assertions follow from this at once. We consider an arbitrary element C of H. This may be written as m
C = Y_ yjAj
(y; E G).
1=I
Moreover I
Y,
i=1
(c1 E F) .
cijai
Therefore, it follows that /
C=
m Y_ c,ja;Aj .
i= L j=1
This representation is unique, for if the right-hand side vanishes : m
l
ciia; Aj=0
then
i=1
cija;=0
(j= 1,...,m)
and further
ci.i=0
(i= 1,...,1; j= 1,...,m).
This completes the proof of the theorem. We now wish to examine simple field extensions. There are two essentially different cases to be discerned according as the adjoined element is algebraic or transcendental. It will later come to light that in the first case the extension field itself is also algebraic so that in these two cases we are justified in
using the terms simple algebraic extension field or simple transcendental extension field. The transcendental case is much easier, therefore we deal with it first.
THEORY OF FIELDS
484
THEOREM 293 (simple transcendental field extensions). Any field F has
only the rational function field F(x) (apart from equivalent extensions) for simple transcendental extension field and all the elements of F(x) - F are transcendental. In order to prove this we consider an extension F(t) with a transcendental element 1. The elements of F(x) may be taken in the form
r(x) = g(x) h(x)
(g(x), h(x) E F [x]; h(x) # 0) .
(127.4)
Since F(O) is transcendental, so h(i9) # 0, thus (127.5)
exists and is an element of F(t9). It is evident, too, that conversely, all the elements of F(9) are furnished by (127.5). The mapping (127.6)
r(x) -+ r(9)
is then one-to-one since, for two elements taken in the form (127.4),
r;(x) =
(i = 1, 2)
rl(x) = r2(x) if, and only if, gl(x)h2 (x) - g2(x)hl(x) = 0, which, by the transcendence of 0, is equivalent to g,(8) h2(9) - g2($) hl(O) = 0, i.e., to rl(t) = r2(0). Further, the mapping (127.6) is obviously homomorphic, and so - since F(x) is a field - isomorphic. Moreover the elements of F are fixed elements,
from which we obtain the relative isomorphism F(x) F s F(t) I F, i.e., the equivalence of the extensions F(x), F(O). It still remains to be shown that an algebraic element (127.4) necessarily lies in F. We may suppose that (g(x), h(x)) = 1. From these suppositions the
existence of an equation (after eliminating the denominator) of the form k
i=0
ci(g(x))`(h(x))k-` = 0
(c1 E F; co, ck 0 0),
follows. Consequently g(x) I h(x) and h(x) g(x), i.e., r(x) E F. Consequently Theorem 293 is now proved.
As a further preliminary to the next case we prove the following easy theorem.
SIMPLE FIELD EXTENSIONS
T.
485
oxns 294. If S is a subring and a an ideal of a ring R with
S fl a= 0,
(127.7)
then in the factor ring R/a the residue classes a (mod a) represented by the
elements a of S constitute a subring S' isomorphic with S : S
S' (a -s a (mod a)),
(127.8)
so that after the corresponding embedding R/a becomes an extension ring of S.
For, it follows from (127.7) that the mapping (127.8) is one-to-one and, furthermore, it is homomorphic, too. This theorem will later prove to be of importance in dealing with the factor ring F[x]/(f(x)), (127.9) where F denotes a field and f(x) a non-constant polynomial over it. Since the condition (127.7) has been satisfied by S = F, a = (f(x)), after embedding (127.9) becomes an extension ring of F, which in future we shall always assume.
THEo1 M 295 (simple algebraic field extensions). Every field F has for every principal irreducible polynomial f(x) of degree n (apart from equivalent extensions) only one extension field F($) which has the property that 0 is algebraic and has f(x) for its minimal polynomial. Moreover F(O) is of finite degree: [F(b) : F] = n, (127.10)
i.e., of the same degree as 1. Furthermore 1, 19, ..., ' i.e., its elements are uniquely given by
ON
is a basis of F (r?)'
(127.11)
where g(x) denotes the polynomials over F of formal degree n - 1. All the elements of F(O) are algebraic and their degrees are divisors of n. In particular,
the elements of degree n are the primitive elements of F(t9). F(t) can be effectively given by F [xJ/(f(x))
(127.12)
For the proof we first suppose the existence of a field F(O) with f(i) = 0, and consider the homomorphism F[x] - F[t]
(h(x) -> h(t9)),
(127.13)
THEORY OF FIELDS
486
where F[z] denotes the ring consisting of the substitution values h('). Thus, in any case, F[t9] S F(t9).
(127.14)
The kernel of the homomorphism (127.13) is constituted by those h(x) for which 0. This condition, according to Theorem 291, is equivalent to f(x) I h(x), therefore the above-mentioned kernel is the ideal (f(x)). Thus from (127.13), we obtain the isomorphism F[x]/(f(x)) & F[i]
(h(x) (mod f(x)) --o. h(ad)).
(127.15)
But since f(x) is irreducible, the ideal (f(x)) is maximal, and so, according to Theorem 130, the left-hand side of (127.15) is a field. This is also true
for the right-hand side. These must contain, together with 0, the field F(a9), i.e. F[a9] ? F(0). Hence and from (127.14) it follows that F(O) = F[0],
(127.16)
and, because of (127.15), that F [x]/(f(x)) s F(a9)
(h(x) (mod f (x)) ->- h(9)).
(127.17)
If, in particular, h(x) = c is a constant, then the class c (mod J (x)) is mapped
onto the representative c. More precisely, this means (because of the embedding) that the elements of F are fixed, i.e., that (127.17) is an F-iso-
morphism. In other words F(8) is an extension field of F equivalent to (127.12).
Furthermore (127.16) means that the elements of F(19) may be given in the form h(3) (h(x)E F[x]). If Euclidean division is carried out on h(x),
f(x), then, because J (O) = 0, it follows that for this purpose one only needs polynomials h(x) of formal degree n - 1. The corresponding h(0) are already different from one another, otherwise 19 would be the zero of a polynomial over F of smaller degree than n. This proves the assertion
in (127.11) and that concerning the basis property of 1, a9, ..., on-1. Hence (127.10) also follows.
We consider another arbitrary element a of F(O). Then F(a) is a subfield of F(0) with F S F(a) S F(O).
According to this, F(a) I F is of finite degree, so a is in any case algebraic. From (127.10) and Theorem 292 we get n = [F(9) : F] = [F(t) : F(a)] [F(a) : F].
(127.18)
SIMPLE FIELD EXTENSIONS
487
Therefore if (127.10) is applied with a instead of 0, this shows that the degree of a is a divisor of n. Since a is a primitive element of F(t4)) if, and only if, F(a) = F(O), i.e., [F(s) : F(oc)] = 1,
it follows from (127.18) that for this it is necessary and sufficient that
[F(a) : F] = n, i.e., that a is of degree n. The existence of F(t9) still remains to be shown. This is easy to prove. We already know that (127.12) is a field, consisting of the residue classes g(x) (mod./(x)) with g(x)E F [x], for which we use the notation -gT;i). Then
f(x) = 0. On the other hand (because of the embedding) f(z) = f(x), so
f(x) = 0.
(127.19)
But since (127.12) (after embedding) is equal to F(.), the required proof of existence is implicit already in (127.19). This completes the proof of Theorem 295.
Because of the existence and uniqueness proposition contained in this theorem the simple algebraic extension fields of F, not equivalent to one another, and the principal polynomials f(x), irreducible over F, are assigned to each other in a one-to-one manner, where to each such polynomial corresponds a field F(t9) in which the primitive element t9 is a root of the equation
f(x) = 0.
(127.20)
Therefore we call (127.20) the defining equation of this field.
In general one should call an equation (127.20) over a field with irreducible or reducible left-hand side an irreducible or reducible equation, respectively. Thus, by Theorem 295, an extension field in which this equation
has at least one solution can be given to every irreducible equation over an arbitrary field. EXAMPLE 1. It is an important part of Theorem 295 that the elements of an algebraic extension field F(i) may be written "denominator free", namely in the form (127.11).
We wish to prove this directly and at the same time study how one can "free" the elements of F($) of their denominators. We denote the minimal polynomial of 61 by f(x) and consider an arbitrary element of F(g): a = g(o) h(ad)
Here h(i})
(g(x), h(x) E F[x]).
0, thus fix) ,f' h(x). Since f(x) is irreducible, we can solve
f(x)fi(x) + h(x) hi(x) = I (fi(x), h,(x) E F[x]). Because f($) = 0 it follows that h(t9)h,(1) = I. thus a = g(a?)hl(a9).
THEORY OF FIELDS
488
EXAMPLE 2. By virtue of Theorem 295 an algebraic field F(9) I F of degree n is an
F-algebra of rank n with the basis 1, 0, ..., 0'1. By the help of the coefficients of the minimal polynomial =x"+alr"_1+...+a" (a,E F) f(x) of 0 the structure constants are computable. For, since 0119 = $1+L
(i = 0, ..., n - 2); 0-10 = -(a" + ... + a119"^1)
the product of two arbitrary basis elements 01, Of is recursively determined. EXAMPLE 3. Skew fields generated by one element are the simple field extensions F,(8) of the prime fields F,, of arbitrary characteristic p (> 0). EXAMPLE 4. The algebraic field F(O) over F is a special case of rings defined by equations. F(t9) is obviously that ring whose generators are the elements of F and the element 0, and whose defining equations are the equations valid between the elements of F, the equations c$ =arc (c E F), and finally f(ad) = 0 ,
(127.21)
where f(x) is the minimal polynomial of 0. It would, therefore, have been more accurate to call (127.21) the defining equation of F(a9) instead of (127.20). Reducible equations are, of course, not suitable for uniquely defining a field.
§ 128. Extension Fields of Finite Degree THEOREM 296. All the fields G I F of finite degree are algebraic and may be obtained from the fundamental field F by adjunction of a finite number of elements; furthermore the degrees of the elements of G are divisors of [G : F]. Conversely, every field over F, which arises from F by adjunction of a finite number of algebraic elements, is of finite degree (and so algebraic). COROLLARY. If a, j9 (j 0) are algebraic elements of an arbitrary field G I F, then (128.1)
are also algebraic elements over F.
In order to prove the first part of the theorem we write
n = [G : F].
(128.2)
Clearly, [F(a) : F] I n follows from Theorem 292 for every cc (E G).
Accordingly a must be algebraic; since, further, according to Theorem 295,
its degree is then equal to [F(a) : F], this is in fact a divisor of n. From (128.2) it also follows that G has an F-basis cot, ..., co,,. Hence G = = F(col, . . ., co). So the first part of the theorem is proved. For the proof of the second part we consider a field
G = F(0,, ..., 0k),
EXTENSION FIELDS OF FINITE DEGREE
489
where 01, ..., O'k are algebraic. These are then also algebraic over the fields
(i = 0, . . ., k; Go = F, Gk = G). = F(01, ..., ,O,) Since furthermore G,+1= G;(l';+I), according to Theorem 295, every Gi
term of the chain of fields
F9Gl9...9Gk_ISG (from the second term on) is of finite degree over the preceding term. Hence it follows, according to Theorem 292, that G I F is of finite degree. Consequently the theorem is now proved. The corollary is similarly true, since the elements (128.1) belong to the field F(«, fi) which is of finite degree over F. THEOREM 297. If the fields H I G, G I F are algebraic then H I F is also algebraic.
We have to prove that every element 0 of H is algebraic over F. On the basis of the supposition an equation such as
,on + al9"-I +
... + a = 0
(at E G)
(128.3)
holds. Since the a.; are algebraic over F, by Theorem 296, the field
is of finite degree over F. Further, because of (128.3), G'(O) is of finite degree over G'. By Theorem 296, G'(0) I F is then algebraic and so 0 is also algebraic over F. This proves Theorem 297. EXAMPLE. If G I F is a field of finite degree, then the rational function field G(x) is of finite degree over F(x), and [G(x) : F(x)] = [G : F].
(128.4)
It is also true that the three algebras G I F,
G[x] I F[x], G(x) I F(x)
have the same rank and every basis of the first is a basis of the remaining two. For the first two of these algebras the assertion is trivial. To prove the full assertion, it is sufficient to show that the elements of G(x) may be written as quotients with numerators from G[x] and denominators from F[xl. For this it is sufficient to prove that every element g(x) (56 0) of G [x] is a divisor of an element (00) of F[x]. Since among the powers (g(x))' (i = 0, 1,. ..) there are only a finite number of linearly independent ones over F(x), an equation of the form r
at(x)t(g (x))t = 0
(ao(x), ..., ar(x) E F[x] , ao(x) 0 0) .
t=o
holds. Hence it follows that g(x) I ao(x), which proves the assertion.
490
THEORY OF FIELDS
§ 129. Splitting Field
Let a field F be given with an arbitrary (finite or infinite number) of non-constant polynomials f1(x),12(x), .. .
(129.1)
over it. An extension field G of F is called a splitting field of these polynomials f(x) or of the equations f;(x) = 0 (over F), when in G [x] these polynomials split into linear factors and no subfield G' (F c G' C G) of G has this property. If, in particular, (129.1) consists of a finite number of polynomials f1(x),..., fk(x), then we may take the single polynomial
1(x) = fi(x) ... A W,
instead of them, since f(x) splits over a field into linear factors if, and only if, this is valid for all the f1(x), . . .,fk(x). Therefore there are only two essentially different cases, where (129.1) consists of only one polynomial or of an infinite number of polynomials. The first case will be studied here,
the second in the next paragraph. (In the trivial case, where the system (129.1) is empty, by the splitting field we have to understand F itself.) As a preliminary we need the following: whenever an isomorphism a -* a between two fields F, F is given, we understand by the image of a polynomial f(x) over F that polynomial f(x) which arises from it by carrying
out the above isomorphism on its coefficients. Of course, the mapping f(x) -+ f(x) is then an isomorphism of F[x] onto F[x]. So two corresponding polynomials f(x)j(x) are simultaneously reducible or irreducible (over F and F, respectively). THEOREM 298. Between two fields F, F let an isomorphism
F ^ F (cc - a)
(129.2)
be given and an irreducible polynomial f(x) over F. Furthermore denote the corresponding (thus likewise irreducible) polynomial over F by Y (A If the equations
f(o) = 0, f(a) = 0
(129.3)
hold for any two elements 9, a from arbitrary extension fields of F and F, respectively, then (129.2) may be continued to an isomorphism of the fields F (a), F(a), in which the elements Lo, a correspond to each other.
Denote by g(x) an arbitrary polynomial from F[x] and by g(x) the corresponding polynomial from F [x]. We establish the isomorphism F(RO)
F(a) (g(e) -* g(a)).
(129.4)
SPLITFING FIELD
491
Since, by (129.3), the fields F(e), F(a) are algebraic, all their elements are given by g(e), g(a), according to Theorem 295. Thus (129.4) maps F(e) onto F(a) in all cases. This mapping is one-to-one, since the following four propositions are equivalent :
f(x) I g(x), A-0 19(x), g(c) = 0. The homomorphism property of (129.4) is evident. If, finally, g(x) = a g(e) = 0,
(a E F) or g(x) = x then g(e) = a or g(e) = e, and therefore the image element g(a) is equal to a or a, respectively. According to this, (129.4) is a continuation of (129.2), mapping a onto a. Consequently Theorem 298 is proved.
We now prove the following: THEOREM 299. For every field F, every non-constant principal polynomial f(x) (E F [x]) has (apart from equivalent extensions) only one splitting field. If we take an arbitrary extension field of F over which then
f(x) _ (x - 191) ... (x -0 ),
(129.5)
G = F(01, . . ., 1n)
(129.6)
.
is a required splitting field, and all the splitting fields of f(x) arise in this way.
(Since, over a splitting field of f(x), because of its definition a factor decomposition of the form (129.5) holds, the first part of the theorem already contains the existence proposition that there are extension fields of F over which an equation (129.5) holds.) First of all we prove that there is an extension field of F over which (129.5) holds. For that purpose we take an arbitrary extension field F (;2 F) of F and consider the prime decomposition (129.7) f(x) = fi(x) ... fl(X) is not linear, then, according to Theorem of f(x) in F'[x]. If, e.g., 295, take an extension field F" = F'(i9) (thus also of F) with f1(1) = 0. It follows that, over F", the polynomial fl(x) has the factor decomposition f1(x) = (x - 1)g1(x) whence, according to (129.7), it follows that the
prime decomposition of f(x) over F" consists of at least r + 1 factors. The assertion follows by induction. This verifies the existence of the field G defined in the theorem. We show that this is a splitting field of f(x). For this purpose we consider a betweenfield G' (F S G' c G) and suppose that f(x) also splits over G' into linear factors:
f(x) = (x - 11) ... (x - OD
(01 E G').
G'. Thus because This factor decomposition then holds over G since G of the uniqueness of the prime decomposition in G [x] it must agree with
THEORY OF FIELDS
492
(129.5). Since, according to this, 19;, ...,19;, differ at most in the order of succession from 191, ...,19,,, so 19,, ...,19 E G'. Hence, and from (129.6),
G 9 G' follows. Because G' S G we then have G' = G. This means that G is in fact a splitting field of f(x). The final assertion of the theorem is trivial. The assertion of uniqueness still remains to be proved. To prove this, we consider, in addition to G, a further splitting field G' of f(x) which is given by
f ( x ) = (x - 01) ... (x - 0. 0,
(129.8)
G' = F(19;, ...,19;,).
(129.9)
(These "new" 0,, ...,19;, are independent of the former ones.) We have to prove the existence of an isomorphism
GIF,:;G'IF(a(129.10) and show at the same time that with a suitable order of 191,
the
additional condition 19,=19;
(i= 1,...,n)
(129.11)
can be satisfied.
We introduce the notations G, = F(191,
..., 9 ),
G, = F(191,
(i=0,...,n; Go=Go=F;
..., ,) G;,=G')
(129.12)
and suppose that for some k (0 a)
holds a fortiori, and, according to (129.12), Gk+t = Gk(ok+1),
Gk +1 = Gk(Ok+1),
so, because of (129.17) and (129.18), the existence of an isomorphism (129.19)
Gk+1 " Gk+1,
which is a continuation of (129.13) and maps tk+1 onto 19k'+, follows from
Theorem 298. This means that the supposition made in (129.13) and (129.14) is satisfied with k + 1 instead of k. Because G,; = G, G , = G' the assertion follows by induction. Consequently, Theorem 299 is proved. EXAMPLE. Let f(x) be an irreducible polynomial over a field F for which
f(x) = (x- 61)... over an extension field of F. The (2 ) fields F(g ,9,) (1 < i < k S n) need not be isomorphic to one another. For example x4 + 2x2 - 2, according to EISENSTEIN'S theorem (Theorem 283), is irreducible over -o and is decomposed over the field of complex numbers into the product
(x+V-I+j
(x-V-1-y3
If the corresponding four zeros are denoted by Y91, ...,194, then F. (61.'82), FO (6u 0a) are evidently not isomorphic.
THEORY OF FIELDS
494
§ 130. Steinitz's First Main Theorem A field F is said to be algebraically closed if every non-constant polynomial f(x) over F has at least one zero in F. This means that every such polynomial
splits into the product of linear factors over F, or that in F[x] only the linear polynomials are irreducible. The prime fields are not algebraically closed, for in the prime field of characteristic 0,
x2 - 2 = 0 has no root, and in the prime field of characteristic p (> 0), x" - x - 1 = 0 has no root, as can be seen immediately.
THEOREM 300 (STEINITZ'S theorem). Every field F has (apart from equivalent extensions) only one algebraically closed algebraic extension field F.
We call this field F the algebraic closure of F. The proof is furnished by the following theorem which is a generalization of Theorem 299 and is obtained from it. THEOREM 301 (STEINITZ'S
first main theorem). For every field
F,
every system of non-constant polynomials over F has (apart from equivalent extensions) only one splitting field and this is algebraic over F. For the proof let us put the given polynomials in the form
f, (x) = xn- a,,
xn_1 +
... + (-1)na,.n (a, E F),
(130.1)
where v runs through the elements of an index set. (Of course n depends on v.) We assign to each f,(x) as many indeterminates xv1, ..., x,n over F
as its degree indicates, and denote the polynomial ring of these indeterminates by
R = Q.
.
., x,.I, .
.
., x , . .
. ].
(130.2)
Let s; denote the ia` elementary symmetric polynomial of xv1, ..., xn (i = 1, . . ., n). Then we form the ideal of R: a = (..., svI - a,,, ..., s,, - am, ...). We prove that a tion such as
(130.3)
R. For, if not, then 1 E a. This means that an equa([30.4)
holds, where we have to understand a finite sum and r denotes an element
of R changing from term to term. Since only a finite number of indeterminates x; occur here, only a finite number of polynomials f,(x) belong to them. By Theorem 299 these polynomials have a splitting field G over F. For these polynomials the factor decompositions of the form
.fv(x) = (x - o1) ... (x - m,.) (z,, E G)
495
STEINITZ'S FIRST MAIN THEOREM
.... a,,,, each time then hold. We establish an arbitrary order for and carry out on (130.4) all the substitutions x,.i = a,.i. Since then [in (130.4)] because of (130.1) every si becomes a,,i, the contradiction I = 0 follows from (130.4). Consequently the assertion has been proved. Hence, and from KRULL'S theorem (Theorem 131), there exists a maximal ideal ao of R with (130.5)
a.
a0
Since moreover R is a ring with unity element, so, by Theorem 130, the factor ring R/ao is a field. Fn ao =0 follows from ao T R. The embedding of F carried out according to Theorem 294 (with F, ao instead of S, a) changes R/a0 into an extension field of F.
By (130.1) and (130.3), for all r.
fv(x) = x" - s1 xn -' + .
.
. + (-
l )"sr"
(mod a),
(x - x,,,) ... (x - x,,,,) (mod a). thus, by (130.5),
fi(x) _ (x - x,,,) ... (x - x,.,,)
(mod ao).
This means that, in the field R/ao, all the f;,(x) are decomposed into linear factors and the residue classes xi (mod ao) are all the zeros of all thef,(x) in R/ao. Since, by (130.2), R/ao arises from F by the adjunction of these residue classes, R/ao is a splitting field of the polynomials (130.1). Since, in addition to this, the adjoined residue classes are algebraic over F, R/a is also algebraic over F, because of the corollary of Theorem 296. Accordingly, only the proposition of uniqueness in Theorem 301 still remains to be proved. Instead of that we prove more generally the following THEOREM 302. If G, G' are two splitting fields of the same polynomials over afield F, every F-isomorphism between two subfields of G I F and G' may be continued to one between G and G'. Since, in particular, the identical mapping of F is an F-isomorphism, the required proof of the proposition of uniqueness in Theorem 301 follows from this theorem. With a view to proving Theorem 302, we call an F-iso morphism between two subfields of G I F and G' I F a partial isomorphism. If, of two different partial isomorphisms g, a the second is a continuation of the first, then we write g < a. Let ,u denote a fixed partial isomorphism and :, the set of those partial isomorphisms e for which a >_ p. It will suffice to prove that C contains an element which maps G onto G'. 17 R.--A.
THEORY OF FIELDS
496
The set G`7 is semiordered. We consider an ordered subset
x 0) we have the very important rule
(x + f,)P = xP + fl"
((x, # E F).
This follows immediately from n
- it
(x +
p aP-L
and from P
pl
(O
p).
f
(132.1)
YIELDS OF PRIME CHARACTERISTIC
499
Furthermore (a - P)P
(XP - #P.
(132.2)
For if p > 2. thus 2 A. p, then we have (- I)P = - 1, therefore (132.2) now follows from (132.1) by applying it to -l3 instead of (l. If p = 2, (132.2) agrees essentially with (132.1), since now in general e = - e (e E F). From (132.2) and the zero-divisor-free nature of F it follows immediately that if a, j3 are different, then cP, NP are also different. This, together with (132.1), means that the mapping z --* a"
(132.3)
is a meromorphism of the module H. If F is finite, then (132.3), of course, must be an automorphism of F+. If, moreover, F is a prime field, then by Theorem 290 (132.3) must be the identical mapping of F+, as is obvious. Conversely, if (132.3) is the identical mapping of F+, then F is a prime field, since in F the equation xP - x = 0 cannot have more than p roots. Since, furthermore, because of the commutativity of F, in addition to (132.1), (cg9)P = aP/9P, i.e., the mapping (132.3) is also homomorphic with respect to multiplication, we have proved the following theorem: THEOREM 305. For every field F of characteristic p (> 0) the mapping
a - x"
(a E F)
is a meromorphism of it. This is an automorphism if F is finite; further it is the identical mapping if, and only if. F is a prime field. The rules (132.1), (132.2) also hold in every commutative ring of prime characteristic p. In particular, in the polynomial ring F[x] over the above field F n
Y- ai x` r=0
_ I aipxP' i -0
If, moreover. F is the prime field of characteristic p, then the important rule
(f(x))P =.f(x)
(f(x) E F[x], 0(F) = p)
follows. EXERCISE. If F is a field and n (> 1) is a natural number for which
a - a"
(aEF)
is an endomorphism of F, then the characteristic of F is necessarily a prime number p and n is a power of p. § 133. Finite Fields
THEOREM 306. Every finite field is of primepower order. Conversely, to every prime power p" (n >-- 1) belongs, apart from isomorphism, only one field F with 0 (F) = p". (133.1)
THEORY OF FIELDS
500
This is (absolute) algebraic of degree n and may be given over its prime field (by a simple adjunction) in the form
F = FP(),
(133.2)
where the primitive element 0 is the zero of an arbitrarily given irreducible polynomial f(x) (E Fp[x]) of degree n. The group F* is cyclic (of order p" - 1), therefore all the different elements of F may be given, by means of a primitive element p of F*, in the form 01110, . . ., op"-:'.
(133.3)
x"' - x = 11 (x - a),
(133.4)
Also aEF
according to which F is the splitting field of the left-hand side of (133.4) and so is normal. The full automorphism group of F is cyclic of order n and consists of the automorphisms
(k = 0, . . ., n - 1).
a --> a°k
(133.5)
Since every finite field must be of prime characteristic, so its module is a finite p-module, thus its order is a power of p. This proves the 'first assertion of the theorem. We want to show that to every prime-power p' (n 1) there is at least one field F which satisfies (133.1). Instead of this we shall now prove that the splitting field F of the polynomial (133.6)
XP" - x (E FP [XD
satisfies the requirement (133.1), where FP is the prime field of characteristic p.
The differential quotient of (133.6) is - 1( 0), so that (133.6) has exactly p" zeros in F. Therefore it suffices to show that these zeros constitute a subfield of F, for then this subfield is the splitting field of (133.6), and so necessarily equal to F. We now consider two zeros a,# (E F; l4 0) of (133.6):
aP"-a=0, pPl-(3=0.
Then we have
-
a#-I
= xj9 ` -
0Cq-1
= 0,
so that a - f, a#-' also belong to the set of zeros. This is, therefores a field. Consequently, the existence of an F with the property (133.1) i, proved.
FINITE FIELDS
501
In the following let F denote an arbitrary field which satisfies (133.1). Because O(F+) = p" the characteristic of F must be p. Denote the prime field of F by FP. Since F+ is an FP vector space, so, because O(FP) = p and by (133.1), this must be of rank n. This means that F is of degree n and algebraic. Since in the finite Abelian group F* (and even in F) every equation xR = 1
(k z 1) has at most k solutions, it follows from Theorem 217 that F* is cyclic.
A primitive element of F* is [according to (133.3)] also a primitive element of F. Accordingly, there exists a 29 which satisfies (133.2). From
[F : FP] = n and Theorem 295 it also follows that the minimal polynomial of t is of degree n; this is irreducible and has .0 as zero. Also the remaining assertion on #9, that to every f(x), in the theorem, there exists a O which satisfies (133.2) and f (z9) = 0, is true on account of Theorem 295, if we prove that all fields of order p" are isomorphic.
This assertion is a consequbnce of (133.4) and the uniqueness of the splitting field (Theorem 299). We prove (133.4) itself as follows. From O(F*) = p" - 1 it follows that al "-I = 1 for all elements a of F*. Thus (XP" - a = 0 for all elements a of F. Since, accordingly, every element of F is a zero of (133.6), (133.4) follows from this and from (133.1). The last assertion of the theorem still remains to be proved. An automorphism of F is determined by the image of its primitive element 0. But t
can only be transformed into a zero of its minimal polynomial, so that there can be at most n automorphisms of F. On the other hand, a -* aP, by Theorem 305, is an automorphism of F. Since (133.5) is the kth power of this automorphism, we have only to show that the n automorphisms given by (133.5) are different. Otherwise there are two integers i, k (0 < i
I)
may be given, according to (133.7), in terms of a primitive element 0 of the group F* by Q', where i runs through those integers 1, .... p" - 2 which are not divisible
p" - 1
(d; n. 1) of polynomials f(x) over finite fields cf. SCHWARZ (1956).
§ 135. Cyclotomic Polynomials
The splitting field of the polynomial
x"-1
(n_>_1)
(135.1)
over the prime field F. is called the nth cyclotonic field (of characteristic p > 0). Of course the cyclotomic field for p > 0 is a finite field, so that the case p = 0 is the most important. When we speak briefly of a cyclotomic field we mean this case. For the determination of cyclotomic fields it is necessary to decompose the polynomial (135.1) into irreducible factors (over F,).
CYCLOTOMIC POLYNOMIALS
509
Here we wish to deal only with this problem, and return to cyclotomic fields themselves more explicitly in § 171. In an arbitrary field we call every zero of the polynomial (135.1) an nth root of unity in this field. It is said to be a primitive nth root of unity when it does not satisfy any equation
xk-1=0
(15k I or (c, n) = 1. So from (135.7) we obtain n-1
/
(x - oc)
F (x) _
(135.8)
(c, n)=1
Accordingly, we have FF(x) E F[x]. On the other hand, by (135.3), we have F,(x) E FF(x). From both of these it follows that F.,(x) E F,,[x]. Because of (135.8) the last assertion of Theorem 315 is true, so completing its proof. THEOREM 316. Over the prime field of characteristic 0 all the cyclotomic polynomials are irreducible.
Let F denote the n' cyclotomic field, i.e., the splitting field of x" - 1. over ."t,. Then F = .70(P)
,
F,(Q)
= 0
where F,(x) denotes the nth cyclotomic polynomial and g a primitive n`h root of unity. Furthermore, we denote byf(x) the minimal polynomial of g
over.. We have to prove that f(x) = FF(x). Let p denote a prime number such that p,f' n. Then g° is a primitive nth root of unity. Its minimal polynomial over .Yo is denoted by g(x). First of all we show that g(x) = f(x). Let us assume that g(x) 96 f(x). Since e, g° are zeros of xn - 1, it follows by hypothesis that f(x), g(x) I xn-1 , and
xn - i = f(x) g(x) h(x)
(135.9)
for some h(x) in .70[x]. Since furthermore a is a common zero of f(x) and g(x°),
g(x") = f(x) k(x)
(135.10)
for some k(x) from .7 [x]. From GAUSS'S theorem (Theorem 207) it follows that the factors of the right-hand side of (135.9) lie in .7[x], and then the same follows from (135.10) for k(x).
THEORY OF FIELDS
512
For every polynomial 1(x) (E --Y [x]) its image corresponding to the homomorphism 3 - 3p is denoted by 1(x). Since then
3 [x] -
[x] (1(x) __> 1(x))
the equations
x" - 1 = f(x)g(x) h(x) ,
(135.11)
g(x") = f(x) k (x)
(135.12)
follow from (135.9) and (135.10), where we now have to think of the lefthand side of (135.11) as in 3p[x]. Because g(x1) = (9(x))"
it follows from (135.12) that f(x), g(x) have a non-constant common divisor. Accordingly, the existence of a multiple factor of x" - 1 (E [x]) follows
from (135.11). But this contradicts the result proved in the paragraph containing (135.4); so we have established the truth of the assertion g (x) =f(x).
We have thus seen that together with e, Np is also a zero of f (x). On the other hand, because p ' n, ep is also a primitive nth root of unity in F. By repeated application it follows that all the
,a
(a > 0; (a, n) = 1)
are zeros of f(x). But since these are all the primitive nta' roots of unity, it follows that F"(x) I f(x). Since the right-hand side is irreducible, f(x) _ = F"(x) follows. Thus Theorem 316 is proved. But, for the case of finite characteristic: THEOREM 317. Over the prime field of prime characteristic p such that p f' n the nth cyclotomic polynomial Fn(x) splits into e 97(n)
(e = o (p (mod n)))
(135.13)
different irreducible factors of degree e and is irreducible if, and only if, n is one of the numbers
n = qk, 2q", 4
(k >_ 0; q an oddprime number)
(135.14)
and the prime number p is a primitive number mod n.
For the proof we have recourse to a splitting field of x" - 1 over J",In this finite field, every zero of F"(x) is a primitive nt root of unity. We have to prove that the degree of o is equal to e. Because of the last
CYCLOTOMIC POLYNOMIALS
513
proposition in Theorem 306 it is sufficient to show that e is the minimum of the natural numbers d with the property
This equation is identical in meaning to the divisibility n I pd - 1, so that
e = o (p (mod n)) is in fact the minimum of these d. This proves the first part of the theorem. In order to prove the last assertion of the theorem we have to examine when the number (135.13) is equal to 1. This is the case if, and only if, e =
= p(n), i.e., o(p (mod n)) = p(n) , i.e., p is a primitive root mod n. In conformity with this, because of Theorem 226, the last assertion of Theorem 317 is also true. EXAMPLE 1. From (135.3) we obtain the special cases
FQ(x)=
xq X
-1
-
x"°- 1
Fq.(x) = x0 -
_
1 =
-X*_1+...+ x(q- 1)q +
+ x4 + i ,
(x°' - 1) (x - 1)
(x°-1)(x'-1)' where q, r are different prime numbers. We have:
Fi(x) = x - 1,
F4(x) = x2 + 1,
F$(x) = x + 1,
Fa(x) = x4 + x8 + x2 + x + 1,
F3(x)=x2+x+l
FB(x)=x2-x+1.
EXAMPLE 2. Since over the prime field of prime characteristic p xmat
- 1 = (xm - 1)'
,
Theorems 315, 316, 317 make the decomposition of x" - I into irreducible factors over every prime field possible. EXAMPLE 3. According to the above theorems the n(" cyclotomic field of characteristic p is of degree p(n) if p = 0, and of degree o(p (mod n)) if p > 0, p ,}' n. EXERCISE. The nth cyclotomic polynomial F"(x) is irreducible over the mth cyc-
lotomic field of characteristic 0 if, and only if, (m, n) 12.
§ 136. Wedderburn's Theorem THEOREM 318 (WEDDERBURN's theorem). Every finite skew field is afield.
In order to prove this, let F denote a finite skew field and Z its centre which is then a field. We write
O(Z) = q,
514
THEORY OF FIELDS
where q is a prime power. Since F+ is a finite-dimensional Z-vector space,
O(F) = q where it = [F : Z]. We have to prove that n = 1. With this end in view we assume that n > 2. It is evident that the group Z* is the centre of the group F*. The class equation (42.32) for F* consequently results in
q" -1=q- 1 +
0q-
(N*)I
O(N
(136.1)
where a runs through a system of representatives of those classes which are formed from the elements of F* - Z* and N* always denotes the centralizer of a in F*. If we add the element 0 to the group N*, then the set N so obtained evi-
dently constitutes a skew field, namely the centralizer of a in F. Hence Z c N c K, and so, by Theorem 292, for the degree r = [N : Z]
rIn, r_ 1) of a rational function
field F(x), the field F(x) I F(y) is algebraic and of degree n. In particular,
ax+b cx±ci
(a. b, c. d
F; ad - be -A 0)
(139.1)
are all the primitive elements of F(x).
For the proof we put y =
J(j)
g(x)
(139.2)
THEORY OF FIELDS
324
with two relatively prime polynomials
f(x)=aox'+...+a, g(x) = bor + ... + b.,
(aorA 0; a.,....arEF) (bo : 0; bo, ..., b' E F)
over F such that n = max (r, s) . We take a further indeterminate z and form the polynomial
F(z) = y9(z) - f(z) (E F(y) [z])
.
(139.3)
Since, according to (139.2), this has the zero z = x, it suffices to show for the proof of the first assertion of the theorem that F(z) is irreducible and of degree n. It is evident that the degree of F(z) is at most n. Since furthermore, in F(z), the coefficient of z" corresponding to the cases r < s, r > s, r = s, is respectively
boy, - ao, boy - ao
and y does not belong to F, this coefficient never vanishes. Consequently the degree of F(z) is exactly n. Since, according to Theorem 293, y is transcendental over F, in the proof of the irreducibility of F(z) it can be regarded as an indeterminate. Then, according to (139.3), F(z) is a linear polynomial in y over F [z], moreover it is primitive, since f(z), y(z) are relatively prime. Accordingly, F(z) is irreducible in F[y, z], and also in F(y) [z] because of GAUSS's theorem (Theorem 207). Consequently the first assertion of Theorem 323 is proved. Further F(y) = F(x) if, and only if, n = [F(x) : F(y)] is equal to I, i.e., y is of degree 1. These y are exactly given by (139.1), which proves the theorem. THEOREM 324. All the F-automorphisms of the rational function field F(x) are given by the substitutions
+b x-- ax cx+d
(ad-bc=0; a,b,c,dE F)
(139.4)
and the group is isomorphic with the linear fractional group of the second degree over F. If y is an arbitrary non-constant element of F(x), then by the substitution
x -> y a mapping
f(x)
f(y)
9(x)
9(v)
(139.5).
SIMPLE TRANSCENDENTAL FIELD EXTENSIONS
525
of F (x) into itself is given, where f(x), g(x) are polynomials over F, such
that g(x) does not vanish, whence, because of the transcendence of y, it follows that g(y) is also non-vanishing. For the same reason (139.5) is one-to-one and so a meromorphism of F(x). It is also evident, conversely, that all the meromorphisms of F(x) are furnished by (139.5). Since F(x) is mapped onto F(y), this meromorphism is an automorphism if, and only if, F(y) = F(x), i.e., if y is a primitive element of F(x). Because of Theorem 323, the first assertion of Theorem 324 is proved. For the proof of the second assertion we consider two automorphisms of F(x) which we give according to (139.4) in the form
x
ax+b
cx+d'
x
a'x+b' c'x+d'
(139.6)
If, of the two linear functions given here the second is substituted in the first, then we obtain
(aa' + bc')x + (ab' + bd') (ca' + dc')x + (cb' + dd')
The comparison with
(aa'+bc' ab'+bd' c d I` c' d',
t ca' + dc' cb' + dd'
shows that a homomorphic mapping of the full linear group of degree two over F onto the automorphism group of F(x) arises if to each regular matrix
ab c dI
(over F) the automorphism furnished by (139.4)
This automorphism is the identical one if, and only if, a i.e., if is a scalar matrix dl
k0 0 k
(k E F,
0),
is assigned.
ax+b = x, cx+a
(139.7)
so that these matrices constitute the kernel of the homomorphism. Consequently, the factor group of the linear group mentioned with respect to the group of matrices (139.7), i.e., the linear-fractional group of the second degree over F, is isomorphic with the automorphism group of F (x). Consequently Theorem 324 is proved. THEOREM 325 (LUROTH'S theorem). Every subfield of the rational function
field F(x) I F has a primitive element. We consider a subfield G of F(x) I F where we may restrict ourselves to the case G F. Since, according to Theorem 323, F(x) I G is algebraic,
THEORY OF FIELDS
526
there is an irreducible polynomial with the zero = = x in the polynomial ring G [z]. After multiplication by an element of F(x) we can turn this polynomial into
F(x, z) = ao(x)z" + ... + so that ao(x), ...,
(ao(x)
(139.8)
0)
are relatively prime polynomials in F[x]. Moreover
[F(x) : G] = n.
(139.9)
Let in denote the degree of F(x, .s) with respect to N.
Because of (139.8) and the fact that ao(x)-I F(x. z) E G[z] all the a,(x)
(i
0.
lie in G, but not all in F, since x is transcendental over F. Of these quotients we take one which does not lie in F and denote it by 0. Then we can write 11 =
g(x)
G),
(139.10)
where numerator and denominator are relatively prime polynomials over F of degree at most m and not both constant. By Theorem 323 we then have [F(x) : F(O)] < m.
(139.11)
The polynomial g(x) f(z) - f(x) g(z) is different from 0 and has the zero _ = x. Since this also holds for F(x, z) and this is a primitive and irreducible polynomial in = over F[x], it follows from GAUSS'S theorem (Theorem 207) that (139.12) g(x)f(z) - f(x) g(=) = G(x, z) F(x, z)
for some polynomial G(x, z) from F[.x, _]. The degree of the left and righthand sides with respect to x is at most m and at least m, respectively, thus
both are of degree m with respect to x. Since, according to this, G(x. _) does not contain the indeterminate x we may write G(x, z) = G(z)
(E F[z] ).
We can even show that G(z) E F. We assume that this is false. According to (139.12) G(z) I g(x)f(z) - f(x) g(z)
The residues arising from the Euclidean division of f(z) and g(z) by G(z) are denoted by.f,(z) and g,(z), respectively. At least one of these
SIMPLE TRANSCENDENTAL FIELD EXTENSIONS
527
polynomials is different from 0, sincef(z), g(z) are relatively prime. Further G(z) I g(x).f1(z) - f(x) g1(z)
But since the left-hand side is of greater degree than the right-hand side, this must vanish. Hence it follows that neither of f1(z), gl(z) is equal to 0 and f1(z)
Ax)
g1(z)
g(x)
lies in F. This contradiction proves that G(z) E F. Since the left-hand side of (139.12) is of the same degree with respect to x
and z, the same follows for F(x, z). But since the latter has degree m and n, respectively, with respect to x and z, we have m = n. By (139.9) and (139.11) [F(x) : G] > [F(x) : F(#)].
Since, on the other hand, we have G Q F ('), it follows that G = F (0). Hence Theorem 325 is proved. § 140. Isomorphisms of an Algebraic Field Since we have practically an unlimited possibility for obtaining isomorphic
structures from a given structure, we never ask for all the isomorphisms of a structure. On the other hand, it is in general of great importance to study isomorphisms which map a structure into a given overstructure. (This is essentially identical with the problem of substructures we have formulated earlier, according to which we look for the determination of all the
substructures of a structure non-isomorphic with one another.) Here we wish to deal with such questions with regard to fields and particularly for relative isomorphisms. At first we admit arbitrary relative fields and later restrict ourselves to algebraic fields and even to those of finite degree. Those isomorphisms of a field G I F which map G into a fixed extension field H of G are called the isomorphisms of G I F into H. First of all we prove the following, almost trivial
THEOREM 326. Over a field F, let a field G and two equivalent overfields H, H' of G over F together with an isomorphism H of H I F with H' I F be given. If S1, S2, ... are then all the different isomorphisms of G I F into H, then HS1, HS2.... are all the different isomorphisms of G I F into H'.
It is evident that HS1, HS2, ... are isomorphisms of G I F into H' and different from one another. So it remains to be shown that all the iso18 R.-A.
528
THEORY OF FIELDS
morphisms of G I F into H' occur among them. For eacn such isomorphism S, H-1 S is an isomorphism of G I F into H, i.e., equal to an Si. Since then
S = HS,, the theorem is proved. THEOREM 327. To every extension field H of an algebraic field G I F there is a normal field N of G I F such that all the isomorphisms of G I F into H occur among the isomorphisms of G I F into N. We denote by y' an image of an element y of G which has been generated
by any isomorphism of G I F into H. Together with y also y' is algebraic over F. We adjoin to F all the images y' of all the y and denote by G' the subfield of H obtained. Since, among the isomorphisms considered, the identical isomorphism of G occurs, we have
G9G'. We denote by N an arbitrary normal field of G' I F and show that N is a normal field of G I F. Since N I F is normal and contains G, it follows from the definition of the normal fields of a field that, in any case, N contains a normal field No of G I F. It suffices to show that necessarily No = N.
We retain the above notations y, y' and consider the minimal polynomial f(x) of y over F. Since No is normal over F, an equation such as
J(x)=(x-yI)...(x-yn)
(y1=y; y1,...,ynE N0)
must hold. Because f(y) = 0, we have f(y') = 0. Because N ? No it follows that y' is equal to one of the yl, ..., y,,, thus belonging to No. Since this holds for all the y', we have G' S No. Therefore we obtain the result that the normal field N of G' I F contains the normal field No
(? G') over F. By the minimal property of N it follows that No = N. Accordingly, N is in fact a normal field of G I F. Since, finally, all the y' belong to G', and so to N, every isomorphism of G I F into H is also an isomorphism into N. Consequently Theorem 327 is proved.
NOTE. On account of the last theorem, of all the isomorphisms of an algebraic field G I F only those into a normal field need be considered further. By Theorems 304, 326 it is mostly unimportant which normal field of G I F
is taken. 1=lowever, we shall often consider isomorphisms of G I F into arbitrary extension fields of G. We shall now consider an algebraic field G I F and its isomorphisms into
an arbitrary extension field H of G. These map G onto subfields of H equivalent to it, which we call the conjugate fields of G over F in H. We similarly denote the images, furnished by the same isomorphisms, of an element y of G as the conjugate elements of y over F in H. It is evident that
ISOMORPHISMS
529
these conjugacies are equivalence concepts. Therefore if G1 (= G), G2, ..., or yI (= y), Y2.... are the conjugates of G and y, respectively, then we can also say that G1, G2, ... are conjugate fields and that yl, y2, ... are conjugate elements, respectively, in H (over F). If we do not put "in H" we always mean the conjugates in an (arbitrary) normal field of G I F. The conjugate fields and elements over F are also called relative conjugate fields and elements, respectively. If F is also a prime field, then we can also talk of absolute conjugate fields and elements. To avoid any misunderstanding it should be noted that if S1, S2.... are all the different isomorphisms of G I F into H and G1, G2, ..., and yl, y2, .. . the corresponding conjugates of G and an element y of G, respectively, then neither the G1, G2, ... nor the y1, y2, ... need be different, for, on the one hand, among the isomorphisms considered, automorphisms (different from 1) of G I F may very likely occur. On the other hand yl, y2, ... are all equal if y E F. The different fields and elements among the G1, G2, ... and V1, Y2.... may be called "all the different conjugates" of G and y in H, respectively. Without specifying "in H" we mean all the different conjugates in a normal field of G I F. As regards the conjugates of an element y of an algebraic field G I F we observe that if f(x) is the minimal polynomial of y over F, then f(y') = 0 for all the conjugates of y in an overfield H of G. Therefore y' is algebraic over F as already stated. But hence it also follows that the number of conjugates of an element is always finite and does not exceed the degree of this element. (For further details see below.) Considering briefly an arbitrary (not necessarily algebraic) extension field G = F(01, 02, ...)
of the field F which arises from this by the adjunction of the elements 491, 02, ..., we see that since the elements of G are expressions in t9I, e 2, .. .
(with coefficients from F), each isomorphism y ->. y' of G F is already
....
On account uniquely determined by the images of the generators 0, t 2, of this, we may also represent these isomorphisms in the form 01 -
;
,
02 __> 02' 9 ...
(140.1)
where i9 denotes the image of 0I. We have applied this type of notation in Theorem 324 already. Of course, homomorphic mappings of an arbitrary structure may also be similarly denoted by giving the images of generators. If G I F is algebraic and (140.1) denotes an isomorphism of it into an overfield H of G, then every t9; must be a conjugate of O; in H. Now, by the above, the number of conjugates of an algebraic field element is finite, and hence, by virtue of Theorem 296 and from the observation
530
THEORY OF FIELDS
above, the same follows for the number of conjugates of a field of finite degree, thus justifying the following two important definitions. By the reduced degree of an element a of an algebraic field G I F we under-
stand the number of distinct conjugates of a. Further, we understand by the reduced degree of afield G I F of finite degree the number of isomorphisms
of G I F (both in an arbitrary normal field of this field). The term "reduced degree" of an element is explained by the fact that by the above
it is at most equal to the degree of this element. A similar statement also holds for fields.
THEOREM 328. If F(i) I F is an algebraic field, N its normal field andf(x) the minimal polynomial of the primitive element 13, then the distinct isomorphislns of F($) I F into N are given by
0-*01
(i= 1,...,n'),
(140.2)
where 01, ..., 13,,, are the different zeros of f(x), i.e., the different conjugates of $ in N. Consequently the reduced degree n' of 0 is equal to the reduced degree of F(13) I F.
If 0 -> 0' is an isomorphism of F(8) I F into N, then, because f(t9) = 0, we also have f(Y) = 0. But since f(x) splits over N into linear factors, it follows that 0' is equal to one of the 01 mentioned in the theorem. Therefore
it only remains to be shown that an isomorphism of the required type is given by (140.2) for every fixed i. Again this is trivial, and, because of Theorem 295 this isomorphism may be given by g(1') -- g(0)
(g(x) E F [x))
Consequently the theorem is proved. A part of this theorem is generalized in the following THEOREM 329. If G I F is an algebraic field, a an element and N a normal field of it, then for every zero a' (E N) of the minimal polynomial f(x) of a there is an isomorphism of G I F into N which carries a into a'. COROLLARY. The reduced degree of a is equal to the number of distinct zeros of f(x) in one of its splitting fields and so depends only on a and F. Since f(x) splits into linear factors over N, N contains a splitting field of f(x). Thus, by Theorem 328, we have the isomorphism F(a) I F
F((x') I F(a -> a')
(140.3)
into N. Now, in conformity with Theorem 303, N is a splitting field over F,
furthermore it contains both the subfields F(a), F(a ). If we then apply Theorem 302 (with G = G' = N instead of G, G'), it then follows that
ISOMORPHISMS
531
N I F has an automorphism which is an extension of (140.3). This automorphism induces an isomorphism of the subfield G I F and maps a onto a'. Accordingly Theorem 329 is true. In order to prove the corollary we have to consider that the reduced degree of a is equal to the number of its conjugates in N. On account of Theorem 329 this number is the same as that of the distinct zeros of f(x) in N. But since N contains a splitting field of f(x), f(x) has in this case the same zeros
as in N. This completes the proof. THEOREM 330. If F, F1, F2 are three fields such that F c F1 c F2 and F2 I F is offinite degree, then the reduced degree of F2 I F is equal to the product of the reduced degrees of F1 I F and F2 I Fl. We denote the relative degrees of F2 I F, F, I F and F2 I F1 by f2, f1 and f12,
respectively. We have to prove that f2 =.fi.fi2 Let N denote a normal field of F2 I F. Then f2 is the number of the isomorphisms of F2 I F into N.
Each of them is of the form
F21Fk F2' IF.
(140.4)
This induces a certain isomorphism,
F'IF
(140.5)
(into N). Conversely, those isomorphisms (140.4) to which a fixed isomorphism (140.5) belongs are so obtained from them that all the isomorphisms of F2'
I
F1'
(into N) are obtained. The number of these isomorphisms [because of (140.4), (140.5)] is the same as that of the isomorphisms of F2 1 F1
(140.6)
into N. Since N contains a normal field of (140.6), the number of these isomorphisms is equal to f12. Since, furthermore, N also contains a normal field of the left-hand side of (140.5), the number of isomorphisms (140.5) is equal to fl. From all this it follows that f2 = J i f12, so that Theorem 330 is proved. THEOREM 331. An algebraic field G I F is normal if, and only if, every iso-
morphism of it into an arbitrary extension field is an automorphism, i.e., G I F has no conjugates except itself. For, if G I F is normal, thus is its own normal field, then it follows from Theorem 327 that all the isomorphisms of G I F into an arbitrary extension field are isomorphisms into G, and so are automorphisms. If, on the other hand,
G I F is not normal, then there is an element a of G such that its minimal
THEORY OF FIELDS
532
polynomial f(x) does not split into linear factors over G. So, if we take a. normal field N of G I F, then N - G contains an element a' such that f(a') = 0. Accordingly, by Theorem 329, G I F has an isomorphism into N which maps a into a' and G onto a field different from it. Consequently it is not an automorphism of G. Thus Theorem 331 is proved.
.,
EXAMPLE. We consider the field 2 = . o(t9) with 04 - 2 = 0. Since x' - 2 is irreducible over 2 is of degree four. The element a = 0E of 2 is of degree two, be-
cause a2 - 2 = 0. The polynomial x2 - 2 also has the zero a' = - a in _T. Now x4 - 2 has only the two zeros 0, -0 in 2, consequently 2 has only the two automorphisms 0 - t9 and 0 - -tA. Both have a as fixed element. Hence we see that Theorem 329 would be false with G in place of N.
§ 141. Separable and Inseparable Field Extensions It is also an intrinsically interesting problem whether a polynomial f(x), irreducible over a given field, can have multiple zeros over a suitable extension field; if this is the case, then its splitting field certainly suffices. This
question is closely linked with the last paragraph. A polynomial f(x) irreducible over a field F is called a separable or inse-
parable polynomial, according as f(x) has only simple or also multiple zeros in an (arbitrary) splitting field. An algebraic element a of a field G I F
is called a separable or inseparable element (over F) according as the minimal polynomial of a (over F) is separable or inseparable. Finally, an algebraic field G I F is called a separable or inseparable field, according as all its elements are separable or inseparable elements also occur among them. The terminology "of first kind" and "of second kind", originally introduced by STEINITZ, instead of "separable" and "inseparable", is rarely used in current literature.
THEOREM 332. Let F denote a field of characteristic p (>_ 0). If p = 0 there are no inseparable polynomials over F. If p > 0 all the inseparable principal polynomials over F are those irreducible principal polynomials which have the form f(x) = g(x")
(g(x) E F[x]) ;
(141.1)
for the irreducibility of a principal polynomialf(x) of the form (141.1) it is necessary and sufficient that g(x) is irreducible and not all its coefficients are pth powers in F. Let us consider an arbitrary principal polynomial f(x) over F. This is, according to Theorem 267, inseparable, if and only if, the condition (1(x), f'(x)) 4z 1
(141.2)
SEPARABLE AND INSEPARABLE FIELD EXTENSIONS
533
is satisfied and f(x) is irreducible. But now condition (141.2) for an irredu-
cible f(x) is equivalent to f(x) I f'(x), thus to f'(x) = 0. This last condition for an irreducible, thus non-constant, f(x) is equivalent, according to Theorem 262, to the fact that p > 0 and (141.1) holds. Therefore only the case in which the last two conditions are fulfilled need be further considered. Only the criterion of irreducibility formulated at the end of the theorem remains to be proved. If g(x) is reducible then, because of (141.1), so isf(x). If
g(x)=x"+aix' +...+a."
(a1.... ,a"EF),
then by (141.1)
f(x)=(x"+alx"-1+...+a,,)P, thus f(x) is again reducible. The case where, in (141.1), g(x) is an irreducible principal polynomial and not all of its coefficients are pth powers in F still remains to be proved. We assume that f(x) is reducible in order to obtain a contradiction by which the theorem will be proved. In the first place we consider the case where f(x) has only one (but then necessary multiple) irreducible divisor in F[x]. So now an equation such as
f(x) = (x" +
atx"-1
+ ... + a")'
(a,, ..., a" E F)
(141.3)
holds for some natural number q different from 1. We may assume that q is a prime. Then, of course, the expression in brackets on the righthand side of (141.3) need no longer be irreducible. The case q = p is absurd, since then, because of (141.1) and (141.3) we obtain g(x)=x"+alx"-I+... +a,,
which contradicts the supposition. Consequently q 0 p. Again because of (141.1), f'(x) = 0, so that the differential'quotient of the expression in brackets on the right-hand side of (141.3) must vanish. Hence, and from Theo-
rem 262, it follows that f(x) _ (h(xP))q
f or a polynomial h(x) over F. This, and (141.1), give the result g(x) _ _ (h(x))q, which is again absurd, since g(x) was assumed to be irreducible.
In the second place we consider the case where f(x) has at least two different irreducible factors over F. Then an equation such as f(x) = f1(x) A W
(141.4)
THEORY OF FIELDS
534
holds with non-constant relatively prime factors in F[x]. Since, according to (141.1), f'(x) = 0, so
fi(x).fi(x) + fi(x).fz(x) = 0 . Since (f1(x), f2(x)) = 1, we have
l(x) I .f; (x)
(i = 1, 2).
This gives f(x) = 0. Thus, by Theorem 262, f(x) = ga(x")
(i = 1, 2),
where gl(x), g2(x) are non-constant polynomials over F. Hence, and from (141.1), (141.4), we have g(x) = g1(x) g2(x), although g(x) was assumed to be irreducible. This contradiction proves Theorem 332. THEOREM 333. An algebraic element at over afield F of characteristic p > 0
is separable if, and only if, F(a") = F(a). Let f(x) denote the minimal polynomial of a over F. According to the definition a is separable if, and only if, f(x) is separable. First, letf(x) be separable. By f1(x) we denote the polynomial determined by
f,(x") = (f(x))" and show that fi(x)
is
(141.5)
irreducible. Let h(x) be an irreducible principal
polynomial and divisor of fl(x) over F. Then h(xP) I f1(xP), thus h(xP) I (f(x))P.
Because of the irreducibility off(x) an equation such as h(xP) = (.f (x))k
(1 < k S p)
follows. By differentiation we get kf'(x) = 0. On the other hand, because of Theorem 332, condition (141.1) is not satisfied, therefore f'(x) # 0. Thus
k = p, h(x") = (f(x))P =fi(xP),
h(x) = f1(x).
Accordingly fl(x) is in fact irreducible. Furthermore because of (141.5)
and the fact that f(a) = 0, f1(x) and f (x) are of the same degree and fl(ap) = 0. It follows that F(a), F(aP) are of the same degree, i.e., that we have F(aP) = F(a).
Next, let f(x) be inseparable. Then (141.1) holds whence g(ap) = 0. Accordingly aP is of smaller degree than a, therefore F((XP) C F(M). Consequently Theorem 333 is proved. By the first part of Theorem 332 inseparability in connection with fields of characteristic 0 is impossible. This is why we have only considered the case p > 0 in Theorem 333. For the sake of generality, however, we some-
SEPARABLE AND INSEPARABLE FIELD EXTENSIONS
535
times admit the case p = 0 in connection with inseparability problems, although this will then be trivial. We shall see that for every algebraic element a of a field G I F of characteristic p (> 0) a separable term occurs in the sequence a, aP, cP°, .... (Of course, all the subsequent terms are then separable.) Let f(x) be the minimal polynomial of a over F and e (>_ 0) the greatest integer for which there is a polynomial g(x) with
Ax)=g(xn.
(141.6)
If p = 0, then e = 0, pe = 1 and g(x) = f(x) trivially. This number e is called the exponent of the element a (over F) and is, as we shall now show, the least integer e (> 0) for which the element cP` is separable. Since, because of (141.6), g(x) is irreducible and, because of Theorem 332, is separable, and also because, according to (141.6), g(cc') = 0, cc' is separable. On the other hand, if e Z 1 the element cc" 1 is inseparable, since it is a
zero of the irreducible and consequently inseparable polynomial g(xP). Hence, the assertion is proved. If G I F is algebraic and the exponents of the elements cc (E G) are bounded above, as they are if G I F is of finite degree, then we call their maximum the
exponent of the field G I F. (This will be considered in the next section.) THEOREM 334. For an irreducible principal polynomial f(x) of degree n over afield F of characteristic p (>_ 0) the factor decomposition
f(x) = ((x - aI) ... (x - an,))" ,
(141.7)
holds over its splitting fields G where aI, ..., an, (E G) are all its distinct zeros and e and n' denote the common exponents and the reduced degree of the latter over F. Hence (141.8) n = pen' , and n' I n; furthermore it follows that f(x) is separable if, and only if, e = 0,
(i.e., n' = n). Let us take f(x) in the form (141.6) with maximal e (? 0), then g(x) is irreducible and evidently not of the form h(xP). Accordingly, because of Theorem 332, g(x) is separable and
g(x)=(x-(3i) ...(x-Nn) over one of its splitting fields with distinct PI, ..., fin, (where n' has, for
the moment, nothing to do with that of (141.7)). Then by (141.6) f(x) = (xPe _ NI) ... (x"e 18/a R.-A.
fl.') .
THEORY OF FIELDS
536
The factors of the right-hand side are each divisible over a proper splitting field of f(x) by a linear polynomial x - al, ..., x - (x.,. Hence it follows that
fi,=a?'
(i= 1,...,n').
(141.9)
and
f(x)=(xre-ar)...(xr`-a.n, for which we can also write (141.7). Since the fi, are different, so from (141.9) the same fellows for the a,. The remaining assertions of Theorem 334 need no further proof. THEOREM 335. Consider an element a and two subfields F c F1 of a given
field of characteristic p (>_ 0) and suppose that a is algebraic over F (thus also over F1). Let e, n, n' and e1, n1, nl, respectively, denote the exponent, degree and reduced degree of a over F and F1, respectively, then e1 < e,
n1 < n, n' < n'. Then if a is separable over F, it is (because el S e) also separable over Fl.
Denote by f(x) and f1(x) the minimal polynomial of a over F and F,, respectively. Because F c F1, we have fl(x) I f(x). Hence, and from Theorem 334, follows Theorem 335. THEOREM 336. A field G I F of finite degree is separable if, and only if, its degree and reduced degree are equal. On account of Theorem 296, take elements a,, . . ., a,, such that
G = F(al, ..., a) , write
F,=F(a1,...,a)
(i=0,...,s; F0= F,F3=G)
and denote by n, n' the degree and the reduced degree, respectively, of G I F and similarly by n,, ni the degree and the reduced degree of F, I F,_1(i = 1, ..., s). By Theorems 292, 330
n=n,...n,, n'=nl...ns.
(141.10)
Furthermore, by Theorems 295, 328. n,, n{ are equal to the degree and reduced degree, respectively, of a, over F7_1. Because of Theorem 334 n; I n,
(i = 1, ..., s).
(141.11)
First, assume that n' = n. From (141.10), (141.11) it follows that nl = n1. This means, according to Theorem 334, that al is separable over (F0 =) F. But since at, can be an arbitrary element of G, we have obtained the proposition that G I F is separable. Next, assume that G I F is separable. Then al, ..., a, are separable over F, whence, by Theorem 335, it follows that a, is separable over F1_1, i.e.,
SEPARABLE AND INSEPARABLE FIELD EXTENSIONS
537
according to Theorem 334, n; = n; (i = 1, . . ., s). Hence, and from (141.10) it follows that n' = n. Consequently Theorem 336 is proved. A theorem similar to Theorem 297 is the following THEOREM 337. If G I F is a separable field, then the separable elements of a
field H I G are also separable over F. We consider an element a of H separable over G. It is to be proved that a is separable over F, too. We denote by f(x) the minimal polynomial of a over G and by al, . . ., as_I the coefficients of f(x). Since f(x) is separable over G, it is separable over the subfield F(at,
.
. ., ac,-j)
of G, and then the same holds for the element a. Furthermore, since al ..., a, _ 1 are separable over F, so ai is, according to Theorem 335, separable over F(al,
.
(i = 1, . . ., s - 1) .
. ., ai -1)
For the sake of uniformity we write
ex =a, and denote by n, n' the degree and the reduced degree, respectively, of the field
Ho= F(aj,...,oc) over F. Again, the equations (141.10) then hold, if nI, . . ., n, and
n'1,
.
. ., n;
are defined as before. On the other hand, according to the above, ni = ni (i = 1, ..., s). Hence, by (141.10), n' = n. This means, by Theorem 336. that Ho I F is separable. In particular, the element a (= a,) of HU is then separable over F. Thus Theorem 337 is proved. THEOREM 338. A field F(a1, a2, ...) I F is separable if a1, a2, ... are separable over F. 0) are separable elements of afield G I F, then so COROLLARY. If a, P (/ are
a + #, a - /3, a.fl, a/-1 .
(141.12)
First of all we prove the special case of the theorem, namely that F(a) I F
is separable if a is. According to Theorem 334 the degree and reduced degree of a over F must be equal. According to Theorems 295, 328, this means that the degree and reduced degree of F(a) I F are equal. This means, according to Theorem 336, that F(a) I F is separable.
538
THEORY OF FIELDS
For the general case it will suffice to prove that F(al, ..., as) I F is separable if al, ..., as are. According to the above special case, in the sequence of fields F, F(a), . . ., F (al, ..., as) every term (from the second on) is separable over the preceding term. By repeated application of Theorem 337 the assertion follows. Consequently Theorem 338 is proved. The corollary is trivial. If G I F is now an arbitrary field, then according to this corollary its separable (thus algebraic) elements over F constitute a field F which we call the separable hull of F in G. An algebraic field G I F is called a pure inseparable field if F = F, i.e., if all the elements of G - F are inseparable over F. An (algebraic) element a of an arbitrary field G I F is said to be a pure inseparable element (over F) if F(a) I F is pure inseparable. An irreducible polynomial f(x) over an arbitrary field F is called a pure inseparable polynomial (over F) if the field defined by the equation f(x) = 0 over F is a pure inseparable field (over F). Notice that a pure inseparable field G I F is in general inseparable. The only exceptions are the fields F I F which are at the same time separable and pure inseparable. Likewise in an arbitrary field G I F only the elements of F are simultaneously separable and pure inseparable. Among the (irreducible) polynomials from F[x] only the linear ones are at the same time separable and pure inseparable. THEOREM 339. In an algebraic field G I F the separable hull F of F is marked by the fact that F I F is separable and G I F pure inseparable. Certainly F I F is separable. Furthermore G I F is pure inseparable; for if
there were in G - F a separable element over F, then this, according to Theorem 337, would be separable over F, and belong to F. Conversely, if Fl is a field between F and G such that Ft I F is separable and G I Fl pure inseparable, then, as above, Fl 9 F. Fl must equal F, since otherwise G - Fl would have a separable element over F, which is impossible. Consequently Theorem 339 is proved.
THEOREM 340. For a normal field N I F the separable hull F of F in N is likewise normal over F. For, since under every isomDrphism separable elements become elements. of the same kind, Theorem 340 follows from Theorem 331.
THEOREM 341. An element a of a field G I F of characteristic p (> 0) is pure inseparable if, and only if, there is an integer e (>_ 0) such that cc" E F,
(141.13)
and the least e with this property is the exponent of a. The pure inseparable elements of G I F constitute a field.
539
SEPARABLE AND INSEPARABLE FIELD EXTENSIONS
We begin the proof by noting that the elements a of G, for which there is an e (>--_ 0) satisfying (141.13), are obviously algebraic, moreover they constitute a field. We consider two such elements e, or (a # 0) of G for which with two integers e, f (> 0) NP`, a' E F.
We may assume that e =f Then (e - a)°` = e" - (IrP` E F ,
(ea-1)' = epe a-' E F ,
whence the assertion follows. Besides, we have established that if
Of`, ...,1 r E F
01,
.,13 E G)
,
then for every element co of G (01, ..., t9s) the same condition c)" E F is satisfied.
We now suppose that (141.13) is true. In order to prove that a is pure inseparable we consider an arbitrary element co of F(a). The polynomial
x"-CDP`(=(x-MY) has co for a zero and lies in F[x]. Hence it follows that the minimai polynomial of co is a power (x - co)k (k z 1). Thus co is separable if, and only if, k = 1, i.e., co E F. Since in conformity with this F is its own separable hull in F(a) I F, this field, and so also a, is pure inseparable. Conversely, we suppose that a is pure inseparable. We denote by e the exponent of a. Then c' is separable. Since, on the other hand, according to the supposition F(a) I F is pure inseparable, (141.13) follows. If (141.13) (for e >_ 1) also held for e - I instead of e, then the polynomial Xpe-1 - av-'
would lie in F[x] and have a for a zero, consequently the minimal polynomial of a would have a degree which is smaller than p This contradicts (141.6) which implies that e is the least number with the property (141.13).
From what has already been proved, and from the above, the only remaining assertion of Theorem 341 follows. THEOREM 342. Over a field F of characteristic p (> 0) all the non-linear pure inseparable principal polynomials are those polynomials xPe - a
in which a is not a p`b power in F.
(e>0;aEF),
(141.14)
THEORY OF FIELDS
540
The principal polynomials concerned are the minimal polynomials of those pure inseparable elements from an extension field of F which lie outside F. First let a be such an element and e its exponent. By Theorem 341
e is at the same time the least non-negative integer with the property (141.13), so that now e >_ 1. Furthermore it follows from (141.13) that
ae=a for an element a from F. Thus the minimal polynomial of a is a divisor of (141.14), and even equal to (141.14), since it must be of the form (141.6). There a cannot be a pth power in F, for otherwise (141.14) would not be irreducible. Conversely, if a is not a pth power in F, then, according to Theorem 332, the polynomial (141.14) is irreducible, and for every zero of (141.14), lying
in a suitable extension field, the condition (141.13) holds. Therefore, according to Theorem 341, a is pure inseparable. Since (141.14) is its minimal
polynomial, this is also pure inseparable. Consequently Theorem 342 is proved. THEOREM 343. A field G I F is pure inseparable if, and only if, all its elements
are pure inseparable over F. Assume at first that G I F is pure inseparable. For every element a of G its exponent e has the property that a' is separable, and so by hypothesis
lies in F. Consequently a is pure inseparable by Theorem 341. Conversely, if every element a of G is pure inseparable, i.e., if the fields F(a) I F belonging to it are pure inseparable, then it follows that F is its own
separable hull in every F(oc), and so also in G. This means that G I F is pure inseparable, so proving the theorem. THEOREM 344. Every pure inseparable field G I F is normal and its only automorphism is the identical one.
From Theorem 342 it follows that a pure inseparable element has no conjugate different from itself. Hence, and from Theorem 343, it follows that every isomorphism of G I F into an arbitrary extension field of G can only be the identical one. Because of Theorem 331, this proves Theorem 344. THEOREM 345. The reduced degree of a field G I F of finite degree and of
characteristic p (> 0) is equal to the degree [F : F] of the separable hull F of F in G. If N is a normal field of G I F, then every isomorphism of G I F into N induces an isomorphism of r F into N. Consequently, if we prove that conversely every isomorphism
FIF;-- F'IF
(p - to)
(141.15)
SEPARABLE AND INSEPARkBLE FIELD EXTENSIONS
541
of F I F into N may be extended in only one way to an isomorphism of G I F into N, then, by Theorem 336, the assertion follows. By Theorems 339, 343 all the elements of G I F are pure inseparable. Since this field is of finite degree, we may write
G = WI, . . ., o,) ,
(141.16)
where the elements
(i= 1,...,s)
(141.17)
for suitable el, ..., e., (>_ 0) belong to F. Now if A -> A* is an isomorphism
of G I F into N which continues the isomorphism (141.15), then from (141.17) it follows that
(i = 1 ,
t*r`' = ai
..., s) .
(141.18)
As at most only one system 6; , ..., 8 (E N) is possible with this property, so, because of (141.16), there can exist, in fact, at most one isomorphism
such as 7, -**.
On the other hand, from the normality of N I F the existence of
elements 0i, ..., tS (E N) which satisfy (141.18) follows, and it is obvious that an isomorphism of G I F into N which is a continuation of (141.15) is defined by s)
E F) .
Consequently, Theorem 345 is proved. Because of this theorem the reduced degree of a field of finite degree is always a divisor of the degree. This also follows from Theorems 328, 330, 334. By the degree of inseparability of a field G I F of finite degree we mean the
quotient of the degree and the reduced degree of this field. This is always equal to the degree [G : F], where F is the separable hull of F in G. It is trivial if the characteristic is 0, and a consequence of Theorem 345 if the characteristic is a prime number p. In this case the degree of inseparability is, because of Theorems 339, 342, a power of p. The degree of inseparability of a field of finite degree is equal to 1 if, and only if, it is separable. EXAMPLE 1. Since in a finite field of characteristic p every element is a pt' power,
it follows from Theorem 332 that there are no inseparable polynomials over finite fields.
EXAMPLE 2. Over the transcendental extension field F,($) of the prime field F, with characteristic p (> 0) the polynomial x° - aA is irreducible and inseparable.
§ 142. Complete and Incomplete Fields
A field F is called complete if only separable field extensions over it are possible; otherwise the field F is called incomplete. In other words, a field is complete if, and only if, every irreducible polynomial f(x) is sepa-
THEORY OF FIELDS
542
rable over it. In particular, all fields of characteristic 0 are complete, therefore we shall consider in this section only fields of prime characteristic. THEOREM 346. A field F of prime characteristic p is complete if, and only if, every element in it is a pth power. If in F every element is a pth power, then, from Theorem 332, it follows that there are no inseparable polynomials over F. If, on the other hand, a is an element of F which is not a pth power in F, then again according to this theorem, the polynomial xP - a is (irreducible and) inseparable. Consequently, Theorem 346 is proved. From this theorem it follows, e.g., that finite fields are complete, but, rational function fields of prime characteristic are incomplete. (Cf. § 141, Examples 1, 2.) All algebraically closed fields are complete.
For an arbitrary field F of characteristic p (> 0) we shall henceforth denote
the set of the pth powers of the elements of F by F". We know that F" is always a field, and that Theorem 305 implies that the meromorphism F
FP (a -).. aP)
(142.1)
holds. The content of Theorem 346 may then be expressed by saying that F is complete if, and only if, FP = F, i.e., if the meromorphism (142.1) is an automorphism.
We now define
(n=1,2,...).
F1=F, FF"=(FP"-')P
(142.2)
Then FP" is that subfield of F which consists of the elements aP" (a E F).
If F D F' then FP" z)
FP'+'
follows by applying the meromorphisrn a -> a"" for every n = 0, 1, .... Accordingly either
F=FP=FP'=
...
(142.3)
or
F:D FP:DP's...,
(142.4)
according as F is complete or incomplete. The intersection
D=FfFPfFP'fl...
(142.5)
is always complete, since obviously D° = D. If, in addition, G is an arbitrary complete subfield of F, then
G= G""cF"", thus G 9 D. This results in the following
COMPLETE AND INCOMPLETE FIELDS
543
THEOREM 347. For every field F of characteristic p (> 0) the subfield F n FP fl Fp' fl ... is complete and contains all complete subfields of F. It should be noted that the field F I FP is pure inseparable, since the p`h powers of the elements of F lie in F". This field F I F" will be important in further discussions, and already in the following fundamentally important .
definition. An element a of a field F of characteristic p (> 0) is said to be p-dependent
on a set (or a system) S 9 F, if (142.6)
e E FP(S).
Otherwise, Q is said to be p-independent of S. We shall show that the fundamental properties I, II, III, formulated at the beginning of § 100 with respect to linear dependence, also hold for p-dependence in the field F. First of all I holds, since o0i E F" (co,, . . ., co.) (i = 1, . . ., n). In order to prove II, we suppose that for certain elements Q, ooi, ... a> of F g E Fp(co1, ..., co.),
F"(col,
..., con-i)
.
We must infer that co,, E F"(o, w1, ..., con-0 .
(142.7)
By hypothesis an equation such as e =
p-1
aiwin i=0
holds for some elements ao, ..., ap_1 from R(col, ..., co,,), _1such that not all a1, . . ., ap_i vanish. Hence it follows that the degree of w on the right-hand side of (142.7) is smaller than p. Since co is pure inseparable over F", and so a fortiori over the right-hand side of (142.7), it follows from Theorem 342 that this degree is a power of p, thus necessarily 1. This implies that (142.7) holds. Finally, III also holds, since for any elements o, el, . ., en:, wi, ... W,
of F, from the assumption
a E F"(ei, ..., em), Lei E F"(w1, . . ., co.)
(i= 1'...m)
it evidently follows that a E F°(co1, ...,
From what has now been proved it follows that the conclusions in § 100 remain valid when p-dependence is understood instead of linear dependence and the definitions are suitably adapted. In conformity with this, we call a subset S of a field F of characteristic
p (> 0) a p-independent set when every element a (E S) is p-independent
544
THEORY OF FIELDS
of S - e. Two subsets S, T of F are called p-equivalent sets when every element of each of these sets is p-dependent on the other set; this simply means that FP(S) = FP(T).
(This definition can also be extended to the more general case, where S, T are systems of elements of F.) Finally, by a p-basis of F we mean a p-independent subset S of F such that F = FP(S). The following theorem is similar to STEINITZ'S second main theorem (Theorem 322). THEOREM 348. Every field F of characteristic p (> 0) has at least one p-basis, and the p-bases are the maximal p-independent subsets of F. Furthermore all the p-bases of F are equipotent. We can omit the proof since it is similar to that of Theorem 322 in that
we have only to replace the notion of algebraic dependence by that of p-dependence. The proof involves no difficulties. For a combined proof of Theorems 322 and 348 see MAcLANE (1938). Cf. KERTtsz (1960).
On account of Theorem 348, we call the cardinal number of an arbitrary p-basis of a field F of characteristic p (> 0) the degree of incom-
pletability of this field. This is an invariant of the field and is 0 if, and only if, the field is complete. If the field F is of finite degree of incompletability k, then F I FP is obviously of finite degree, and then [F
:
FP] = pk.
THEOREM 349. If G I F is a pure transcendental field of characteristic p (> 0) and degree of transcendence n, and if F is of degree of incompletability k and both of these degrees are finite, then G has degree of incompletability
n + k. The union of a transcendence basis of G I F and a p-basis of F then yields a p-basis of G.
It is sufficient to prove the theorem for the case n = 1, for then the general case is proved by induction. Let G = F (0), and let S be a p-basis of F. Then we have G = FP(S)(b) = F1(t91)(S, e) = GP(S, 0).
Since 0 does not belong to F, still less to S, it is sufficient to prove, for the first assertion, that S U 0 is a p-independent set in G. Instead of this, because of Theorem 238, it will suffice to show that S constitutes a p-independent set in G and 0 is p-independent of this set.
Instead of the first of these two assertions we prove more generally: if, for an element a and a subset A of F, a is p-dependent on A in G, then
545
COMPLETE AND INCOMPLETE FIELDS
this also holds in F as well as G. By hypothesis, a is an element of GP(A) = FP(8P) (A) = FP(A) (&")
and is algebraic over FP(A). But since & is transcendental over FP(A) (S F), by Theorem 293 a E FP (A). This means that at is p-dependent on A in F, which was to be proved. If, further, is p-dependent on Sin G, then 0 E GP(S) = FP(OP)(S) = FP(S)(9P).
This means that an equation such as P
$
AM
(f (x), g(x)
E
FP(S)[x])
holds. This contradiction completes the proof of the first assertion of the theorem.
The truth of the second assertion of the theorem follows from G = F(s) = FP(S)(0) = FP(SP)(S)(0) = GP(S, 0). THEOREM 350. If G I F is a separable field of characteristic p (> 0), then every p-independent subset of F is also p-independent in G. Further every p-basis of F is a p-basis of G. COROLLARY. The algebraic extension fields of a complete field are complete.
For the first assertion of the theorem it suffices, as in the proof of the last theorem, to prove that if an element a (E F) is p-dependent in G on a set A (c F), then the same also holds in F instead of G. We put G = F (S). Then a E GP(A) = FP(SP) (A) = FP(A)(SP).
(142.8)
On the other hand, the application of the meromorphism P -->. PP of G shows that GP I F° is separable. Since, then, the elements of SP are separable over FP(A), it follows from (142.8) that GP(A) I FP(A) is separable. But now a is pure inseparable over FP, and so also over FP(A), consequently a E FP(A). This establishes the assertion. In order to prove the second assertion of the theorem we denote a p-basis of F by B. Then we have F = FP(B) and so F C GP(B).
Consequently G I GP(B) is separable. Since this field is also pure inseparable,
it follows that G = GP(B). Hence, and from the first assertion already proved, it follows that the second assertion of the theorem is true.
THEORY OF FIELDS
546
If F is a complete field and G its algebraic extension field, then G I F is separable according to the definition of complete fields. Since further the only p-basis of F is now the empty set, the same follows for G from Theorem 350. This proves the corollary. THEOREM 351. If G I F is a field of characteristic p (> 0) and of finite degree, and F is of finite degree of incompletability, then F and G are of the same degree of incompletability. Since G I F, F FP are of finite degree, [G : F] [F : FP] = [G : FP] = [G : GP] [GP : FP]
from Theorem 292. The application of the meromorphism e
eP of G
results in [G : F] = [GP : FP].
Hence [F : FP] = [G : GP], i.e., the theorem follows. THEOREM 352. Every field F of characteristic p (> 0) has an extension field G with
GP = F.
(142.9)
If F is complete, this implies that G = F. We take a field F such that F'
and F, F are disjoint. Because F F'P
F
F", F.
Accordingly F may be embedded in F' so that the subfield FP is replaced by F. After embedding, the field obtained from F is denoted by G. Then equation (142.9) holds as required. If F is complete, then, because GP' = FP = F = GP, G is also complete, from which G = GP = F follows. This proves the theorem. By (142.9) the field G is, to within isomorphism, uniquely determined and, since G - GP = F must hold, we denote it by FP-'. FP-' is uniquely determined even up to equivalent extensions, for if G, G' are two fields such that GP = G'P = F and we assign to each A (E G) that A' (E G') for which AP = A'P, then we have the isomorphism
GIFsrG'IF(A--*A'). Moreover we define FP-° =
(FP-A+')P
(n = 2, 3, ...)
COMPLETE AND INCOMPLETE FIELDS
547
by recurrence. Then the terms of the ascending chain of fields
F9FP--'cFp_2c...
(142.10)
are uniquely determined up to equivalent extensions. Moreover each term is the pth power of the term following it. Of course, everywhere in (142.10)
either "=" or "e" holds [cf. (142.3), (142.4)]. THEOREM 353. To every field F there is a complete extension field F which
has the property that the between fields G such that F c G c F are incomplete. This field F is, apart from equivalent extensions, uniquely determined and algebraic over F.
Let p denote the characteristic of F. We may suppose that p > 0, otherwise. the theorem is trivial. As in Theorem 47 we form [as a counterpart to (142.5)] the union field of the chain (142.10) which we denote by F and prove that for this the requirements of the theorem are fulfilled. Evidently F° = F, therefore F is complete. Since each term of the chain is algebraic over F, the same holds for F. If G is a complete field between F and F, because of Theorem 346 it must contain all the terms of the chain (142.10), so that G = F. It still remains to prove the uniqueness part of the theorem. If F is a complete extension field of F, we can form a chain of fields in it as in (142.10).
If the union of (142.10) is again denoted by F, this is a complete subfield of V. Consequently, if we postulate that the between-field G such that
F c G c F is incomplete, then, necessarily, F = F. This completes the proof of the theorem. EXAMPLE 1.
The degree of incompletability of the rational function
field
.-9:",(xt, ..., x") is equal to n. EXAMPLE 2. In every complete field F of characteristic p (> 0) define a°-` (a E F;
n = 1, 2, ...) as the (only) root of the equation x" = a. For every k (= 0, ±1, ...) a - a°" is then an automorphism of F. EXAMPLE 3. If, for a subset S of a field F of characteristic p (> 0), F = F'(S),
(142.11)
then
F=P"(S)
(n=1,2,...)
(142.12)
also holds. This assertion is true if n = 1. We assume its truth for an arbitrary n. If we denote by S'" the set of elements a' (a E S), then, by applying the meromorphism
e - e°", the equation
F°" =
F
= F""+`(S"") (S) = F""+`(S°" , S) = FP' +`(S),
follows by induction. This proves (142.12) in the general case.
THEORY OF FIELDS
548
§ 143. Simplicity of Field Extensions THEOREM 354. The set of fields between a field F and its extension field G is finite if; and only if, G is a simple algebraic extension of F. COROLLARY. If an overfield G of a field F is a simple algebraic extension of F, then the same holds for all fields between F and G.
To prove the theorem we first suppose that G is a simple algebraic extension of F. Let 19 denote a primitive element of G and f (x) its minimal polynomial over F : G = F(t9), f(19) = 0.
If H is a field between F and G, then the minimal polynomial g(x) of 0 over H is a divisor of f(x). Since, according to this, there are only a finite number of possibilities for g(x), the finiteness of the number of fields H will follow if we show that different g(x) belong to different H. For this purpose we suppose that the same polynomial g(x) belongs to two such fields H1, H2. We put
H12=H,f1H.,. Because g(x) E H1[x], H2[x], it follows that g(x) E H12[x]. Since g(x) is irreducible over H1, it is also irreducible over H12. Consequently we have
[G:H12]-[G:HI]=[G:H2], that is, H 1= H 2.
Now suppose, conversely, that between F and G there are only a finite
number of fields. If G I F had a transcendental element e, then F(e), ... would be an infinite number of different fields between F and G, consequently G I F must be algebraic. Consider a chain F(e2),
F c F(O) c F(#1, 02) c
... c G
of fields between F and G. Since this chain is finite, by hypothesis, there must be an n with
G=
r9n)
It is sufficient to show that, if n = 2, G I F is a simple extension, then the same will follow by induction for all n. Therefore we assume that G = F(+91, '02),
where we may restrict ourselves to the case 01, 192 o 0. If F is finite, then also
G is finite. Consequently the assertion is now true, according to Theorem
SIMPLICITY OF FIELD EXTENSIONS
549
306. If F is infinite, there are an infinite number of different elements 01 + c62 (c E F). Thus, there are two different c, d (E F) such that F((91 + (1#2)= F(01 + (1192).
This field contains the elements 01 + co, #1 + d092, and so also their difference (c - d)-92. Consequently it also contains 192, i91 and so the whole
field G, to which it must be equal. Accordingly G is a simple extension of F. This proves Theorem 354 and the corollary follows. For the proof of Theorem 354 cf. WILKER (1951 - 52). THEOREM 355. An algebraic field extension F(a1i . . ., ak) I F is simple if the adjoined elements al, ..., ak are, with at most one exception, separable. Since the theorem is trivial for k = 1 we have only to consider the case k Z 2. It will suffice to prove it for k = 2, since if for k >_ 3 we suppose
it to be true for smaller k, there is a
such that
F(al,
..., ak--I) = F(P);
F(al,
..., ak) =
since then F(e, ak)
and, according to Theorem 338, of the elements e, ak at least one is separable,
the general validity of the'theorem follows. Now let F(oc, fi) I F be the given field, where a, fi are algebraic and, e.g., is separable. We denote the minimal polynomials of a and P over F by
f(x), g(x), respectively. We take a splitting field G of the polynomial f(x) g(x) over F(a, ,9) and denote by al
QQ
QQ
//
QQ
a), a2, ..., am; Y1 (= N)+ #22 ..., Nn
all the different zeros of f(x) and g(x), in G. If F is finite, then F(oc, 9) is
also finite. The assertion is, therefore, true. Then we suppose that F is infinite. Since every equation such as
(X(+ fijx = al + N1x
(j # 1)
has at most one solution x (E F), there exists an element c (E F) such that
a(+C9f0al+A
(i= l,...,m; j=2,...,n).
We show that for the element
0 =0cl+ c#1=a+cP the equation F(a, j9) = F (0)
holds.
THEORY OF FIELDS
550
The left-hand side contains the right-hand side, therefore we have only to prove that a, f3 E F (s). Because a = 0 - ci, it suffices if we show that E F(s).
(143.1)
Now
g(fl) = 0,
f(1' - e fl) = 0.
According to this, f is a common zero of the polynomials
g(x), f(( - cx).
(143.2)
No further common zero exists (in G), since for every other zero f1, (j 9& 1) of g(x) the inequalities 0 - c fl t
a;
(i = l , ..., in)
all hold, and so f(O - c f) 0 0. Furthermore, since g(x) is separable, f is a simple zero of it. We have now shown that the polynomials (143.2) have greatest common divisor x - fl. But since the coefficients of these polynomials lie in the field F(t), the same follows from Theorem 203 for x - f3. This means that (143.1) holds, and completes the proof of Theorem 355.
We now formulate an important special case with a slight improvement. THEOREM 356. Every separable field extension G I F of finite degree is simple, and an element 0 is primitive if, and only if, the number of its conjugates in a normal field of G I F is equal to the degree of G I F. The first assertion follows from Theorem 355. In order to prove the second assertion we denote the degree of G I F by n. This is now, according to Theorem 336, equal to the reduced degree of G I F, therefore this field has, in its normal field, exactly n isomorphisms. Hence, and from Theorem 328, the second assertion of Theorem 356 follows. THEOREM 357. A field extension G I F of finite degree and of characteristic
p (> 0) is simple if, and only if, for the exponent e and the degree of inseparability pf of G I F the equation e = f holds. We denote by F the separable hull of F in G. Then we have [G
If G I F is simple, i.e., G =
F]'= pf for a (primitive) element r9, then G = F(g)
and pf is the degree of 0 over F. But, since G I F is pure inseparable, this degree, by Theorem 341, is equal to pe, whence e =f.
SIMPLICITY OF FIELD EXTENSIONS
551
If, conversely, e = f, then we take an element a from G with exponent e. Then we have [F(a) : F] = pe = pf = [G : F], and so G = F (a). Since, on the other hand, F I F is separable, it follows from
Theorem 356 that F = F(p) for a separable element j9. From
G=F((x)=F(a,fi) and Theorem 355, it follows that G F is simple, whence Theorem 357 is proved. THEOREM 358. If a pure inseparable field extension F(aI, ..., ak) is simple,
then there is a primitive element among the adjoined elements al, ..., a,,.
We put G = F(al, ..., ak) and suppose that, e.g., the exponent of al, which we denote by e, is the greatest possible. Then e is at the same time the exponent of G I F, whence, according to Theorem 357, it follows that pe is the degree of inseparability of G I F. Since this field is pure inseparable, this means that [G : F] = pe. On the other hand, by Theorem 341, [F(a1) : F] = pe, so that G = F(a1). Theorem 358 is now proved. EXAMPLE. Theorem 355 is, in general, false if among the al, . . ., ak two inseparable
elements occur. Indeed, if G = Sr-, (x, y) is the rational function field of the inde-
terminates x, y over P-,, p > 0, and F = .g', (x°, y°) then G = F(x, y) with x, y° E F, whence we see, because of Theorem 358, that G I F cannot have a primitive element. This is also obvious, since [G : F] = p2 and every element of G I F has degree < p. According to Theorem 354, there are infinitely many fields between F and G. These are as follows: F(x + ay) (a ( F), F(y).
§ 144. Norms and Traces in Fields of Finite Degree We consider a field G I F of finite degree. Since this is a finite-dimensional
F-algebra, so for every element a its characteristic polynomial f(x), its norm N(a) and its trace T(a) (over F) are defined (see §§ 74, 75). These are called the characteristic polynomial, the norm and the trace of a with respect to G I F and denoted if necessary, more precisely by fGIF(x; a),
NGIF(a),
TGjF(a).
(144.1)
These last two terms are called the relative norm and relative trace of the element a (E G). First the following simple theorem holds:
THEORY OF FIELDS
552
THEOREM 359. If H I G, G I F are two fields of finite degree with
r = [H:G], then
(144.2) (a E G), fHIF(x; a) = (GjF(x; a))r (a))r THIF(a) = rTGIF(a) (a E G). (144.3) NHIF(a) = (NG,F
For the proof, assume an F-basis cot, ..., co, of G and a G-basis P1, of H. Then, according to Theorem 292,
..., er
(i = 1, ...,1; j = 1, ..., r)
w1ej
is an F-basis of H. Thus I
acol = E a,kwk k=1
(i = 1, ...,1)
(144.4)
with uniquely determined elements aik of F. Hence I
a(oleej =
k=1
alkCokQj
(i = 1, .
.
.,
1; j = 1, ..., r) .
(144.5)
Temporarily we arrange the basis elements co1LOj as follows:
wleh
, C'01,01, ...; co wr,
, Wlt-Or
The matrices connected with the linear mappings defined by (144.4), (144.5)
are denoted by A and A', respectively. Then A = (a11) is a square matrix of type la, and
A' =
is a square matrix of type (Ir)a in which r matrices A are placed diagonally and in the empty places are only zero elements. The corresponding characteristic matrices are
xU - A, xU' - A',
(144.6)
where U, U' denote the unity matrix of type 12 or (1r)a, respectively. Between
the determinants of the matrices (144.6) we have the equation
IxU'-A'I=IxU-Ajr. Hence, and from (74.4), (75.4), (75.5) equations (144.2), (144.3) follow. Consequently the theorem is proved.
NORMS AND TRACES
553
For a more detailed study of the concepts introduced in (144.1) we use the minimal polynomial (144.7) + ... + c,. 9(x) = xn + of the above element a of G I F, where n = [F(a) : F]. We prove that c,_1xn-1
(144.8)
.fF() F(x) = 9(x).
To do this we use in F(a) the F-basis 1 , a, ..., an-1. Because of (144.7) we have
(i = 0, ..., n - 2),
acct = (x`+1
... -
aan-1 = __co - Cla -
Cn_lan_'-
By comparing these equations with (144.4), we see that the above matrix is now specialized to 0
1
0
0
0
1
0
...
A=
(144.9) i
0
0
0
- CO - Cl
1
- Cn_2 - Cn_1
ii
1
This is called the concomitant matrix of the polynomial (144.7). Its characteristic matrix is x
-1
0
0
x
-1
0
xU - A=
(144.10) 0
0
x
CO
Cl
Cn_2
-1 X + Cn_1
If we add, in order, for j = n,.. ., 2, x times the jth column to the (j - 1)`h column, and place the first last, then we obtain a matrix with diagonal
-1, ..., -1, g(x) and only zero elements above the diagonal. By suitable elementary row transformations this becomes the diagonal matrix 1
-1 (144.11)
-1 9(x) 1
THEORY OF FIELDS
554
(of type n2). The determinant of (144.11) is (- 1)n-1 g(x). On the other hand, the determinants of the matrices (144.10), (144.11) differ from one another by the factor (- 1)n-1 because of the column permutation, so that the determinant of (144.10) is equal to g(x). (144.8) is now proved. This result, and Theorem 334, yield the following THEOREM 360. If a is an algebraic field element over a field F, then its characteristic polynomial is, with respect to F(a) I F, equal to its minimal polynomial over F and may be represented as a)In n ' (x fFtOF (x; a) = ( (ai = a), (144.12) i ('I
where n, n' denote the degree and reduced degree, respectively, of F(a) I F, and II, . . ., an the different conjugates of a in a normal field of F(a) I F. The norm and trace of a with respect to F(a) I F are: NF(-)IF(a) _ (al ... an.)nn'-"
(144.13)
TF(a);F(I) = n n'-I (II + ... + an). We prove the following THEOREM 361. A field G I F of finite degree is inseparable if, and only ij; always TGIF(a) = 0
(a E G).
If G I F is inseparable, then its degree of inseparability is a power of the characteristic p (> 0) of F. Hence, and from (144.32), (144.132) the "only if" of the theorem follows. In order to prove the "if" part we suppose that G I F is separable. According to Theorem 356, we can write G = F(a), where m ;A 0 may be assumed.
Let g(x) be the minimal polynomial of I. By hypothesis
g(x)=fl(x-ai) i=1
(a1=a)
holds [as a special case of (144.12)], where 21, ..., an are the distinct conjugates of a in a normal field of G I F. Now g'(x)
0.
According to Leibniz's rule we have g'(x) _
g(x)
(144.14)
Consider the identities n
xn - ain = (x - a,)
a;'-kxk-1
k=1
(i = 1, . . ., n).
NORMS AND TRACES
555
. ., a are different from 0, so by cancelling the first term of the left-hand side and dividing by - a/' the congruences
Since at, .
n 1
( x - ai)
(i = 1, ... , n)
a; kxk-1 (mod xn) k=1
are obtained in the polynomial ring F(a1, .
.,
.
an) [x]. After multiplication
by g(x) and addition, on account of (144.14) we obtain
x - at
g'(x)
g(x) > TF(-)IF (a/ k)xk`(mod x"). k=]
Since the left-hand side is different from 0, the same holds for at least one TF(,,)IF (ai k). Consequently Theorem 361 is proved. EXAMPLE. Since (144.11) is obtained from (144.10) by elementary transformations,
(144.11) is the normal form of the characteristic matrix (144.10). The elementary divisors are - 1,.. , - 1, g(x).
§ 145. Differents and Discriminants in Separable Fields of Finite Degree We take a separable field G I F of finite degree. The concept of the conjugates of the elements of G can often be easily modified as follows: we take a given fixed normal field N of G I F and denote by s1 (= 1), s2i ..., sn all the different isomorphisms of G I F into N, where n is the degree of G I F. By the conjugates of an element a of G I F (in N) we mean the elements a; = s,a
(i = 1,
.
. ., n),
(145.1)
which, unlike the "customary" conjugates of a, are not necessarily distinct. Usually we assume the order of the isomorphisms s2, ..., s,, as arbitrarily fixed, and then we call al the i' conjugate of a. (In particular, al = a.)
If we denote by f(x) and g(x) the characteristic polynomial of a with respect to G I F and the minimal polynomial of a over F, respectively, then by Theorems 359, 360
f(x) = g(x)n-- "
(145.2)
where m is the degree of a (over F). The isomorphisms sl, . . ., s,, are continuations of isomorphisms of F(a) I F and every isomorphism of F(a) I F into N has exactly nm-1 continuations among the s1, . . ., s,,. This means that in (145.1) all the different conjugates of a occur with the same multi-
THEORY OF FIELDS
556
plicity nm'1. Hence, and from (145.2), it follows that
f(x) = (x - (xl)
... (x - an).
(145.3)
Notice that the elements (145.1) are all the different conjugates of cc (in N) if, and only if, m = n, i.e., a is a primitive element of G I F. Now we call 6GIF((X) = f'(a) = (a - a2) ... (a - an) (145.4) the different and DGIF (a) = k-
1)(E) NGIF
IbGIF (a))
(145.5)
the discriminant of the element a (E G). The more precise terms relative
different and relative discriminant or - if F is the prime field of G absolute different and absolute discriminant, are also used. Clearly SGIF(a) E G,
DGIF(a) E F;
moreover we see that the different and the discriminant are invariant with respect to the choice of the normal field N of G I F. Both the different and the discriminant of a are unequal to 0 if, and only if, a is a primitive element of G I
F.
Henceforth we shall denote different and discriminant by b(a) and D(a). Because of (145.1) and (145.4), we have b(a) = (s1a - se(X)
... (sla - S.00-
Hence, and from (145.5), it follows, from Theorem 360, that
D(a) _ (- 1)(1)
i=1
s,b(a) _ (- 1)")
(sac - ska) . t. k=1
t#k
Consequently, by (145.1), we have
D(a) = fl
1si 0) of S there is a natural number n such that
not > .
(149.1)
Otherwise we speak of a non-Archimedean ordering. The above terms are derived from the axiom of Archimedes according to which.
for two intervals AB, CD, there is always a natural number n such that nAB > > CD. THEOREM 375. An Archimedean ordered ring is necessarily commutative.
and if it has a unity element then it has no non-identical order preserving automorphism.
ARCHIMEDEAN AND NON-ARCHIMEDEAN ORDERINGS
581
Let R denote an Archimedean-ordered ring. We consider two positive ) there is, according to (149.1), an elements a, /3 of R. To every m (E
n (E 4'') such that (n - 1)o c::!9 m(3 < na.
(149.2)
Hence
m(af - floc) = a m# - mil a < a na - (n - I )o:
of. = a2 .
Since this holds for all in (E -t'), it follows, again from (149.1), that
a/3-floc 0) of an ordered ring R with unity element has the
property that for all n = 1, 2.... either a > n or na < 1 holds, then we call a an infinitely large or infinitely small element, respectively. We show that the ordering of R is Archimedean if and only if, R has neither infinitely large nor infinitely small elements.
The assertion -only if" needs no proof. In order to prove "if" we con-
sider two positive elements a, /3 of R. If there are elements in, n (= 1, 2, ..
such that
a l ,
then a < mnf follows. This proves the assertion. The propositions a > n, na-1 < 1 are identical for an arbitrary element
a (> 0) in a field. Thus if a is infinitely large, then x-1 is infinitely small and conversely. Hence it follows that the ordering of a field is Archimedean if it has no infinitely small elements.
The extension of a non-Archimedean ordering can, of course, be only non-Archimedean. On the other hand, the following holds: THEOREM 376. In the case of an algebraic field G I F the extension of an
Archimedean ordering of the field F to an ordering of G is necessarily Archimedean.
ORDERED STRUCTURES
582
COROLLARY. An absolute-algebraic field can only be Archimedean-ordered.
in order to prove the theorem it suffices to show that to every positive element a of G there is a k (= 1, 2, ...) with a < k. Let n-t 1=0
be the minimal polynomial of a.. By hypothesis there is a k (= 1, 2....) such that
k> 1 -a, if a >_ k, then a
(1 =0,....n- 1).
1-a;, i.e.
a,> 1 -a
(i=0,...,it - 1),
from which it follows that n-t f(a) ;> an + (I - a)
57,
a' = 1.
i=0
Since this contradicts f(a) = 0, we have a < k. Consequently, we have proved the theorem and, on account of Theorem 374, the corollary also follows. EXAMPLE 1. Let V be the.3-vector space with basis w1, ..., w,. The ordering of V given in § 148, Example 3, is evidently non-Archimedean. We obtain an Archimedean
ordering of V if we take a real transcendental number 0 and define the positivity domain as the semimodule of those elements C1w1 + ... + Cn(un
(Cl, ..., c,, E .7)
of V, for which we have c1+c211
(This example relies, of course, on the usual ordering of the field of real numbers with which we shall only deal later on.) EXAMPLE 2. Let R = .fo[x]. The semiring consisting of the polynomials f(x)(r R)
with positive leading coefficients is a positivity domain of R. The corresponding ordering of R is non-Archimedean (although it is the extension of the Archimedean ordering of .? ). On the other hand, we obtain an Archimedean ordering of R if we take a real transcendental number 0 and define as positivity domain the semiring of those f(x) for which f($) > 0. EXAMPLE 3. The last two orderings of R = .$-0[x] obviously furnish, by virtue of Theorem 373, a non-Archimedean and an Archimedean ordering of the quotient field .7,,(x). In the first case an element a0x" + ... + any
b of T (x) is positive if ao b > 0. Moreover, all the order preserving automorphisms of .7-.(x) are then given by
x + cx + d, where c, d (e > 0) are elements of .?-,,. (For the second case cf. Theorem 375.)
583
ARCHIMEDEAN AND NON-ARCHIMEDEAN ORDERINGS
EXAMPLE 4. We construct a non-Archimedean-ordered skew field as follows: let
(r,(x)) = (..., r0(x), r1(x), ...) be an infinite sequence in both directions with terms from P-,(x) which, however, vanish to the left from a certain place on. in the set of these sequences we define addition and multiplication by (r,(x)) + (s,(x)) = (r;(x) + s,(x)), (ri(x)) (si(x))
rk(x) s,-k(x°k)), k
where we have to sum over k = 0, ± 1, ..., and a denotes a non-identical automorphism of , 0(x). It is easy to show that is a non-commutative skew field. Now order ."o(x) in a non-Archimedean way as in Example 3, and take an order preserving automorphism of .F'0(x) for ,s. Let an element of be positive if its first nonvanishing term is positive. A positivity domain is now defined in i. As for the rest, we can take the elements of it in terms of a new indeterminate y as a formal power series (cf. § 167)
The construction (without ordering) also holds if we take a field of prime characteristic p instead of .7-0, and leads in this case to a non-commutative skew field of the same characteristic p. EXERCISE 1. A homomorphism S - S' (fin -- o') of ordered structures is said to be order preserving if from e < a (o, a E S) o' < a' always follows. Show that the kernel of such a homomorphism, especially in the case of rings S, S' (0 0), consists only of 0 and infinitely small elements. (POLLAK.)
EXERCISE 2. Every ordered ring without infinitely large elements admits of an order preserving homomorphism, where the image of the ring is different from 0 and Archimedean-ordered. (POLLAK.)
§ 150. Absolute Value in Ordered Structures If S is a module, ring, skew field or field, then we obtain a very simple
classification of S by taking any two elements a, - a (a # 0) to form a class and taking 0 alone as a class. If, moreover, S is ordered, then every class contains exactly one non-negative element by which the classes are then represented. This happens in the following important definition: by the absolute value I a I of an element a of an ordered structure S we understand the non-negative member of the pair a, - a. The absolute value has the following properties:
10I=0,IaI>0(a0 0), ja.+
-a,
(150.1) (150.2)
(150.3)
(In modules only (150.1), (150.2) are to be considered.)
584
ORDERED STRUCTURES
Of these only (150.2) needs proof. Since
it follows that
x+ji.-a-rg_ilal-Iii
follows.
(150.4)
We apply (150.2) to a - f7 instead of x: i.e.
Similarly
Ia:_!9a-h'"+
Ia-i
flIlaI-
These may, because of (150.13), be summarized in
(lal - IfD
la! -
From this and (150.14), (150.4) follows. As a ccnsequence of (150.Q and (150.3) should be noted :
-'=x'".
(150.5)
NOTE. If S = .70 the absolute value I a I (a E .. T) is, because of Theorem 374, uniquely determined. Accordingly, the absolute value in an arbitrary S is
to be regarded as a generalization of the absolute value defined in S. An extensive further generalization in fields will be the subject of the next chapter. EXAMPLE. An ordered ring S is homomorphically mapped by at -- I a I, with respect to multiplication, onto the semigroup of non-negative elements of S.
CHAPTER X
FIELDS WITH VALUATION Soon after STEINITZ created the general theory of fields, KURSCHAK (1913) guided by the p-adic fields of HENSEL (1908) made a far-reaching discovery that the concept of absolute value in fields is capable of generalization. From this he developed the theory of valuation for fields, one of
the most important chapters of algebra with the widest possibilities of application. In this theory, among many others, the pure algebraic foundation of the notion of the field of real numbers was developed, where it came to light that the p-adic fields are analogous concepts. OsTxowsiu (1918, 1935)
has contributed largely to the further development of the theory of valuation. Concerning the treatment of the theory of numbers on a valuationtheoretical foundation cf. HASSE (1949). § 151. Valuations Let a field _9" and an ordered field F be given. The ordering relation in F is denoted by " 0) there is a natural number N such that
4p(ai - ak) < a
(1, k > N).
(152.1)
We say that a (E.9') is the cc-limit of a sequence [ai] if, for every s (E F, > 0),
there is a natural number N such that
(i > N).
,p(ai - a) < a
(152.2)
We denote this by 92-lim ai = a.
(152.3)
(If necessary, the more precise notation' lim instead of lim is used. If 92-lim a; = 0
,
(152.4)
then we call [ail a p-zero sequence. This means, in other words, that for every
s (E F, > 0) there is a natural number N such that
rp(ai)<s
(i > N).
(152.4)
Further (152.3) is equivalent to the fact that [ai - a] is a q,-zero sequence. For a fixed valuation, we omit the attribute "q2" in the notation now introduced (and also in those to be introduced later on) when there is no
588
FIELDS WITH VALUATION
possibility of ambiguity. If Yis an ordered field and unless otherwise indicat-
ed, then in place of q' we have to understand the absolute valuation co. This will always be related to the ordered field F (instead of -7). The above concepts are well known in analysis for the case when 'P- (= F) is the field of real numbers, or of complex numbers, and ' means the "absolute valuation" in those fields. We shall not take this into consideration, however, but we shall obtain these fields as special cases of the present investigations.
We can write every element of .7as a limit, for a = lim [a]. We shall show that the limit of a sequence, if it exists, is always uniquely determined. To prove this we suppose that the sequence [a;] has two limits, a and b. Then, not only (152.2) holds, but, with a suitable natural number N'. also
q(a,-b)N').
For i > N, N' we have
9,(b-a)=q#a;-a)-(a;-b)) 0) suitable r, s (> i) with gq(a,, - a,) > 2e. Since we have 2e < q(ar
- a) < q(af - a,) +
cp(ai
- a,)
589
CONVERGENT SEQUENCES AND LIMITS
the first or the second term of the right-hand side is at least equal to e; we put ni = r - i or ni = s - i, respectively. Since, accordingly, we always have 99(ai - ai+R,) z e, it follows that [ai - ai+,,,] is not a zero sequence. Consequently the theorem is true, from which the corollary follows immediately. THEOREM 380. To every convergent sequence [ai] from Jr there is a p (E F, > 0) such that 92(ai) < i
(i = 1, 2, ...);
(152.5)
moreover, if [ai] is not a zero sequence, then there is also a I' (E F, for which, with a suitable natural number N, we always have cp(a) > v
(i > N) ;
0)
(152.6)
further, in the special case." = F, q' = co, this N may be chosen so that
either ai > r (i > N) or ai < - v (i > N).
(152.7)
From (152.1), it follows for k = N + I that q2(a,) < e + p(aN+i)
(i > N).
Accordingly none of the T (al), 97(a9).... is greater than ,u = max ((9'(a>.). .... q.(aN), e + (p(aN+I)) ,
so that (152.5) is proved.
If the assertion in (152.6) were false, then to every pair r, N there would be a k (> N) such that 92(ak) < r .
According to (152.1). N may be so chosen that for all i (>N) we have Ip(ai - ak) < I'. Hence 9,(a,) < 92(a, - ak) + 97(ak) < 2v .
If we apply this with v = - , then we see that [ai] is a zero sequence. This contradiction proves the assertion in (152.6). For the special caseY= F, 99 = w, (152.6) implies that for every i (> N) either a, > r or -a, > I'. Consequently, if (152.7) is false, for every N there
is a pair i, k (> N) such that a, > v, -ak > r.
Since
I
ai - ak I > 2r
follows from this, [ai] is not convergent. This proves Theorem 380. THEOREM 381. If [ai] converges so does [q,(a)]. If [ai] has a limit, then [92(a)] has too, and then lim 92(ai) = cp(lim a).
(152.8)
590
FIELDS WITH VALUATION
Since, according to (151.4), we always have I 9v(u) - 90) I < 9w(u - v) , from the definitions (152.1) and (152.2) it follows that the theorem is true. We define the sum and the product of two sequences [ai], [bi] by
[ai] + [bi] = [ai + bi], [ai] [bi] = [aib,] In this way we obtain a commutative ring which is isomorphic with the complete direct sum of a countably infinite number of fields.7. From here on we consider the sequences [ai] as elements of this ring. It is obvious that here [0] and [1 ] are the zero element and the unity element, respectively. The additive inverse of [bi] is [-b,] so for subtraction we have
[ai] - [bi] = [a, - bi] The multiplicative inverse of [bi] is [bt 1] provided that all the bl, b.2, .. . are distinct from 0; so for division we have [a,] [bi]-1 = [a, b, 1] THEOREM 382. LetO denote each of the four fundamental compositions. If the
sequences [a,], [bi] converge, then [a,Obi] converges; if they have a lim-
it, then [a,Obi] also has a limit, and lim (a,(Dbi) = lim a,Olim b, ,
provided that for division b1, b2, . . . are different ,from 0 zero sequence.
(152.9)
and [bi] is not a
We shall prove the theorem first for addition and multiplication. Let [ai], [bi] be two arbitrary sequences. Because of the identities
(u+v)-(u'+v')=(u-u')+(v-v') and uv - u'v' = v(u - u') + u'(v - v')
we have
9q((ai + b) - (at + bk)) < 92(ai - ak) + 9p(bi - bk) ,
(152.10)
92(aibi - akbk) < 9p(br)9v(a, - ak) + V(ak)99(b, - bk).
(152.11)
If [ai], [bi] are convergent, then for every s (E F, > 0) there is an N (> 0) such that E
92(ai - ak), 49(bi - bk)
N) .
From this, and from (152.10), it follows that [ai + bi] is convergent. Further, by (152.5), there is a p (E F, > 0) with 99(bi), 99(ak) < p p
(i, k = 1, 2, ...)
CONVERGENT SEQUENCES AND LIMITS
591
and to every s (> 0) an N (>0) such that 4v(a, - ak) , 92(b, - bk) < 2µ
(i, k > N).
Then the right-hand side of (152.11) is less than e, thus [a, b,] is convergent.
We now suppose that lim ai = a, lim bi = b. The above reasoning may be repeated if ak is everywhere replaced by a and
bk by b. We obtain as a result lim (a, + b,) = a + b, lim a,b, = ab. This proves that part of the theorem which relates to addition and multiplication.
Since if [b,] converges then so also does [-b,], and if lim b, = b then lim (-b,) = -b, the assertions relating to subtraction are reduced to the case of addition. Finally, in order to prove the assertions referring to division we suppose that the conditions stipulated at the end of the theorem are satisfied. We shall show that if [b,] converges then so also does [b, 1] and that if Jim b, = b then lim b, 1= b - '. This reduces that part of the theorem which refers to division to the case of multiplication.
If [b,] converges, then according to (152.6) there are certain v and N such that always (i > N). p(b) > v For every e (E F, > 0) N may be chosen so large that 99(bk - b) < v2e
(i, k > N).
From these it follows that 4'(bi 1 - bk 1) = 99(b,)-19P(bk)-199(bk - b) < e ,
therefore [bi 1] converges.
If, in addition, lim b, = b (# 0), then we may similarly conclude that lim bj 1 = b-1. For this, we have only to write everywhere b instead of bk and to subject v to the further condition 92(b) > v. Consequently Theorem 382 is proved. THEOREM 383. The convergent sequences [a,] constitute a ring in which the set it of zero sequences constitutes a maximal ideal. This it is called the ideal of zero sequences. The first assertion and the ideal property of it are a consequence of Theo-
rems 380, 382. It remains to be proved that the ideal it is maximal. For this it is sufficient to show the following: if [a,], [b,] are convergent se_ 20 R.-A.
592
FIELDS WITH VALUATION
quences, but [ai] is not a zero sequence, then [bi] is an element of the ideal
(n, [ai]). This assertion is equivalent to the following one: There is a convergent sequence [xi] such that [ai] [xi] _ [b,] (mod n)
.
(152.12)
Now, because of (152.6), the sequence [ai] contains only a finite number of vanishing terms, consequently, after the addition of a suitable zero sequence, a sequence [ai ] is obtained from [ai] with only terms different from 0. Then [a'] = [ail (mod n) . Since furthermore, according to Theorem 382, [a; ] is convergent and is not
a zero sequence, it follows, again from Theorem 382, that the equation
[ail [xil = [bil determines a convergent sequence [xi]. Thus (152.12) holds and Theorem 383 is true. We understand by a perfect field a field with valuation in which every convergent sequence has a limit. Sometimes the valuation of a perfect field is called a perfect valuation.
It is evident that a field with the trivial valuation is always perfect. We shall become acquainted with other perfect fields later. In perfect fields we shall introduce the concepts of infinite series and infinite products as follows: let. rbe a perfect field with respect to the valuation 99. To every countably infinite sequence [ai] formed of the elements of.Twe can assign, temporarily but formally, i.e., without any more closely defined sense, an infinite sum (i.e., consisting of an infinite number of terms) 00
i=1
ai=a1+a2+...
(152.13)
and an infinite product (i.e., consisting of an infinite number of factors) 00
ai = a1a2...;
(152.14)
q=
these are in general called an infinite series and an infinite product, respectively. The finite sums and products
si=a1+... +ai, pi=a1...a,, formed from them, are called the partial sums of (152.13) and the partial products of (152.14). More precisely, si is called the i`s partial sum and pi the ith partial product. We now call (152.13) a WW-convergent infinite series,
CONVERGENT SEQUENCES AND LIMITS
593
if the sequence [s,] formed from the partial sums is 97-convergent, i.e., because of the perfectness ofSf it has a 99-limit; then we say that this infinite series "exists" and is equal to this limit:
at = a1 + a2 + ... = 92-lim sl = 92-lim (a, + ...a)
.
1=1
Otherwise we call the infinite series 97-divergent. For this, it is usual to say that the infinite series does not exist. Furthermore, we call (152.14) a 92-convergent infinite product, if [pi] has a 92-limit different from 0; for this we say that the infinite product "exists" and is equal to this limit: OD
fl of = a1 a2 ... _ 92-lim pt = 92-lim a1 ... a; .
t=1
Otherwise we say that the infinite product is 92-divergent or "does not exist". From the definition it immediately follows that every existing limit may
be turned by the formula 92-I'M c, = c1 + (c2 - c1) + (c3 - c2) + ...
(152.15)
into an infinite series. Similarly we have the formula 92-lim c, = c1(c2cl 1)(c3e2 1) ...,
(152.16)
provided that the left-hand side does not vanish and no ci is equal to 0. Obviously, for the convergence of a1 + a2 + ... and a1a2..., the conditions lim a, = 0 and lim a, = 1, respectively, are necessary but, in general, not sufficient conditions. If necessary (152.13) and (152.14) are written more precisely as and
92-s al
99-11 a; , i=1
i=1
respectively.
Following the corollary of Theorem 379 we define two valuations of a field with the same ideals of zero sequences as equivalent.
NOTE. Let F be an ordered extension field of F whose ordering is an extension of the ordering of F. If 99 is a valuation of a field Y with the value
field F, then by the statement
(a) = 99 (a) (a E Y) a valuation 0 of. F is
defined with the value field F. Although these valuations at first sight appear
to have the same status, it may very easily happen that they are not equivalent. For example, if [al] is a 92-zero sequence whose terms are different from 0, and if F has a positive element a such that there is no element
sequence. (Cf. the e with 0 < e < E, then [a,] is evidently not a Example at the end of § 153.) However, we shall deal with the equivalence of valuations later in § 157.
594
FIELDS WITH VALUATION
§ 153. Perfect Hull It is of great importance that every field .7 with valuation - as will be seen - may be extended to a perfect field Y in which all the sequences, convergent in3, have a limit. Moreover we can subject .Y to certain further
conditions by which we attain (in a certain sense) uniqueness, but Y will depend essentially on the valuation of.''. This last fact affords many possibilities for the construction of overfields of a given field. By way of introduction we give some definitions. We say that a valuation q/ of an extension field 3' of the field 3, with valuation q', is an extension of the valuation 9), if the value field of 3' contains that of Jr in an order preserving way and, for the elements a of 3, q' (a) = 99'(a). If a fixed extension of 99 is considered, we often likewise denote it by 99.
A perfect extension field .3' of a field with valuation _9' is called a perfect hull of 3, if the valuation, of ' is an extension of the valuation 9) of .7 and consists of the qi-limits of the 92-convergent sequences formed from
the elements of 3. By a (bicontinuous or) topological isomorphism of a field .7 with a field c.P,
both with valuation, we understand an isomorphism of .7 with c.P which transfers the zero sequences, formed from .5r, just into the zero sequences, formed from cP. If there is such an isomorphism we call the fields Jr, cO (continuous or) topologically isomorphic. If between two fields 3, cP an isomorphism a --> a° holds and 99 is a valuation of .i , then a valuation 99° (with unaltered value field) is evidently defined in c.P by 99°(a°) = 97(a), such that a is now a topological isomorphism between _" and 4t.
Two extension fields c0, 4g' of a field .7 (all with valuation are called topologically equivalent (over ..3"), if their valuations are extensions of those of 3 and there is a topological isomorphism c.P 1.7 -- dt' 13. From now on let a field.7 with the valuation 99 and the value field F be given. (The notations are governed by the conventions made at the beginning of § 151, 152.) We shall construct a perfect hull of -7 and show afterwards that it is, in the sense of topological equivalence, uniquely defined.
To do this, we denote the ring of 97-convergent sequences [a;] by R and the ideal of zero sequences of ll by n. Since according to Theorem 383, n is maximal,
3 = T/n
(153.1)
is, by Theorem 130, a field. We denote its elements by [a1 ]' = residue class [a1] (mod n).
(153.2)
The special elements [a]' evidently constitute a subfield .7" of .7 such that
.7
.7' (a -# [a]') .
(153.3)
PERFECT HULL
595
We suppose that in .}' the corresponding embedding of Y instead of .g' is carried out, while we retain this notation for Y after the embedding, so that Y is now an overfield of Y. We define a certain valuation - (with a suitable value field) in Y, which will be an extension of q.'
We first deal with the case J r = F, q' = a). For this case we write F instead of .}'. In conformity with (153.2) we can take an arbitrary ele-
ment of F in the form [a;]' = residue class [a;] (mod n) , where it now denotes the ideal of zero sequences of F and [a;] is an w-convergent sequence. We call this element positive and write it as [aj]' > 0,
if there is a S (E F, > 0) and an N (E J) such that (i > N). a; > b (153.4) We have to show that the positive elements of F are uniquely determined by this. For this purpose we consider an element [fi]' of F equal to [a;]'. It has to be proved that the condition corresponding to (153.4) is also satisfied for fl;. The assumption implies the congruence [a;] _ [fli] (mod n). Since accordingly [(3; - aj] is a zero sequence, we can choose N above to be so great that besides (153.4) we also have
< Hence
a;
- 2s
,
(i>N).
then, because of (153.4),
(i>N), by which we have shown that in F the positive elements are uniquely defined.
Evidently these constitute a subsemiring of F which we denote by P. We show that P is even a positivity domain of F. For this purpose we consider an element [a;]' other than 0, of F, for which the sequence [a;] is therefore convergent, but not a zero sequence. It has to be shown that of the elements [a;]', - [a;]' exactly one belongs to P. According to Theorem 380, (152.7), there are a v (E F, > 0) and N (E 4") such that we have
either aj > v (i> N) or aj < - v (i > N). For the latter we may write -a; > v, so that condition (153.4) is satisfied,
for exactly one of the a,, - aj, i.e. of the elements [a;]', - [a;]' (= [-a,]') exactly one is positive. Accordingly P is in fact a positivity domain of F.
FIELDS WITH VALUATION
596
We put
P=Pf1F.
(153.5)
Since F is embedded in F, P consists of those a (= [a]') for which a > 0 (in F). This implies that P is the positivity domain of F. Since, because of
(153.5), P e P, the ordering of F according to the positivity domain P is an extension of the ordering of F. According to this the corresponding absolute valuation of F, which we denote by 11 11 or by w, is likewise an
extension of the absolute valuation of F. This will be taken as the required valuation of F, which we shall retain in what follows. Before passing to the general case we wish to show, as a preparation for it, that every co-convergent sequence [at] in F has an 6-limit and for it w-lim at = loci], .
(153.6)
We consider an arbitrary positive element [et]' of F. According to the definition in (153.4) there are then a 6 (E F, > 0) and N (E such that we always have et > 6 (i> e"). On account of the assumption, we may choose N so that
ai - akI
J )
also holds. Accordingly 6
6
et+ai - ak> 2 , et- at +ak> T. Again by (153.4), we have from this [et + at]' > ak, [ei - at]' > - Mk i.e.,
Vi]' + [ail' > ak,
[et]' - [ai]' > - ak ,
so that finally 11 Mk - [at]' 11 < [et]' for all k (> N). This implies (153.6).
We call attention to the consequence of (153.6) that every w-zero sequence is an c-o-zero sequence, which we shall often apply without reference.
We now consider the general case. We wish to define in .3r a valuation (P which is an extension of qq. (The preceding special case is not excluded. The
valuation to be given for this special case - as will be seen - turns out to be 161.)
Let [at]' be an arbitrary element of Y. We shall show that by 0([ai]') = w-lim q'(ai) (=[m(a,)]')
(153.7)
PERFECT HULL
597
a valuation 0 of . 1 ' (with the value field F) is defined with the desired properties.
For this, we must first of all show that the right-hand side of (153.7) exists and is uniquely determined by [ai]'. Since [ai] is qr-convergent, [ip(ai)] is, according to Theorem 381, an ca-convergent sequence. Hence, and from
what was proved above, the existence of the right-hand side of (153.7) follows.
It still remains to be shown that the equation cw-lim 9' (a) = w-lim 9' (b)
(153.8)
follows from [ai]' = [bi]' . By hypothesis [ai - bi] is a 99-zero sequence, consequently [q9 (ai - bi)] is, according to Theorem 381, an co-zero sequence. Because 99 (a) - q' (b,) I < 9, (ai - bi), 1
[9, (a) - 99 (b;)] is then an co-zero sequence. Hence, and from Theorem 382, (152.9) follows the assertion (153.8). What we have proved so far may
also be expressed by saying that a mapping 99 of Yinto F is defined by (153.7).
Further we have to show that ? is a valuation of Y. If [a,]' = 0, i.e., [ai] is a 9,-zero sequence, then [q:(ai)] is, according to Theorem 381, an co-zero sequence, thus, according to (153.7), ([a,]') = 0. If, on the other hand [a,]' 0 0, i.e., the p-convergent sequence [ai] is not a 9,-zero sequence, then (153.7) and (152.6) show that 99([a,]') > 0. Accordingly, 0 has the property given in (151.1). If 0 had not the property given in (151.2),
then there would be two elements [a,]', [bi]' of .i ' such that
([a,]' + [bi]')
- ([a,]') - 9'([b,]') > 0.
Hence, by (153.7), the existence of an s (E F, > 0) and N (E d) such that
,(ai + b) - 9,(a) - 9p(bi) ? s
(i > N)
would have followed. Since this is absurd, according to (151.2), it follows from this contradiction that Q' has the property given in (151.2). From (153.7) and Theorem 382, it follows that, together with 99, also has the property
given in (151.3). Accordingly, qi is in fact a valuation of Y. It is also an extension of q', for since Yis embedded in Y, it follows from (153.7), that 9'(a) = m([a]) = co-lim q,(a) = qu(a).
FIELDS WITH VALUATION
598
It still remains to be shown that in the special case Y = F, q = w we have 0 = w. For this case the definition (153.7) becomes: ik[ar]') = w-lim w(a;) . The right-hand side is, according to (153.6), equal to [w(a1)]'. This, according to the definition of tv, is equal to Hence the assertion has been proved. For the following we fix for the valuation 9 defined by (153.7), which in the case Y = F, 99 = w becomes the absolute valuation w. We now prove the following theorem: THEOREM 384. Every field J r with valuation has, to within topologically equivalent extensions, one, and only one, perfect hull, and that is the field Y defined above by (153.1) and (153.7).
First of all we prove that 3' is a perfect hull of Jr. (We continue to use the previous notation.) We begin with the proof that for every 99-convergent sequence [at], (153.9) -lim a, = [a,]'. (In the special case." = F, 99 = w, (153.9) becomes (153.6), which has already
been proved, but the following proof is also valid for it.) Take an element
is (> 0) of.}'. From (153.4) there exists a 6 (E F, > 0) such that
b<E. For every term a of
it follows from (153.7) that
([a,]' -
w-lim 99(a; - a.).
But there is an N (E,4) for which we always have
4'(ai-
a 2
N).
Because of the preceding work we may choose N so that also
From these we get
N),
by which (153.9) is proved. This means that Yconsists of the -limits of the q-convergent sequences [a;].
In order to complete the proof that.3is a perfect hull ofYwe still have to verify only that
' is perfect. For this purpose we consider a qi-convergent
PERFECT HULL
599
sequence [ai] (formed of elements ai of 7), in order to prove that it has a -limit, which will verify the assertion. (We deal with a generalization of one part of the former statement involved in (153.9).) Now, first of all, we consider the case where we may form an (5-zero sequence [St] of the elements of F consisting only of positive terms. (There
is no need that this should always happen;
cf. HAUSCHILD-POLLAK
(1965).) Since according to (153.9) every element of.}' is the -limit of a 99-convergent sequence, there is for every i an ai (E 5) such that
,&i - ai) < bi
(i = 1, 2, ...) .
Further there is for every s (E F, > 0) an N(E- ) such that E
0(a-i - ak) , bi
N).
Since it follows from this that
9'(ai - ak) = 0(ai - ak) = m(Cai - ak) - (a, - ai) + (ak - ak)) < E (i, k > N), [ai] is a 99-convergent sequence. We write
a = t-lim ai . Then, for every s (> 0) from F there is an.N' such that
T(ai-a)
We can choose N' so great that also 1
a; N1), n2 (> NO ...
(153.12)
92-lim a,,, = 0 .
(153.13)
with the property r
First, we suppose that ri] is a -zero sequence and N1, N2.... are any natural numbers. To every positive element a of F there is then an N (E.4*) such that always
j(ai)N).
FIELDS WITH VALUATION
602
Also we may chose n1,, n2, ... satisfying (153.12) and such that IG (a,,,, - ai) < Hence 99 (a,n)
(t = 1, 2, ...) .
2
(ai,,,) < e, and we have proved (153.13).
Conversely, we suppose that for arbitrary N1, N2, ... the n1, n2, ... with the property (153.12) may be chosen in such a way that (153.13) is valid. Let is be an arbitrary positive element from F. According to Theorem
385 there is a positive element E of F such that 2e < E. We choose the N1, N2, ... in such a way that we always have &en, - ai) < e (ni > Ni, i 1, 2, ...) . Because of the supposition the nI, n.2,
. .
. may then be chosen in such a way
that for a suitable N (E J") we always have faint) _
faint) < e
(i > N).
Hence 45 (a) < 2 e < 9 (i > N), i.e., 15-lim a, = 0. Consequently we have shown that the assertion concerning the f)-zero sequences formulated in (153.12) and (153.13), formed from ", is true.
For the same reason a similar statement holds also for the
TP-zero
sequences formed from 4. From both these, and (153.10), it immediately follows that by (153.11) every q-zero sequence is in fact mapped into a w-zero sequence. Thus Theorem 386 is proved. EXAMPLE. In Theorem 385 it is necessary to suppose that F" is not perfect. Suppose
that the field.. has the trivial valuation, so that it is perfect and consequently equal to its perfect hull S r, and let F = i and F = .moo (x) be the value fields of .7" and X. respectively, where F, as in § 149, Example 3, shall be ordered in a non-Archimedean way. (See Theorem 374.) Now x-1 is a positive element of F, and there is no element
eofFsuch that 0<e<x-'.
§ 154. The Field of Real Numbers
As one of the most important applications of valuation theory first of all arises the field of real numbers which we define as the perfect hull of the field 5 with the absolute valuation and denote by o). The elements and subfields of 9( 'o) are called real numbers and real number fields, respectively. By Theorems 384, 385, o; is an Archimedean-ordered field provided with the absolute valuation. We always denote the ordering, the absolute value and the limit by 0) of
0?,
there is a further element b (> 0)
from 30) such that the condition
I1(c+h)-f(c)I <e
(IhI l,c> Ia,I +... + Then f (c) > 0, f (- c) < 0. This, by Theorem 388, proves Theorem 389. Tm?OREM 390. Every equation
x"-a=0
(n>-1; aE.5o),>0)
(154.1)
has exactly one positive root in -7(o. Since the left-hand side of (154.1) is negative for x = 0 and positive for x = 1 + a, so by Theorem 388, (154.1) has at least one positive root. There
cannot be two positive roots since b" < c" follows from 0 < b < c. The theorem is now proved.
THE FIELD OF REAL NUMBERS
605
Because of this theorem we introduce for the positive root of the equation (154.1) the notation
a(154.2)
We define more generally the power a' for a positive real basis a and for a rational exponent ,bmy putting amn-1
= (a n-I
(a E _7(o), > 0; m E 7, n E -44'
.
(154.3)
It is now necessary to show that the right-hand side depends only on a and mn-1. In other words, we have to show that we always have (am)"-' = (ak)'-' (m, k E .7; n, I E -t'; mn-1 = kl-1). The left-hand side and the right-hand side are positive roots of the equations
x" = am and x = ak , respectively. Because of Theorem 390, we may replace these equations, without altering their positive roots, by xn' = ank.
xn1 = am' ,
Now, because ml = nk, these equations are identical, whereby the uniqueness of definition (154.3) is proved. We shall prove that
(ab)' = arbr
,
arts = eras,
(a')s = ars
(a, b E (o), > 0; r, s E Yo) .
(154.4)
First of all, it follows from Theorem 390 that (154.43) is true ifs is a natural
number. Let us therefore take two natural numbers m, n such that rm, sn are integers, then ((ab)r)m = (ab)rm = a'mb'm
= (ar)m
(b')m
= (arb')m,
(ar+s)mn = a(r+s)mn = armn asmn = (ar)mn (as)mn = (ar a3)mn ,
((a)s)mn = (ar)smn = arsmn = (ars)mn .
From these equations, and again from Theorem \\390, it follows that (154.4) is true. Further, we shall prove the two properties of monotonicity
0N').
Then
a`k - ac' 15 a'(d`k`h' - 1) < a'(a"-I - 1) < g. Since accordingly [a`"] is convergent, the right-hand side of (154.7) exists.
In order to prove the uniqueness of the definition (154.7), we take a further sequence [d"] from .ro such that c = lim d,,. We denote the "mixed" sequence c1, d1, c2, d,., ... by [e"]. According to the preceding work, all three limits lim a'R , lim a" d , lIm a`"
THE FIELD OF REAL NUMBERS
607
exist. Since the first two limits separately must be equal to the third, they are equal to each other. Accordingly, a` is uniquely defined by (154.7). We leave to the reader the simple proof that (154.4), (154.5), (154.6) also remain valid for real numbers r, s and then a` is a continuous function of both variables a, c. The definition of an ordered module will henceforth also be related to the multiplicative notation. The meaning of an ordered Abelian group is now obvious. We have to interprete correspondingly order preserving automorphisms and isomorphisms, in connection with Abelian groups, and we may speak, in the corresponding sense, of the order preserving isomorphisms between two structures, one of which is an Abelian group, the other a module. We denote the positivity domain of .7jo) by P. This is the subsemifield of .7"(o) consisting of the positive elements. Correspondingly, the group of positive real numbers is denoted by P". This is ordered and has for its positivity domain the semigroup of those real numbers which are greater than 1. THEOREM 391. All the order preserving automorphisms of the group of positive real numbers are given by
a-->a`,
(154.9)
where c can be any positive real number. (This theorem may be regarded as a definition of the power a` for a, c > 0. Since a° = 1 , a = (a`)-1, we can now arrive at the complete definition.) For the proof of the theorem we denote the group of positive real numbers, as above, by P'f(a)
(154.10)
f(ab) = f(a)f(b) ,
(154.11)
a < b = f(a) < f(b) .
(154.12)
of P". Then for a, b E P"
We first show that
f(d) = (f(a))`
(a E P", I E J) .
(154.13)
608
FIELDS WITH VALUATION
For t = 0 it is trivial. Further, according to (154.11), f(1) = 1, f(a-t) = = (f(at))-1, while (f(a))-t = ((f(a))t)-1, so that it is sufficient to prove (154.13) for t > 0. We put t = kl-' for two natural numbers k, 1. Because of (154.11) we then have ((d ))h' = f(a) = f(ak) = (J (a))k = ((f(a))t )f .
Hence, and from Theorem 390, (154.13) follows. We shall also show that every equation
a"=b
(a,bE. (0,; a>0,b>0; a
1)
(154.14)
is solvable with a real number x. For, it is obvious that there are integers k, l such that ak < b < ar. Hence, and from the continuity of ax, according to BoLZANO's theorem (Theorem 388) the assertion follows. We now take a fixed element b (# 1) and an arbitrary element a of P" . According to what has been proved for (154.14),
f(b) = b` , a = bd
(154.15
for two suitable real numbers c, d. We take two sequences [r"], [s"] from .moo such that
r1, r2, ... < d < sI, s2, ... ,
lim r" = lim s" = d.
(154.16)
(154.17)
From (154.152), (154.16) we get
b'" 0 and (155.1), holds), and for the degree n* of Fk(x) the factor decomposition
(155.2) holds, it follows from the induction hypothesis that every Fk (x) has at least one zero in F(i). Now the number of pairs r, s occurring in (155.3) is equal to n*, whereas the number of polynomials Fk(x) equals n* + 1. Consequently there is a pair r, s (1 < r < s - - i. Since here a + bi becomes a - bi (a, b E Y(o)), we call these numbers conjugate complex numbers (more precisely "conjugate complex numbers over .Fto,"). Here two conjugate complex numbers are denoted by a, &, and we say that a is the (complex) conjugate of a.
The field Y(O)(i) is not orderable since i2 = - 1, so that we can not define any absolute value in it in the usual sense of the word. But we call the non-negative real number I
a I = (acac)1 = (a2 + b2)1 ,
(155.7)
by long-established custom but very incorrectly, however, the absolute value of a for a complex number a = a + bi (a, b E to)) We shall show that this defines a valuation which we (again incorrectly) call the absolute valuation of Jr(o)(i). [Since we are dealing with an extension of the absolute valuation of Y(o), the absolute value in _7 (o) and Y(o)(i) can be similarly denoted.]
It is evident that I a = 0 only for oc = 0. For two arbitrary complex numbers a, i we have aP6i4 = oca(3B, whence, because of (155.7), the homo-
THE FIELD OF COMPLEX NUMBERS
613
morphy property I ap I = I a I I P I follows. It still remains to be shown that
I«+flI 1. Such a c always exists,
since q' is not the trivial valuation and 9(c)9(c-1) = I for c 0. Because of (157.2), +p(c) > 1. We consider an arbitrary element a (0 0) of.9rand put 93(a) = (q'(c))" ,
V(a) _ (+Y (c))"'
for two suitable real numbers a, a'. For these we show that a = a'. We take two integers in, n (n 0 0) with
Then (92(c))mn -' < (92(c))"
= 92(a) ,
and so 99(c'") w(a) of the group F* into the module . '(,,,) which also satisfies condition (158.7).
If necessary we suppose that the definition of w (a) is extended to include
the case at = 0 by putting w(0) = oo, which means that w(0) is "greater" than all the other w(a). Here (158.7) remains true. (158.6) remains also true if we put oo + w(a) = Co. We emphasize once more that non-Archimedean valuations and exponent valuations differ from each other only immaterially, so that the previous definitions and statements, so far as they concern non-Archimedean valuations, may also be applied directly to exponent valuations. Similarly, future statements about one of these valuations will be taken as valid for the others. We say that the (non-Archimedean) valuation 99 and the expo-
nent valuation w correspond to each other, if (158.4) [or (158.11)] holds for a fixed g (> 1). For instance, a sequence [a;] from F is a w-zero sequence if, and only if, to every real number C there is an N (E -/") such that w(a;) > C
(i > N) .
EXPONENT VALUATIONS
621
On account of Theorem 398, the exponent valuations equivalent to an exponent valuation w(a) are given by ctii'(a) ,
where c can be any positive real number. THEOREM 400. Instead of (158.7) for w(a) # w(fl) we even have
w(a + p) = min (w(a) , w(p))
(158.12)
.
Let us assume, e.g., that w(a) > w(8). If the assertion is false, then, because
of (158.7), w(a + fl) > w(8). On the other hand, from (158.7) and (158.9), w(fl) =
a + a + p) > min (w(oc), w(a + 6))
.
From both results w(fl) Z w(a). This contradiction proves the theorem. Exponent valuations are, in several respects, simpler than Archimedean valuations, as we shall see later. First of all, in addition to Theorem 379, the following also holds : THEOREM 401. A sequence [a;] from afield F with the exponent valuation
w is convergent, if [a; - ai+i] is a zero sequence. For this, it is necessary that the terms of the sequence [w(a;)], except for finitely many, are equal, unless [a;] is a zero sequence.
If [a; - a,+,] is a zero sequence, then, for every real number C there is a natural number N such that
w(a;-(zi+I)>C
(i>N).
Because 1-1
ak - al = X (ai - a! +I) !=
(0 N).
The first assertion of the theorem is now proved. In order to prove the second assertion, we suppose [a;] to be convergent and not a zero sequence. To every real number C there exists a natural number N such that we always have w(a; - (xk) > C (1, k > N). (158.13) If the assertion is false, then for every i (> N) there is also a k (> N) with w (a;) # w (ak) .
FIELDS WITH VALUATION
622
For such a pair a;, ak, because of (158.9) and Theorem 400, (158.13) becomes min (w(00, w((Xk)) > C C.
From this we obtain w(a;) > C so that [a;] is a zero sequence nevertheless. This contradiction proves Theorem 401. THEOREM 402. If [a;] is a convergent sequence, but not a zero sequence, from a field F, which is perfect with respect to an exponent valuation w, then
w(lim a) = w(aN) = w'(aN+1) =
...
(158.14)
for a suitable N.
In order to prove this, we put a = lim a,. By Theorem 400 we have: w(M) 0 w(a)
w(a, - a) = min (w(at), w(a)) .
But since [a; - a] is a zero sequence, Theorem 402 follows. We now consider an arbitrary field F with an exponent valuation w, in order to define the following important notions. By the value module of F we understand the submodule of Y o) consisting of all the values w(a). When it is denoted by %, then according to (158.6) we have the homomorphism
F* -
(a - w(a))
.
(158.15)
Further we take into consideration those elements a of F for which w((X) z 0.
(158.16)
[Thus, these elements are 0 and the elements of F which are mapped onto the non-negative elements of fin the homomorphism (158.15)]. We shall show that these elements a constitute a subring of F, which we denote by R = R,4, and call the valuation ring of F (with respect to the exponent valuation w). We notice at the same time that, because w(l) = 0, R contains the unity element of F. In order to verify the ring property of R, let us take two elements a, fi from R. Because w(a), w(f) >- 0, it follows from (158.6), (158.7), (158.9) that w(a - A), w(afl) >_ 0 ,
and then a - fl, afi E R. Accordingly, R is, in fact, a ring. We shall show that the elements a with w(a) = 0 are all the units of R.
(158.17)
EXPONENT VALUATIONS
623
For, the units of R are those elements a of F for which both w(a) Z 0 and w(a-') >_ 0 are satisfied. Because of (158.10), the assertion follows. We shall show that the elements a of F with the property w (a) > 0
(158.18)
i.e., the elements of R other than the units constitute a maximal ideal of R, which we denote by p = pw and call the valuation ideal with respect to the exponent valuation w. We shall also show that p is a prime ideal. The ideal property of p follows at once from (158.6) and (158.7). The maximal property of p follows from the fact that every ideal of R containing
a unit must be equal to R. Since, by (158.6), if w(a/4) > 0, then either w(a) > 0 or w(fl) > 0, so p is a prime ideal. (This also follows from Theorem 186 as well as from the above.) is a field, From Theorem 130 it follows that the factor ring R/.p = which we denote by 2a = tSW and call the valuation factor field with respect to the exponent valuation w. THEOREM 403. Two exponent valuations w, w' of afield F are equivalent if, and only if, the valuation rings R, R, are equal. It is clear that Rw = F if, and only if, w is the trivial valuation. Hence the theorem follows for the case where one of w, w' is the trivial valuation. Therefore we exclude this case in future. If now w, w' are equivalent, then according to Theorem 398, w'(a) = cw(a)
(a E F)
(158.19)
for a positive real constant c. Since the propositions w(a) z 0, w'(a) > 0 are then equivalent, R,, = R,,.. Conversely, let us assume the latter. From this it follows that ,pW = ipW therefore
w(a) > 0 . w'(a) > 0. We consider two elements at, # such that w(a), w(f) > 0, whence w'((x), w'(fl) > 0. For arbitrary integers m, n, according to (158.6), w(a'"(3") = mw(a) + + nw(fl). The same holds for w' instead of w, thus
mw(a) + nw(f) > 0 a mw'(a) + nw'(f) > 0. Hence it evidently follows that w(a)(w(fl))-' = w'(x)(w'(f))-'. This implies the existence of a positive real number c which has the property that (158.19) holds for all a such that w(a) > 0. Because of (158.10), the same is true for
w(a) < 0. Since, finally, the units a of Rw = Rw are characterized both by w(a) = 0 and by w'(a) = 0, (158.19) is true generally. Thus Theorem 403 is proved. 21 IL-A.
FIELDS WITH VALUATION
624
According to this theorem the determination of the possible (non-equivalent) exponent valuations of a field F is reduced to the examination of all its valuation rings. THEOREM 404. All the valuation rings of a field F are F itself and those maximal subrings of F with unity elements which are not fields. First of all we suppose that R (A F) is a valuation ring of F. Let w denote
the exponent valuation of F belonging to it. Because w(1) = 0, we have 1 E R. From the definition in (158.16) it follows that for every element a of F other than 0 at least one of (x, ac-' belongs to R. Since then F is the quotient field of R, R itself cannot be a field. We still have to show that for every element a of F - R the ring {R, a) is equal to F. By hypothesis we have w(a) < 0. Consequently for every element # of F -nw(a). Hence there is a natural number n such that w(/3a - ") ? 0, fa " E R, # E { R, a},
therefore {R, a} = F. We now consider, conversely, a maximal subring R of F containing the unity element which is not a field. We have to prove that R is a valuation ring of F. First of all we show that, for every element a of F other than 0, one of a, a-1 lies in R. With this aim we suppose that a is not an element of R, so that it follows that the ring {R, a) is equal to F, and hence a
I =f(a)
for a polynomial f(x) over R. Consequently there is a principal polynomial F(x) over R with
F(a-I) = 0.
(158.20)
Let n denote the degree of F(x). Since, by hypothesis, F is the quotient field of R, there is an element P (A 0) of R with e,
ea_I....
Pa-("-1) E R
.
(158.21)
Here we may postulate that e is not a unit in R. Hence it follows, because
PP-' = 1, that e-' does not lie in R. If e-' lay in the ring {R, a-1}, this would imply that 9 e- = g((X-' )
for a polynomial g(x) over R, whose degree may be taken as n - 1 at most, because of (158.20). But, because of (158.21), it follows from this that
P-I = PP 2 = Pg(a-')
EXPONENT VALUATIONS
625
belongs to R, which is impossible. Since, according to this, lie in the ring {R, a-1} then
e_2
does no
R c {R,a-1} c F. Now by hypothesis, " = " must hold on the left-hand side, so that it follows that a-1 E R, as was asserted. We now denote by U the group of units of R and by that module which arises from the factor group F*/U after passing over to the additive notation. Then, f consists of the cosets ocU (a E F, 0); further the rule aU -I- 19U = ajU
(158.22)
holds inffor the addition of elements. a relation < by We define in ocU < flu cs a-'# E R - U .
(158.23)
This is well defined, since if ocU = a1U, 19U = fl1U then the propositions " fl E R - U, al-'til E R - U are equivalent. Since, for a, fl, y (E F, 0) the rule
a
f-'y ER-U=a-')'ER-U
holds, this relation < is transitive. For any two elements ocU, flu of4', of the three propositions
aU = flu, aU < flu, flu < aU exactly one holds, for if a laFof does not belong to U, then according to what has been stated exactly one «_ Ill,
,q_la ( =
(a-1N)-1)
belongs to R - U. Hence, we have defined an ordering relation < in 44f by (158.23).
Addition is monotone since, if
aU U), where we have to take into consideration that U is the zero
element of.
We show that the ordering of
is Archimedean. For this purpose
let us consider two elements aU, 9U (> U) of Lam. Since a'1 does not belong to R, the ring {R, a-1} is equal to F, and so 14-1 E {R, a-1}. Accordingly
14-1 = h (a-') for a polynomial h(x) over R. Therefore it follows from a E R that fl- Ian E R for a suitable natural number n. This implies that 14U < naU, so that f is in fact Archimedean-ordered. For any two elements a, fl of F with a, fl, a + j rA 0 the relation (a + fl)U >_ min (oU, 14U)
(158.24)
also holds. Let aU min (w(a), w(8))
(a, fi E F; a, /3, at + f 96 0).
DISCRETE VALUATIONS
627
Again of is called the value module, which occurs instead of the value field. If f is Archimedean-ordered, then we return to the exponent valuations Accordingly, general valuations essentially include all the exponent valuations, but not the Archimedean valuations. Cf. also PICKERT (1951) and FucfLS (1951).
ExAMPLE. In a field which is perfect with respect to an exponent valuation, an infinite series a, + a$ + . . . is convergent, by Theorem 401, if, and only if, lim a, = 0.
§ 159. Discrete Valuations A non-trivial exponent valuation w of a field F with cyclic value module
is called a discrete valuation. So we may now take the value module in the form {c}, where c is a positive real number. Because of the order preserving isomorphism .Y* ti { c} (n -+ nc) it is possible in the case of a discrete
valuation, by passing over to an equivalent valuation, to take,7+ as a value module. In this case we call the discrete valuation normed. THEOREM 406. A non-trivial exponent valuation is discrete if, and only if,
among the positive elements of the value module there is a minimal one. We have to prove only the assertion "if". We consider a non-trivial expo-
nent valuation w such that among all the positive values w(a) there is a minimal value w(b). To every w(a) with a # 0 we may assign an integer n such that nw(b) 5 w(a) < (n + 1)w(b) . Then
0 5 w(a) - nw(b) = w(ab-") < w(b). Now w(a) = nw(b) follows from this because of the assumption. Hence the theorem is proved. THEOREM 407. The valuation ring belonging to a discrete valuation is a
principal ideal ring with (prime decomposition and to within associates) only one prime element which is a generator of the valuation ideal. Conversely, if a principal ideal other than 0 belongs to an exponent valuation as valuation ideal, then the valuation is discrete. In order to prove this, let us consider an exponent valuation w of a field
F with the valuation ring R and the valuation ideal p. First of all we suppose w to be discrete. Let a be a proper ideal of R. Since the elements of a have non-negative values, a contains an element
a (# 0) such that for all the elements P (# 0) of a, w(f) > w(a) > 0. Hence w(fla-1) Z 0, i.e., floc-' E R. This means that P lies in the principal ideal (a), i.e., a = (a). Thus R is a principal ideal ring. In the following, for the sake of convenience, we suppose that w is normed. Let n denote an element of p such that w(n) = 1. For every element a (00) of
R, w(a) = k is a non-negative integer. From w(aaa-k) = 0 it follows that
FIELDS WITH VALUATION
628
an-k = e is a unit of R. Because a = nnk, R has the single prime element n. Asp consists of 0 and the a such that w(a) = k > 0, sop = (n). Conversely, let us suppose that the valuation ideal p is a principal ideal (n) other than 0. For every element a of p, an-1 E R, thus w(an-1) Z 0, w(a) Z w(n). According to this w(n) is the least positive element of the value module. This implies, by Theorem 406, the validity of Theorem 407.
THEOREM 408. Let F be afield, perfect with respect to the nonmed discrete
valuation w, with the valuation ring R and the valuation ideal p. Take fixed elements ..., n_1i no, nI,... from F with
(i = 0, ± 1, ...)
w(n;) = i
(159.1)
and a system of representatives 91 of R mod p, containing the element 0. Then the infinite series a = amnm + am+lnm+1 + ...
(am,
am+1,
.. E 91; am }o 0)
(159.2)
are all the different elements (# 0) of F, where m can be any integer, and w(a) = m .
(159.3)
(If, on account of Theorem 407, we choose an element n such that p = (n), we may then write n; = n'.) We already know (cf. the Example at the end of the preceding paragraph) that the infinite series (159.2) is convergent. In order to prove (159.3) we denote the i`h partial sum of the right-hand side of (159.2) by a;. Because of (159.1) and Theorem 400, w(i) = w(amnm) = m
(i = 1, 2, ...) .
Since, on the other hand, a = lim at,
lim (aj - a) = 0. Hence, and from Theorem 400, it follows that w(a) m is impossible and so (159.3) is true. Then we show that every element a (# 0) of F may be written in the form (159.2). For if we put m = w(a), then w(an;,f) = 0 follows, thus anml is a unit in R. Accordingly am (mod p)
for some am (
0) from 91. Since w(mn
- xm) >_ 1, so
w(a - amnm) >_ m + 1.
629
DISCRETE VALUATIONS
We suppose that for an s (> m) with any an, ..., as (E R) whatever, W (a - an,n,n - ... -- a ac) >- S -1-
1.
(159.4)
This is, according to the above, in fact the case particularly for s = m. If we
now temporarily denote the expression in brackets on the left-hand side s + 1, and so w (fins+1) >_ 0. Accordingly of (159.4) by fl, then w
s+l = a:+l (mod p) for an as+1 from RJR, thus we have
w(f -as+1n.,+1)Zs+2. This means that (159.4) is satisfied for s + 1 instead of s. By induction it follows that elements am (s' 0), am+l, ... of ll exist such that (159.4) holds for all s = m, m + 1, .... This implies the validity of (159.2). Lastly, in order to prove the uniqueness of the representation of a in the form (159.2), it should be noted that, by (159.3), the number m on the righthand side of (159.2) is uniquely determined by a. Consequently every further representation, similar to (159.2), of a is of the type : a = I9mnm + 1m+1nm+1 +
where Ym, m+t,
are elements of M. After subtraction we obtain
0=(01m-Ym) am +((Xm+1-Nm+1)nm+1+ If a,, - fl, were the first non-vanishing coefficient on the right-hand side, then the right-hand side would have the value s which is, however, false. This completes the proof of the theorem. § 160. ,p-adic Valuations
So far we have become acquainted with only one application, though a very important one, of valuation theory which consists in constructing the field (o) as the perfect hull of the field .5 with the absolute valuation and, proceeding from here, the field,) (i) as the algebraic closure of Y(o). Now we shall explicitly give certain further valuations which similarly
lead to important applications. Let a field F and a subring R of F with unity element and with prime decomposition be given. Suppose also that F is the quotient field of R.
FIELDS WITH VALUATION
630
Let n be a prime element of R. The elements a (96 0) of F may then be represented in the form
a=
aw(")
(K, A E R; n X K, A)
T,
(160.1)
where the exponent w (a) indicates an integer uniquely defined by a. (Notice
that if a E R, 0 then Tcw(") I I a.) We shall show that w is a discrete and normed exponent valuation of F, which we call the z-adic valuation of F and denote by w,,. Further, we shall show conversely, that every discrete valuation is equivalent to a FI-adic valuation. It should be noted that w,, (for fixed F and R) depends only on the prime ideal ,p = (n), therefore w,, is also denoted by w, and called a p-adic valuation. We repeat that the normed discrete valuations and the (n-adic or) p-adic valuations are identical.
For the proof we consider a further element fi (s 0) of F and write this as in (160.1) in the form
vER;aXµ,v).
_IFIw(") V
(160.2)
Since
Av
w(aj9) = w(a) + w(fl). If further, e.g., w(a) < w(p), then from (160.1) and (160.2) we obtain
a+#=
(rcv + Ap.,k),w(a)
(k = w(fl) - w(a) Z 0) .
vv
This gives w(a + i) w(a). Thus, w(a + fi) Z min (w(a), w(,B)) always holds. According to this, w is an exponent valuation. It is also discrete and normed, since the value module is evidently equal to 7+. Let us consider, conversely, a discrete valuation w of F. Let the valuation ring and the valuation ideal be denoted by R and ,, respectively. By Theorem 404, F is the quotient field of R. Furthermore, R, according to Theorem 407 is a ring with prime decomposition and ac is the unique prime element for which 4 = (ac). Thus the ,p-adic valuation w,, of F certainly exists. We even show that wand w , are equivalent. For this purpose let us consider an arbitrary element a (A 0) of F. This may be represented in the form
x= TK nwv
,
where K, A are units of R. Thus w(a) = w(9c) wp(a), so that W. wy are, in fact, equivalent.
P- ADIC VALUATIONS
631
Now, in general, we understand by a p-adic field any field which is perfect with respect to a p-adic valuation. (In the literature, it is usual to call only certain special cases of ?these "p-adic fields". Cf. HASSE (1949). Further we call the perfect hull of a p-adic valuated field the p-adic hull of this field. The p-adic hulls of algebraic number fields are called p-adic number fields. The elements of a p-adic number field are said to be p-adic numbers. We know from Theorem 408 that the elements of a p-adic field F (since this is just a field perfect with respect to a discrete valuation) may be uniquely given as infinite series
a = an,'G"' + am+1 arm+I +
(am, am+l, ...E R; am # 0)
,
(160.3)
where n is a generating element of the valuation ideal p and 91 denotes a system of representatives mod p of the valuation ring R of F, containing the element 0. We call (160.3) a normed p-adic series. (Hence the term p-adic field. Of course (160.3) is convergent and represents an element of the field,
are arbitrary elements of the valuation ring. Then when am, am+l, (160.3) is simply called a p-adic series.) _ In particular, we have to consider the p-adic hull F of a field F. Let w denote the p-adic valuation, R the valuation ring of F belonging to it and p itself the valuation ideal of R. We denote by w the valuation of F (in exponent form). Every element (-A 0) of r arises as a = w-lim ai , where [a;] is a w-convergent sequence from F. According to Theorem 402,
w(x;)=c for a suitable N (E 4") and an integer c for all i > N, whence w(a) = lim w(a,) = c .
Consider a further element P (# 0) of F : # = w-lim 9, , where [i4;] is a suitable w-convergent sequence from F. If, moreover,
a+#
0, then w(ac + 0) = lim w(a; + 9,) .
Hence, it immediately follows that the valuation w (together with iv) is non-Archimedean. We have also found that the value module of w like 21/a R.-A.
632
FIELDS WITH VALUATION
that of w is .7+, thus w is a discrete valuation of F. Let the valuation ring R and the valuation ideal belong to it. Then w itself is the -adic valuation of F and consequently is a P-adic field. Let n be a generating element of ,p, i.e., an element from F such that w(n) = 1. Since w is a continuation of w, w (7c) = w (n) = 1, thus 7L is also a generating element of . Furthermore, since the elements ( : A 0) of F are the w-limits of w-convergent sequences from F, this means, in other words, that the infinite series (160.3)
formed from F furnish all the elements a (0 0) of F. Since an, elements (
0, the
0) of R and P are given by m
0 and m > 0, respectively. We wish to apply this to the case F = . With this end in view we
proceed from the subring 7 of .9' ; this has . for quotient field and is a ring with prime decomposition. Consequently, a p-adic valuation of 3 belongs to every prime number p, which we call the p-adic valuation of 5 and denote by wp. wp (a) is, according to the general definition (160.1), that integer for which an equation of the form
a=
pw,(a)
(k, I E J"; p X k, 1)
(160.4)
holds, where a can indicate any element of .50 other than 0. The corresponding p-adic hull of 0 is denoted by p) and called the p-adic number field.
(Of course p1 has characteristic 0.) Its elements are the p-adic numbers. Since in J' the numbers 0, . . ., p - 1 constitute a system of representatives mod p, the p-adic numbers are given by the p-adic series a = am pm + am+l pm+l + .. .
(ann am+1, ... = 0, ..., p - 1) , (160.5)
where m can be any integer. The cases where am # 0 in (160.5) are the normedp-adic series, which furnish all the different p-adic numbers. Ap-adic number a is called a p-adic integer, if a may be represented in the form (160.5) with m Z 0. The p-adic integers constitute a ring, which is called the ring of p-adic integers. This is the valuation ring of pl with respect to wp. The units of this ring are called the p-adic units. These are the special cases of (160.5) given by m = 0, am 0. From (160.5) it follows that the elements (0 0) of 3 p) may be uniquely given in the form pop, where @ is a p-adic unit and m an integer. It is evident that the (infinitely many) p-adic number fields .21, Y(3), . . . are not pairwise isomorphic. For an easy application of p-adic number fields see HASSE (1935). EXAMPLE 1. The p-adic field contains non-rational algebraic elements, i.e., algebraic elements lying outside F.. For instance, let us take an equation such as
x"- a =0
(a E J; p,f'n, a),
(160.6)
633
P-ADIC VALUATIONS
which has no solution in .f`-,. This means that a is not an nth power in J. Furthermore, let us suppose that in 3 the congruence
x" - a = 0 (mod p)
(160.7)
has a solution at, (= 1, ..., p - 1). We shall show that (160.6) then has a solution a in .",,, for which moreover a = ao (mod p) . (160.8) Since ao - a = 0 (mod p), and because of the polynomial theorem, we may determine
a sequence of integers a,, a2, ... (= 0, . . ., p - 1) so that all the congruences
(a,+alp+...+a,p'?-a=0 (modp'+') (i=0,1,...)
...
hold. It is evident that the p-adic number a = ap + a1 p +
is then a solution
of (160.6), for which (160.8) also holds. EXAMPLE 2. The field .t',,, is not algebraically closed, since, e.g., the equation xE - p = 0 has no solution in it. This follows simply from the fact that p is a prime element in the valuation ring of . ,,,. We call the algebraic elements of . ",,, algebraic p-adic numbers. EXAMPLE 3. The field .3r",,, is not countable, as is obvious from (160.5). Hence, by the same inference as in § 155, Example 7, it follows that the degree of transcendence
of F(,) is not countable. EXAMPLE 4. The field .97,,, is not orderable. For the case p 96 2 we see this as follows: Since the congruence x2 + p - 1 = 0 (mod p) is solvable in .7, there exists, according to Example 1, a p-adic number a such that a2 + p - 1 = 0. Because -1 = = aQ + p - 2 the assertion follows from the ARTIN- SCHREIER theorem (Theorem 371). If p = 2 prove as in Example I that in ' (2) the equation as + 7 = 0 is solvable. Because - I = a2 + 6 the former inference again holds. EXAMPLE 5. The polynomial x°-1 = 0 splits into linear factors over the field
.-%,,) [i.e., -2-&) contains p - 1 (p - 1)`h roots of unity ]. This follows from Example 1,
since the congruence x" = 0 (mod p) has the solutions 1,
.
. .,p - 1.
§ 161. Ostrowski's First Theorem THEOREM 409 (OSTROWSKI's first theorem). All the real valuations of the fields F are (apart from equivalent valuations) the trivial valuation, the absolute valuation and the p-adic valuations. In the first place we know that the valuations listed above are not equi-
valent to one another. We now consider an arbitrary non-trivial real valuation q' of .5. At first we suppose 92 to be Archimedean. Since p(1) = 1, we have qq(n) 5 n (n = 0, 1,
. .
.). Because q7(- a) = qu(a)
,p(n) I for all integers a (> 1). Thus, according to (161.2) p(b) 5 q,(a)loeab = bloga O(a)
i.e., 1ogb q ( ( b ) 5 log. p (a)
(a,b = 2, 3, ...) .
Since, a, b may be interchanged, "=" holds here. Consequently
log. p(a) = K
(a = 2,3,...)
OSTROWSKI'S FIRST THEOREM
635
for a fixed positive number K, because q,(a) > 1. Hence, because q,(-a) _ = q,(a) and 99(0) = 0, q7(1) = 1,
9,(a)=IaII` This then holds because of the homomorphy property of 99 for all a from .9. According to this, and Theorem 398, 99 is equivalent to the absolute valuation .70. Secondly we suppose q, to be non-Archimedean. Now (p (a) 5 1
(a E J) .
We shall show that especially the a such that p(a) < I constitute an ideal
of 7. This follows from the fact that for two elements a, b of ..7 in the case q, (a) < 1 we always have ,(ab) = q,(a)q,(b) < 1
and if p(a) < 1, p(b) < 1 q,(a - b) 5 max(q,(a), q,(-b)) < 1
.
This ideal is prime, since for q,(ab) < I at least one of p(a) < 1 and p(b) < 1 must hold. Let (p) be this ideal, where p is a prime number. Then
p(p) < 1 .
(161.7)
Furthermore q,(a) = 1 for all the integers a prime to p. For every element a it follows that ( 0) of . 99(a)
= (9, (p))wP(a) ,
where wp denotes, the p-adic valuation of .,. Thus if we pass from q, to an exponent valuation according to the rule 9'(a) = g-w(a)
where g denotes a fixed real number (> 1), then w(a) = Kw,(a) for a fixed real number K, which, because of (161.7), must be positive. Consequently, according to Theorem 398 the proof of Theorem 409 is completed. § 162. Hensel's Lemma
We prove the following theorem, important in itself, which at the same time is preparatory to the following paragraphs. THEOREM 410 (HENSEL'S lemma). Let F be afield, which is perfect with respect to a non-trivial exponent valuation w, with the valuation ring R and
FIELDS WITH VALUATION
636
the valuation ideal p. Then, if f (x), go(x), ho(x) are three polynomials over R such that the leading coefficients of go(x), ho(x) are prime to P and
f(x) * 0 (mod p),
(162.1)
fl x) - go(x)ho(x) (mod i,) ,
(162.2)
(4,, go(x), ho(x)) = 1
(162.3)
hold, then there exist two further polynomials g (x), h (x) over R such that g(x), go(x) are of the same degree,
f(x) = g(x)h(x) ,
(162.4)
and
g(x)
go(x) (mod ip)
,
h(x) _ ho(x) (mod ,p).
(162.5)
COROLLARY. For every irreducible polynomial
f(x)=aco+ar1Y+...+ax" over F
min (w(arn), ..., w(a)) = min (w(acJ, To prove this we denote the degrees of the polynomials f(x), go(x), ho(x)
by n, r, s, respectively, (n > r + s; r, s >_ 0). We may assume that r since otherwise the theorem is true. By (162.2),
f(x) - go(x) ho(x) - 0 (mod ,) .
(162.6)
On account of (162.3) there exist polynomials G(x), H(x) over R such that
go(x)G(x) + ho(x)H(x) - I - 0 (mod ,p) .
(162.7)
We take from p an element n (# 0) for which w(n) is not greater than the least value of the coefficients of the left-hand sides of (162.6) and (162.7). Then
J (x) - go(x)ho(x) (mod a) ,
go(x)G(x) + ho(x)H(x) - I (mod a)
(162.8) .
(162.9)
We prove that there are polynomials uj(x), vj(x) (i = 0, 1, ...) over R
of degrees 5 r - 1 and 5 n - r such that for the polynomials gj(x), hj(x) defined recursively by
gj+i(x) = gj(x) + ni+luj(x) ,
(162.10)
hj+1(x) = hj(x) + nj+IVj(x)
(162.11)
HENSEL'S LEMMA
637
the congruences
f(x) = g.(x)h,(x) (mod n'+1)
(i = 0, 1, ...)
(162.12)
hold.
According to (162.6), (162.12) holds for i = 0. We suppose that for an m (> 0) the ui(x), v;(x) (i = 0, . . ., m - 1) are determined as required so that (162.12) is satisfied for i = 0, . . ., m. Then we wish to determine a suitable pair of polynomials um(x), v,,,(x) such that (162.12) is also satisfied
for i = m + 1. This requirement is that f(x) = gm+t(x)hm+](x) (mod n"'+2)
for which we can write, because of (162.10), (162.11), f(x) - gm(x)hm(x) °- am+1(gm(x)vm(x) + hm(x)um(x)) (mod n'"+2). (162.13)
The left-hand side is, according to the case i = m of (162.12), equal to nm+'Fm(x), where the second factor is a polynomial of degree 5 n over R. Thus because of (162.10) and (162.11), (162.13) is identical with go(x)vm(x) + ho(x)um(x) - F,(x) (mod n) .
(162.14)
On the other hand, it follows from (162.9), on multiplication by Fm(x), that go(x) G(x) Fm(x) + ho(x) H(x) Fm(x) _- F,(x) (mod n).
In order to obtain from this a suitable solution of (162.14), we determine two polynomials qm(x), um(x) by Euclidean division so that H(x)Fm(x) = go(x)gm(x) + um(x)
and um(x) is of degree 0. Because of the irreducibility of f(x), the validity of w(aI) >_ 0 (i = 1, ..., m) follows from the corollary to HENSEL'S lemma (Theorem 410). Accordingly,
w(1 -al+...±am)z0. This means that the right-hand side of (163.5) is at most 1, by which (163.2) is proved. Secondly, let us consider the case of an Archimedean valuation q'. According to Theorem 412 of the following section we may restrict our-
selves - as has been already noted - to the case n = 2. We again denote by y an element of G for which qq(y) 5 1. Similarly, as above, it is sufficient to prove that q,(y + 1) < y,(y) + 1.
(163.6)
We may suppose that y lies outside F. Because n = 2, the minimal polynomial of y over F may then be put in the form
f(x)=x2+ax+b. As before V(V) = (cv(b))'
+V(y + 1) = (,p(1 - a + b))*.
,
Accordingly, assertion (163.6) means that
T(1 - a + b) < ((p(b))* + 1)2.
Since qq(1 - a + b) 4q((b).
(163.7)
REAL PERFECT VALUATIONS
641
Since f(x) is irreducible, VA 0. Hence, and from (163.7), m :A 0. We form
a sequence [c,] from the elements of F such that a
c1=2, c,+1=-a-
b c
(i=1,2,...).
(163.8)
This is possible, since the c1, c2i ... turn out to be different from 0, indeed, we shall show that
(i = 1, 2, ...).
99c. ? 2 99a)
(163.9)
Because (p(2) 5 2,
If (163.9) is then assumed for some i, then, because of (163.7) and (163.8), we obtain 99(c1+L) = 4'(a +
Al
> 9?(a)-
9%(b) > 4'(a) - 2
Vi(a) >
2 9'(a)
Hence (163.9) follows by induction. We now show that our sequence [c,] is convergent. We have
c,+s - c1+1= -
b c1+1
+
b c,
= b c1+1 - c, c, c,+I
Therefore from (163.9) it follows that m(c1+2 - c,+1) < (99(a))2
9'(c,+1- c)
If the first factor of the right-hand side is denoted by q, then according to (163.7) 0 < q < 1, and we obtain 99(c,+1 - c,) 5 q'-1 9'(c2 - c1)
(i = 1, 2,. ..).
This gives
,-1 99(c, - Ck)
]
are (p-convergent, and then
o-lim a; = a(i'col + ... + a(")COn
(163.11)
where
a(k) = 97-lim a(k)
(k = 1, ..., n).
(163.12)
Since from (163.10) and (163.12) for every i (= 1, 2, ...) y!(act -a(' (j), -
... -
n
a(' W") :!9
k=]
p(a ") - a(k)) 'P(Wk)
both the "if" assertions of the proposition are true. In order to prove the "only if" assertion, we call the sequence [at] an m-sequence, if a;"'
a(") = 0
(i = 1, 2, ...).
For the 1-sequences the assertion is true, since for these
We denote by m one of the numbers 2, . . ., n, and assume that the assertion for the (m - 1)-sequence is true. We consider a p-convergent m-sequence [act]. It suffices to prove the assertion for this sequence, since from this the assertion will follow by induction for the case m = n, i.e., for the general case.
REAL PERFECT VALUATIONS
643
if the sequence [aJ')] is 99-convergent, then the sequence loci - aim) CJm] _ [all)wl + ... + a(m-1)wm_1J 77
is p-convergent and at the same time an (m - 1)-sequence. For this case it follows from the induction hypothesis that the assertion is true. If, on the other hand, the sequence [a(im)] is not 99-convergent, then we infer from this a contradiction and so prove the proposition.
By hypothesis there exists a positive number s and natural numbers n1, n2, ... such that g9(aim) - a;+Rr) > s
(i = 1, 2, ...).
(163.13)
We put (a; - ai+", )
f'i = (asm) -
(I= 1, 2,. ..).
(163.14)
Since now, according to Theorem 379, [a; - a;+";] is a 'N-zero sequence, so also, because of (163.13), (163.14), [,;] is a si-zero sequence. A glance
at (163.10) and (163.14) shows further that [f; - wm] is an (m - 1)sequence. This is, according to the above, p-convergent, and W-lim WI - wm) = -wm . i
By the induction hypothesis we obtain from the formula (163.11) an equation of the form -co. = b1w1 + ... + bra-lwm-1 with elements b1, . . ., bm_1 from F. Since, because of the basis property of w1, ..., w", this is impossible, the proposition is proved. Hence the assertion of Theorem 411 that G is perfect with respect to V follows immediately.
We still have to prove the uniqueness assertion of the theorem. For that purpose we anticipate a special inference from the proposition. If p-lim a; = 0, then, from (163.11), i
a(k) = q'-lim aik) = 0 i
(k = 1, ..., n).
On the other hand, by (163.10) and Theorem 360, NGIF(a,) is a homogeneous polynomial in the ail), ..., o(,) of degree n with coefficients from F, which
depend only on the wl, ..., w" and their conjugates. Hence there follows for the sequences [a;] considered in the proposition, the rule: y,-lim a; = 0
qi-lim Nc i F((X;) = 0.
(163.15)
644
FIELDS WITH VALUATION
We now suppose that vp is any real valuation of G and an extension of (p. We have to prove that (163.1) necessarily holds for this +p. We suppose
that for an element a of G (163.1) is false. Then we have either 1P((X") < < 99(N(a)) or V(an) > q)(N(a)), where we have put N = NGIF. Corresponding to these cases we write N(a)
Mn
which implies that +'(Q) < 1.
(163.16)
N(e) = 1.
(163.17)
Further evidently
From (163.16) +p-lim e'= 0. Then, according to (163.15), 4p-lim N(e) = 0, i.e., 99-lim (N(CO))' = 0, contradicting (163.17). Consequently we have
proved Theorem 411. EXAMPLE. On account of Theorem 359, definition (163.1) may be replaced by (163.18)
VW = T (NF(-) F(a))IF(a):F1-1.
Hence we see that rp((x) depends only on F, p, a. Thus Theorem 411, apart from the statement '°G is perfect with respect to +p", holds for all algebraic field extensions G I F of a field F which is perfect with respect to gyp.
§ 164. Ostrowski's Second Theorem THEOREM 412 (OSTROWSKI's second theorem). The Archimedean-valued perfect fields are topologically equivalent to one of the fields .5o), 5o)(i) valued by the absolute valuation.
Let F denote a field which is perfect with respect to an Archimedean valuation 92. Since the characteristic of F must be equal to 0, it may be supposed that .9 is a subfield of F. From OsTROwsKI'sfirst theorem (Theorem
409) it follows that the valuation of ..5 must be equivalent to the absolute valuation. Thus we may assume that F contains the perfect hull 5o) of Y0 and that (p is an extension of the absolute valuation of J 57(o). First, we distinguish two cases which we shall, however, soon consider together. If x2 + I is reducible over F, then it may be supposed that 5o)(i), too, is a subfield of F. Since furthermore [J9( '0,(!) : 5r(o)] = 2, it follows from Theorem 411 that for the elements a + bi of o)(1)
99(a + bi) _ 1(a + bi)(a - b 1)11 = (a2 + b2)'
a + bi
OSTROWSKI'S SECOND THEOREM
645
i.e., ois also valued by the absolute valuation. For this case we shall show that F = ob(i). If, on the other hand, x2 + I is irreducible over F, then the complex field
G =F(i) over F exists and we may suppose that .5o)(i) is a subfield of G. Because [G: F] = 2 it follows from Theorem 411 that G has one, and only one, real valuation, which is an extension of 99, and that it again agrees with the absolute valuation for the elements of.$o)(i). Here G is, according to Theo-
rem 411, perfect. It means that for G the former case holds, whence, anticipating the result, it follows for that case that
Jp(i) = G D F
.off,
[7o)(i) :. 0)] _ [G : F] = 2.
Since F = .Yjo) follows from this, it suffices to prove for both cases the following :
If F is an extension field of o}(1) and (p a real valuation of F, which is an extension of the absolute valuation of 90;(i), then F = .jo)(i).
For the purpose of the proof we suppose that F D .o)(i). Let a be an element of F - .7(0,(i). We denote by a the Weierstrass lower bound (cf. § 154, Exercise 1) of the set of all values (p(a - z) for z E .7()(i). (This means that for these z we always have q^(a - z) >_ a and that there is for every h (> 0) a z such that q(ac - z) < a + h.) In any case, a Z 0. First of all we show that there exists a zo (E Jo)(i)) such that q:(a - zo) = a.
(164.1)
According to the definition of a we may form from .9o)(i) a sequence such that lim 97(u -
a.
(164.2)
Since, on the other hand, we have
5 97(a - zj + qu(a), there is a real number C (> 0) such that I Z. l = 4i(zn) < C. According to the BOLZANO - WEIERSTRASS theorem (§ 154, Exercise 2) we may
therefore form a convergent partial sequence from [z.]. If we retain the for this, then (164.2) remains valid, and notation
limz=zo,
646
FIELDS WITH VALUATION
for a zo from -o) (i). It then follows that
a=lim p(oc by which we have proved (164.1).
Together with the preceding a, a - zo also lies in F(o)(i). Therefore, if we write a instead of a - zo then (164.1) goes over into qu(a) = a. Since a # 0 here, it follows that a > 0. Thereby we have the result that there is an element a of F and a positive real number a, for which the following always holds :
T(a - z) = qu(a)
(z E F(o)(1), I z I < a).
(164.3)
We shall show that for every a with the property (164.3) 'AM - z) = qq(a)
(z E 5o)(11, I z I < a).
(164.4)
For this purpose we take a natural number n and a primitive nth root of unity a from Y0)(i). Then, according to (164.3), Pn_IZ)
9p(a - z)9:(a - aaZ) ... 92(a -
=
p(an
- Zn) S
s q)(an) + p(z") = a"+ I z I".
If we apply (164.3) to the factors of the left-hand side from the second on, then we get 4'(a - z)an-1 < an + I Z I., i.e., n
rr
9'(a-z)Sa1+ [
(n=1,2, ...).
a 1
J
For I z I < a this gives (p(a - z) < a. Hence, and from (164.3), follows (164.4).
We generalize (164.4) to 99(a - nz) = qu((X)
(z E
z I < a; n = 1, 2, ...).
(164.5)
For n = 1, (164.5) is the same as (164.4). Let (164.5) be assumed for some n. If we then apply (164.3) to nz + z instead of z, (164.3) follows
from the assumption, thus also (164.4) for a - nz instead of a. Thus, again from the assumption,
tq(a - nz - z) = tp(a - nz) = p(a) (z E .(o)(i); I z I < a). Hence (164.5) is proved by induction. Now, from (164.5)
tP(a - z) = T(a) = a
(z
E'o)(')).
OSTROWSKI'S SECOND THEOREM
647
So, for any two complex numbers z1, z2,
Iz1-z2I=9q(z1-Z2) for perfect hull, so we must determine the irreducible factors of f(x) over 5 ol. We proceed most easily from the factor decomposition
f(X) = (X - 01) ... (X -
(&'1 = ) Let r1 and 2r2 be the number of real or non-real conjugates of 0, respectively (r1 + 2r2 = n). Then f(x) splits into the product of r1 linear and r2 quadratic irreducible factors over ,. If 19k is real, then a valuation of G is given by over
(166.1)
V)(9(6)) = 19('0k) I
where g(x) denotes the polynomials over F of degree < n. If '9k is not real, then we have to put _ V(9(0)) = 19(O,) 9(9k) 1
,
REAL VALUATIONS OF NUMBER FIELDS
649
where Ok denotes the complex conjugate of 11k. The right-hand side is again equal to I g(0k) 1, therefore all the Archimedean valuations of G are given by formula (166.1), where ?9k has to run through the conjugates 01, ..., of 0, however, so that of each pair of conjugate-complex ele-
ments only one is admitted. The r1 + r2 Archimedean valuations so obtained are evidently distinct. Secondly, we define the non-Archimedean valuations of G. We give these more easily as exponent valuations. Let cop denote the p-adic valuation
of.$, where p may be an arbitrary prime number. We also denote the extension of w, to ible factors:
0)
by wp. We decompose f(x) over 5p) into irreduc-
f(x) = f1(x) ... f,(x) All the exponent valuations w of G, which continue the valuation wp of .7o, are then given by
w(g(o)) =
WVp(N(9(0k)))
(166.2)
k
where Ok denotes a zero of fk(x) in a suitable extension field ofpl, nk the degree of f(k)(x), and N the norm with respect to f(p)(0k) 1 -7;pl, while g(x) means the same as in (166.1). In connection with the above see § 170, Example.
§ 167. Real Valuations of Simple Transcendental Field Extensions Let a simple transcendental extension field F(x) be given, i.e., the rational function field of an indeterminate x over a fundamental field F. We wish to determine those real valuations of F(x) which are extensions of the trivial valuation of F. Since the required valuations of F(x) are non-Archimedean, we may take them in the form of exponent valuations. Therefore, let w denote an exponent valuation of F(x) such that w(a) = 0 for a E F, 0 0. We distinguish two cases, one when we have always
w(f(x)) > 0
(167.1)
for the non-constant polynomials f(x) (E F[x]), the other when
w(f(x)) < 0
(167.2)
holds, too, but we note in advance that these two cases will not be essentially distinct. When wv(f(x)) = 0 always holds, all the polynomial quotients have
FIELDS WITH VALUATION
650
w-value 0, thus F(x) is then trivially valued as the quotient field of F[x]. As we wish to disregard this, we suppose in the first case that ">" occurs in (167.1).
In the first case letp(x) denote a polynomial over F such that w(p(x)) > 0. Among the irreducible factors of p(x) at least one must occur with positive value, therefore we may suppose that p(x) itself is an irreducible principal polynomial. Now, if we replace w, if necessary, by an equivalent valuation, we may also assume that w(p(x)) = 1. (167.3)
If f(x) is now a polynomial non-divisible by p(x) over F, then we shall show that w(f(x)) = 0. Let p1(x), f1(x) be such that 1 = P(x)Pj(x) + Ax) ft(x)
(P,(x),.fi(x) E F [x])
Hence, because of (167.1),
o = w(1) z min (w(p(x)) + w(pl(x)), w(f(x)) + w(fl(x))) Z min (w(p(x)), w(f(x))).
From this w(f(x)) < 0, and so w(f(x)) = 0 follows from (167.1).
We now consider an arbitrary polynomial g(x) over F. This can be written uniquely as
g(x) = (P(x))k4(x)
(9(x) E F[x],P(x) f' 9(x))
From what has been proved so far and from (167.3) it follows that
w(g(x)) = k . Because of the rule w
g(x) f h(x)
w(g(x)) - w(h(x))
(167.4)
the values of all the elements of F(x) are now known. Further, we see [cf. (160.1)] that w is that .p-adic valuation of F(x) which belongs to the prime polynomial p(x). Since, conversely, all the irreducible principal polynomials
p(x) (E F[x]) are admissible and yield non-equivalent .p-adic valuations, the present case has been entirely explored. In the second case we take a principal polynomial p(x) over F of minimal
degree with the property w(p(x)) < 0 .
(167.5)
We shall show that p(x) must be linear. For let us suppose that
p(x) = x° + PA(x)
(n>_2),
(167.6)
SIMPLE TRANSCENDENTAL FIELD EXTENSIONS
651
where p1(x) aenotes a polynomial of degree < n, then by hypothesis we have
0,
w(pt(x)) >_ 0, w(x)
w(x")
0.
w(p1(x)))
0.
Hence, from (167.6), w(p(x)) > min (w(x")
,
Since this contradicts (167.5), p(x) is in fact linear. Then
p(x)=x+a for some element a from F. But, because of w(f) = 0 and Theorem 400, it follows from (167.5) for every element 9 (# 0) of F that w(p(x) + j9) = w(p(x)).
This together with (167.5) means that all the linear polynomials of K [x] have a common negative value. When we thus replace w, if necessary, by an equivalent valuation, then we get
w(x) = -1. For every polynomial
9(x)=aoxm+... +am
(am00)
over F we have w(g(x)) _ -m as a result of Theorem 400. If we again apply rule (167.4) we obtain for the value of an arbitrary element of F(x) other than 0, the formula wl
x
h(x) j= n - m,
(167.7)
where m and n denote the degree of the numerator and denominator, respectively. (Moreover (167.7) also holds for g(x) = 0, since then, according
to our previous convention m = - oo, while on the other side w(0) = Co.) Conversely, we shall show that a (discrete) exponent valuation of F(x) is defined by (167.7), which we call the degree valuation of F(x).
If gf(x) is a polynomial (5& 0) over F of degree m, (i = 1, . . ., 4), then 91(x) w(---
93(x) I 92(x) 94(x) 1
_ m2+m4-ml-m3=w
91(x) 92(x)
+w I
and W
93(x)
91.(x) l
92(x)1 + 194(x) I =
91(x)94(x) + 92(x)93(x)
w 1
92(x)94(x)
93(x) 9a(x)
,
FIELDS WITH VALUATION
652
On the right-hand side of the last equation the degree of the numerator is at most max (m1 + m4, m2 + m3), but that of the denominator equals m2 + m4, thus the right-hand side of this equation is at least
m2+m4-max(ml+ m4, m2+m3)= = min (m2 - m1, m4 - m3) = min w 91(x)
,
x 9a(x)
92(x)1
19.I(x)1
According to this, (167.7) is in fact an exponent valuation. To sum up we have the following THEOREM 414. All the real valuations of the rational function field F(x), which extend the trivial valuation of a field F, are (apart from equivalent valuations) the ,p-adic valuations belonging to the irreducible principal polynomials p(x) (E F [x]) and the degree valuation of F(x). (Clearly the above valuations are not equivalent to one another.) It is important, as we have already suggested, that the exceptional position of the degree valuation of F(x) as opposed to the .p-adic valuations of F(x) (in a sense to be stated precisely below) is only apparent. It is well known, but is worth stressing, that the indeterminate x does not
have an invariant meaning, for the field F(x); it may in fact be replaced by any element Y
_ ax+ ;tx +
(z, "'
y, b r-- F; ab -
0) .
(167.8)
Then, according to Theorem 323, all the primitive elements y of F(x) for which F(y) = F(x) are given by (167.8). We choose, in particular,
y_x 1
(167.9)
and denote by wj, the .p-adic valuation of F(y) (= F(x)) corresponding to the prime polynomial p(y) = y. Let g(x) h(x)
be an arbitrary element (0 0) of F(x). In order to determine its wy value, we denote by m and n the degree of the numerator and denominator, respectively. We have g(x)
9(y-1)
y
y"h(y-1)
653
SIMPLE TRANSCENDENTAL FIELD EXTENSIONS
Both numerator and denominator of the second factor of the right-hand side are polynomials of y with a constant term other than 0, therefore they are prime to y. Consequently x
tip,,
h(x) = n - m.
We now see the correspondence with (167.7), so that the p-adic valuation xY of F(y) (= F(x)) is identical with the degree valuation of F(x). Thus we have seen that the degree valuation of F(x), after we have introduced the primitive element y =
Ix
,
is converted into the p-adic valuation corrrespond-
ing to the prime polynomial y of F [y].
The p-adic valuation of F(x) corresponding to the polynomial x - a is also called the valuation (of F(x)) belonging to the place a. Correspondingly,
the degree valuation is also called the valuation belonging to the place 00
(since, after we have introduced the primitive element
I,x this becomes
the valuation belonging to the place 0). The above valuations are called the point valuations of F(x). If, in particular, F is algebraically closed, i.e., only linear polynomials are irreducible over it, then point valuations are all the valuations of F(x). Let w,, denote the point valuation of F(x) belonging to the place a (a E F or a = oo). If, for an element r(x) of F(x), w.(r(x)) = n or
-n
(n = 1, 2, ...)
,
then we say, using the terminology of function theory, that x = a is an n -fold zero or an n -fold pole of r(x), respectively. As to every valuation in general, so also to every point valuation wa there belongs a perfect hull of F(x). Its elements, according to (160.5) may be uniquely given by the (p-adic) infinite series f(x) _= am(x - oc)m + an:+1(x - 00"' + ...
am+I, ... E F) ,
(167.10)
where m can indicate any integer. These are similar in form to the power series in the theory of functions, therefore they are called formal power series. For the degree valuation (x = co ), we have to replace x - a in (167.10) by
-.xI
In function theory one is generally interested only in the substitution values f(E) for 5 E F in connection with a power series f(x), i.e., we understand f(x) as a function of the variable x. Therefore the notion of power series in algebra and that in function
654
FIELDS WITH VALUATION
theory are related in the same way as the polynomial notions of both theories. (Cf. Exercise 2.) EXERCISE 1. Quite generally, if 99 is a valuation and a -- a' an automorphism of a field F, then tp(a) = p(al) defines a valuation ip of F. Prove that all the place valuations of the above-considered field F(x) can be obtained from one of them in this way by means of F-automorphisms. EXERCISE 2. The above notions introduced with respect to F(x) are meaningful even when F is arbitrarily valued (but now Theorem 414 is no longer valid). Let e.g.,
F = F-(,) be the p-adic number field. We take into consideration a formal power series (167.10) where now a, am, am+1, ... are elements of JF,," . With an element of F-(," we carry out the substitution x = :
!l') = am( - ex), + am+l(E - (z) '+1 + .. . However, this equation has a meaning only if the right-hand side is convergent, and then represents an element of .F'(,). The , for which this is the case, constitute the domain of convergence of the formal power series (167.10). Show that if p 96 2 the domain of convergence of ap
f(x) =A E
1l
(E) x"
n
which are not units, and that then
consists of those p-adic integers
(f($))2 = 1 + $ . For further investigations of this kind cf. HASSE (1963). EXERCISE 3. Take for basis the rational function field F = .70(x1, ..., x"), form the rational function field G = F(x) and denote by G the perfect hull of G belonging to the degree valuation. Put
AX) = (x - x1) ... (x - xn) = x" + g(x)
In G we have I
X - x,
_
I
a0
x
X= "
xi
1
x'r , wx)
_
1
x" i
gx) ''
00
(
X" )
If these are inserted into
"
1
f' (x)
x - x,
f(x)
we obtain a short proof of WARING'S formula (Theorem 279).
CHAPTER XI
GALOIS THEORY The main object of the theory of GALOIS is to obtain a detailed survey of the subfields of a separable normal field of finite degree. The applications
of this theory far surpass its objective and are deeply involved in many questions of algebra, number theory and geometry. It was one of the earlier
established branches of knowledge of modern algebra and is admirably neat. The twenty years old GALOIS, killed in a duel to the great sorrow of posterity, far exceeded with his life-work the stage of development of the mathematical sciences of that time. As to generalization cf. CARTAN (1947) and NAKAYAMA (1949-50).
§ 168. Fundamental Theorem of Galois Theory A separable normal field N I F of finite degree is said to be a Galois field and the group of its F-automorphisms is called the Galois group of N I F. If F is a prime field, then N is called an absolute Galois field. In this case the Galois group is the full automorphism group of N. We can express the definition of Galois fields in two other forms: Galois fields over an arbitrary field F agree with those separable algebraic fields F(ad) for which the minimal polynomial of 0 splits into a product of linear factors over F(O). The validity of this transformation of the above definition follows from Theorems 299, 303, 356. Furthermore, it follows from Theorems 331, 336 that a field N I F is a Galois field if, and only if, its degree [N : F]
is finite and equal to the number of automorphisms of N I F. Galois fields constitute the object of the theory of GALOis, or briefly Galois theory, and in this the Galois groups as equally important topics will also be considered. More precisely, we shall mostly deal with a fixed Galois field and its Galois group only. We shall show that two isomorphic Galois fields N I F, N' I F have isomorphic Galois groups. We shall even show that from every isomorphism
NIF;zt N' I
(a->. soc)
follows the isomorphism
d ^ 4'
(a - sas-1)
between the corresponding Galois groups c0, cdi'. 22 R. -A. 655
GALOIS THEORY
656
It is evident that a->sas-1 is an isomorphic mapping of N' onto itself for every a E 44, i.e., an automorphism of N'. The elements of the fundamental field F remain fixed here, thus sas-1 belongs to 4'. Accordingly s4ts-1 e_ C cp'. Likewise s-1c4 s S 4, i.e., 4' c s4s-1. Therefore sc4s-1 = 4t'.
This means that a -* sas' is a mapping of cP onto 4'. This mapping is obviously one-to-one and homomorphic, so proving the assertion.
Galois theory is concerned with certain other fundamental notions in connection with a Galois field N I F and its Galois group 4. In order to introduce these we next consider a quite arbitrary field N and its full automorphism group cP. If G is a subfield of N, we denote by G the set of those elements a of 44 for which the equation as = a
is satisfied by all a E G. If furthermore V is a subgroup of c0, then we denote similarly by - the set of those elements a of N for which the given equation
is satisfied for all a E A. Evidently G is a subgroup of 4, furthermore is a subfield of N. We call G and X the invariance group of G and the invariance field of, respectively. (In other words : G is the group of G-automorphisms of N, furthermore A' is the field of those elements of N which
are fixed under the automorphisms contained in X.) According to the above we have defined a unique correspondence
X->X where X runs through the subfields of N and the subgroups of 4. We call this the correspondence in the sense of Galois theory. Substantially it means a certain mapping by which the set of subfields of N is mapped into the set of subgroups of co and this into the former set. We say that G is the subgroup of c belonging to the subfield G (c N) and the subfield of N belonging to the subgroup ° (9 c.4). From the definitions we obtain the following rules immediately: G1 c G2
G,
G2
1719- Ir2 Ir1X2
(G1, G2 S N) .
G 1,'
2C(4)
(168.1)
(168.2)
Further we shall prove that GDG
(G S N),
(168.3)
Since, for every element a of G, the condition as = a is satisfied by all the elements a of G, we have a E G, whence (168.3) follows. (168.4) is proved similarly.
FUNDAMENTAL THEOREM OF GALOIS THEORY
657
We prove further that
(G c N),
G=G (
c c4)
(168.5) (168.6)
.
From (168.1) and (168.3) it follows that the left-hand side of (168.5) is contained in the right-hand side. Also the converse of this follows from (168.4) by application to = G. Hence, (168.5) is proved and (168.6) may be proved similarly.
TinoREM 415. For a group Re of automorphisms of a field N, the relative field Nla' is of finite degree if, and only if, W is finite, and then (168.7)
[N : Jr] = O(M).
(168.8)
is Thus, if N The automorphism group of N I is of finite degree, then l° is finite, consequently according to (168.4) X, too, is
finite. Conversely, we suppose that M° is finite. We next show that N I A' is then of finite degree and
[N:?]s0(
(168.9)
.
For this purpose we write n
(168.10)
It is sufficient to show that any n + I elements w1, ..., can+1 of N over Jr are necessarily linearly dependent. We consider the system
(awl) x1 + ... + (awn+l)xn+1 = 0
(a E -7)
of n homogeneous linear equations with n + 1 unknowns x1,. .., xn+1. This has, according to Theorem 248, a non-trivial solution. There is thus a least natural number m (_< n + 1), for which the equation system
(awm),n = 0
(a E M°)
(168.11)
is satisfied by any elements 1, ..., $n, of N, where m does not vanish. Of course, we may assume that (168.12) $m E X. We take an arbitrary element b of X, replace a in (168.11) by b-1a, and then
carry out the automorphism b: (acal) (b$1) + ... + (awm)
0
(a, b E a°)
CALOIS THEORY
658
Hence, and from (168.11), (168.12), we obtain by subtraction,
(awl) ($t - b$t) -I- ... + (awm-i) (em-i -
0.
Because of the minimal property of m
bpi = i
(i = 1, ..., m - 1; b E)
must then hold. Therefore fit,
- , m-i E .
. This and (168.12) imply,
for the special case a = 1 of (168.11), that oli.... , alm, and a fortiori are linearly dependent over X. (168.9) is now proved. oh, ..., Since, accordingly, the field NIA' is of finite order, so, from Theorem 335, for the order of its automorphism group the inequality
O(,YC) 5 [N : C]
(168.13)
follows. Since, moreover, from (168.4) 2C ', equality must hold here as well as in (168.9) and (168.13). This implies the validity of (168.7) and (168.8), and consequently Theorem 415. THEOREM 416. A field N I G is a Galois field if, and only if, there exists a finite group c° of automorphisms of N such that
G = X.
(168.14)
For the proof we suppose, first, that for afinitegroup l° of automorphisms of N the condition (168.14) is fulfilled. Hence, and from (168.8), [N : G] _
= O(. Moreover, it follows from (168.7) that G = T. Then [N : G] = O(G).
(168.15)
Since G is the automorphism group of N I G, so, according to the above, (168.15) means that N I G is a Galois field. Conversely, we suppose N I G to be a Galois field, i.e., that (168.15) is satisfied. We write A' = G. Since ° is finite, [N : G ] _ [N : G] follows G, i.e., = G. from (168.8) and (168.15). This and (168.3) give Consequently Theorem 416 is proved. THEOREM 417 (fundamental theorem of GALOIs theory). Let N I F be a Galois field and 4 its Galois group, and let X --> X be the correspondence in the sense of Galois theory. Then the set of subfields of N I F is mapped oneto-one onto the set of subgroups of c4 by
G->G
(F9G9N)
(168.16)
and
[N : G] = O(G)
(168.17)
FUNDAMENTAL THEOREM OF GALOIS THEORY
659
for elements assigned to each other. The inverse mapping of (168.16) is
r-. jr
( ° c Co.
(168.18)
SUPPLEMENT 1. For arbitrary subfields G1, G2 of N I F and for arbitrary
subgroups -71, -72 of cp
621
X1 c 4°2 a dr1 {Gl, (:;2} = Gl (1 G2, {DL
i'
2}_
1fl
2,
`2
{G1 n G2 = {Gl, G2},
1n
(168.19) (168.20) (168.21)
2= ['?'I, I2). (168.22)
SUPPLEMENT 2. For the conjugates aG (a E cot) of a subfield G of N I F
aG =aGa 1.
(168.23)
SUPPLEMENT 3. A subfield G of N I F is (normal i.e.) a Galois field over F if, and only if, G is normal in c.P, and then the Galois group of G I F is isomorphic
with thefactor group c4/G (= F/G). If the elements of a system of representatives mod G of cog act on the elements of G, all the different automorphisms of the field G I F are obtained, i. e. the elements of its Galois group. NOTE. Apart from (168.17) the contents of the fundamental theorem and of Supplement 1 may be so stated that the first or second lattice of the subfields of N I F is isomorphically mapped by (168.16), onto the second or first lattice of the subgroups of cot, respectively. In Galois theory, particularly
in Supplement 3, lies the justification for the similarity of the terms "normal field" and "normal divisor". The fundamental theorem may be applied to an arbitrary separable field G I F of finite degree as one has recourse to its normal field N I F which is evidently a Galois field and is called the Galois field of the field G I F; then the subfields of G I F are assigned, because of (168.21), to those subgroups of the Galois group of N I F which contain the invariance group G of G. In order to prove Theorem 417 we denote the full automorphism group of N by Since cP is the Galois group of N I F, F = d, .
(168.24)
On the other hand, there is, according to Theorem 416, a subgroup coo of vL' such that F=coo.
GALOIS THEORY
660
By Theorem 415, c o =o. and so F = 4io. Hence, and from (168.24), it follows that 40 = 4i, consequently
F=4.
(168.25)
Further, from Theorem 415, [N : cQ] = 0(c4). Hence, and from (168.24), (168.25),
[N : F] = O(F) .
(168.26)
We now consider an arbitrary subfield G of N I F. Since both N I F and N I G are Galois fields, we may apply (168.24), (168.25), (168.26) instead of F. Consequently (168.27) G=G and (168.17) holds. Since G ? F, according to (168.1), G c F, thus, according to (168.24), G c co. Accordingly, by (168.16), the set of subfields of N I F is mapped into the set of subgroups of cQ.
In order to show that this mapping is onto the latter set, we consider 9 4 it then an arbitrary subgroup 2l° of c.P. We put G = R. Because follows from (168.2) and (168.25) that G ? F, therefore G is a subfield
of N I F. Since V is finite, it follows from Theorem 415 that ' = 'V. Since then
AT, the mapping (168.16) is in fact onto the set of subgroups
of c.4.
In order to verify that this mapping is one-to-one, we suppose that for two subfields G, H of N I F the assigned groups are equal: G = R. Since, besides (168.27), VI = H also holds, it follows that G = H, i.e., that the mapping (168.16) is one-to-one. The mapping G
G, inverse to this, may be given, according to (168.16),
as G -* G. Since here G runs through all the subgroups A' of 4, this mapping agrees with (168.18). Consequently Theorem 417 is proved.
From the fact that (168.16) and (168.18) are one-to-one, and from (168.1), (168.2) we obtain the rules (168.19), (168.20).
In order to prove (168.21) we bear in mind that {G1, G2} Q G1, G2. According to this and (168.1), {G1, G2} c Gl, G2 ,
thus
{G2} S G1 fl G2 . On the other hand, the right-hand side is evidently a part of the left-hand side, whence (168.211) follows.
Similarly (168.221) is obtained from (168.2). (i = 1, 2). In order to prove (168.212) we apply (168.221) to By the subsequent application of the mapping X -k X, because X = X, we obtain (168.212).
FUNDAMENTAL THEOREM OF GALOIS THEORY
661
Similarly we obtain (168.222) from (168.211) by application to G. = Xt (i = 1, 2). Consequently Supplement I is proved.
In order to prove Supplement 2 we denote the elements of G by a. Then aG consists of the elements am. Consequently, the group aG consists of those elements s of 4 for which the condition
saa = as
(168.28)
is satisfied for all or E G. For (168.28) we may write a-1saa = or. Thus the required s are those elements of d for which a'sa E G, i.e., s E aGa-1. Hence Supplement 2 is proved. Lastly, in order to prove Supplement 3, we must notice that, because of Theorem 331, G I F is normal if, and only if, G has no conjugates in N over F except itself. This implies that
aG = G
(168.29)
for all a E c.0. Since (168.29) means the same as aG = G, for which we can
write, according to (168.23), aGa-1= G, the above-mentioned condition consists in the fact that G is normal in 4. Hence the first assertion of Supplement 3 is proved.
For the following statements we assume that G I F is normal, i.e. a Galois field. For every a (E c4) let a1 denote the mapping
a-* ax of G. This is, according to the above, an automorphism of G I F. It is again evident that all the different al form a group c01. For this the homomorphism 4 '" X11 (a-* a)
holds trivially. Its kernel consists of the elements a of cP for which the equation as = a is satisfied for all a E G. Since, accordingly, this kernel is equal to G, the isomorphism
4 /G
c.01 (a G -* a)
(168.30)
follows. This results in
[N : F] = 0(4) = O(G) O(4l) whence, according to (168.17) and Theorem 292,
O(Q = [N : F] [N : G]-1= [G : F].
(168.31)
Now cPl is a subgroup of the Galois group of G I F. Both are of the same order because of (168.31), and so equal. This, with (168.30), completes the proof of Supplement 3.
GALOIS THEORY
662
THEOREM 418. Let N I F1, N F2 be two Galois fields and (41, CP2 their Galois groups. With respect to F = FI fl F2 as fundamental field, N is a Galois field if, and only if, the group c44 _ {c4l, 4t2} is finite, and then this is exactly the Galois group of N I F.
If N I F is a Galois field, then F is its Galois group. On the other hand, according to (168.212),
F = F1 fl F2={FI,F2}=(4I,c02)=4. Hence it follows that co is finite and is the Galois group of N I F.
If, conversely, c4 is finite, then, according to Theorem 416, the field N 14 is a Galois field. Since, moreover, according to (168.221), at = {41, (^ti2) = 41
n42= F1 fl F2 = F ,
Theorem 418 is proved. THEOREM 419. Let N I F be a Galois field and P I F a further field, both with a common overfield. Then the field {N, P} I P is a Galois field and every
automorphism of it is the extension of one, and only one, automorphism of N I F. Furthermore the Galois group of {N, P} I P is isomorphic with a subgroup of the Galois group of N I F.
For the proof we put N = F(O)
.
Then {N, P} = P((9).
Letf(x) and g(x) denote the minimal polynomials of $ over F and P, respectively. Then g(x) I f(x).
(168.32)
Since f(x) is separable, the separability of g(x) follows from (168.32). Since furthermore f(x) splits into linear factors over N, it follows from (168.32) thatg(x) splits into linear factors over {N, P}. Accordingly {N, P} I P is, in fact, a Galois field.
If 0 -* 0' is an automorphism of the last mentioned field, so that g(X) = 0, then, because of (168.32), f(O') = 0; we are therefore concerned with the extension of an automorphism of N I F. Since both automorphisms are uniquely determined by 19', one is uniquely determined by the
other. If we assign the first to the second, then evidently we obtain an isomorphism of the Galois group of {N, P} I P with a subgroup of the Galois group of N I F. Theorem 419 is thus proved.
FUNDAMENTAL THEOREM OF GALOIS THEORY
663
TmOREM 420. Let N I F be a Galois field with the Galois group 4 and
(F =) Go c G1 c
... c Gr (= N)
(168.33)
a chain of subfzelds of it, and let
(C4=)
'oD'r1D...Z) -'r(=1)
(168.34)
be the chain of the related invariance groups. All the fields
Gi I Gi_1 (i = 1,
..., r)
(168.35)
are Galois fields if, and only if, (168.34) is a normal series of 4. Furthermore, the Galois groups of the fields (168.35) are then isomorphic with the factor groups
,:77i-1/ri
(i= 1,...,r).
(168.36)
COROLLARY. The Galois group c.P of N I F is solvable if, and only if, there is a
chain of fields (168.33) in which all the fields (168.35) are Galois fields of prime degree. The theorem follows immediately from the Supplement 3 of Theorem 417. From Theorem 138 it then follows that the corollary is also true.
In certain cases we name the Galois fields according to the properties of their Galois group. Thus a field N I F is called cyclic, Abelian or solvable, respectively, if it is a Galois field and its Galois group is cyclic, Abelian or
solvable, respectively. Of course a Galois field of prime degree is always cyclic.
§ 169. Stickelberger's Theorem on Finite Fields
According to Theorem 306, the number of automorphisms of a finite field is equal to its degree, thus finite fields are absolute Galois fields and, according to the same theorem, cyclic. Although the fundamental theorem of the Galois theory is almost trivial for them, nevertheless its application yields valuable results. An example of this is the following important theorem). If f(x) is a polynomial of degree THEOREM 421 n without multiple factors over a finite field F of characteristic p A 2 and m the number of irreducible factors of f(x), then its discriminant is a square if, and only if 2 1 n - m. For another proof cf. the Example at the end of § 174. For the proof we may suppose that f(x) is a principal polynomial. We first consider the case wheref(x) is irreducible, i.e., m= 1. Then we have recourse to the finite field G = F(a) where f (a) = 0. Since G is (absolute) cyclic, 22/a R.-A.
GALOIS THEORY
664
so G I F is cyclic of degree n. Let s be a primitive element of its Galois group. Then
f(x) = (x - a) (x - sa) ... (x - sn-1a) We write S=
11 (sta - ska) 051_ 0),
where q is a prime number other than p and the fundamental field F contains the qth roots of unity, then there exists an element a of F such that G = F(.,/a). Let a denote a qth root of unity (in F), 0 a primitive element of G, for which G = F(d)
therefore holds. Also let s be a primitive automorphism of G I F. Then we call every expression (e, 8') = 0 + e(s8) + e2(s'o) + ... + eq-I(3q-I8,)
(172.2)
GALOIS THEORY
674
a Lagrange resolvent. It will be shown that among these Lagrange resolvents occurs a suitable element a of G. Because se = e we have
S(0, ,O) = s8 + e(s'0) + ... + eq-1(sgo) .
Since further sq = 1, comparison with (172.2) results in the equation
s(e, t) = e-1(e, ) . Then (172.3)
s ((e, o')q) _ (e, 0)q
On the other hand, it follows from (172.2) after summation for all the that
E (e, ) = q8 . e
The right-hand side is not 0, because p r q, therefore it lies outside F. Consequently, for at least one e, (e, t9) lies outside F and therefore is a primi-
tive element of G. We take such a e and put co = (e, 0) .
According to what has been said, G = F(co). Since s denotes a primitive element of the Galois group of G I F, it follows from (172.3) that coq belongs to F. Hence, Theorem 426 is proved. Theorem 426 even holds, according to FRIED (1956), for all natural numbers q such that p }' q. As regards the generalization of Theorem 426 for an arbitrary n (= 1, 2, ...) instead of q, see BoORBAKI (1939) and LUGOWSKI-WEINERT (1960).
THEOREM 427 (ABEL'S theorem). A polynomial (172.4)
of prime degree q over afield F is reducible if, and only if, a is a q`b power in F.
In this case (172.4) splits into linear factors over F if, and only if, F contains the q`s roots of unity. For the proof we suppose that (172.4) is reducible over F, i.e., there exists
a principal polynomial f(x) (E F[x]) of degree n (1 < n < q - 1) such that
f(x)Ixq-a.
(172.5)
Over a suitable extension field of F,
xq - a = (x - 13) (x - elI) ... (x -
eq-
)
(172.6)
675
CYCLIC FIELDS
with 994 = a,
e4 = 1
(172.7)
.
We take the constant term of f(x) in the form
(8 E F). From
(172.5) and (172.6),
P=
(172.8)
'0k0"
for some integer k, which we do not determine more precisely. According to (172.7) and (172.8),
N4=a"
Now the equation
qy + nz = I has a solution y, z (E J) because (q, n) = 1R.zIt follows that
a = qy+nz = (ayY)"
thus a is a qts power in F. From now on we suppose that a = 994 for some 99 (E F). Then
x4- (X =x4-994 has the factor x - 99. What has been stated so far proves the first part of the theorem. Since x4 - a = 994((99-Ix)4 - 1) ,
we see that x4 - a splits over F into linear factors if, and only if, the same holds for x4 - 1. This proves the second part of Theorem 427. The generalization of the first part of this theorem is the following THEOREM 428 (VAHLEN-CAPELLI theorem). Over afield F a polynomial
x" - a
(n ? 2; a E F, 0 0)
(172.9)
is reducible if, and only if, a = YRd
(d l n, > 1; j9 E F)
(172.10)
or
4 In,
a= -4y4
(y E F).
(172.11)
The proof is founded upon the following PROPOSITION. If f(x) is a separable principal polynomial and g(x) a nonconstant polynomial over a field F, and the irreducible factor decomposition
g(x) - 99 = fl gi(x) 1=1
(172.12)
676
GALOIS THEORY
holds over the extension field G of F, defined by
G = F(s)
(f(z9) = 0) ,
(172.13)
then the polynomial
h(x) = f(g(x))
(172.14)
has the irreducible factor decomposition over F:
h(x) _ [J N(g;(x))
(N denotes NG(X),F(X))
.
(172.15)
Thus the degrees of the irreducible. factors of h(x) are multiples of the degree
of f(x) and, in particular, h(x) is irreducible over F if, and only if, g(x) - 0 irreducible over G.
First of all we prove the Proposition. Because of (128.4), f(x) _ = N(x - 0). If x is substituted here for g(x), equation (172.15) follows from (172.12), (172.14). Since in this the factors of the right-hand side belong to F[x], we have only to show that they are irreducible over F. With this end in view we denote an irreducible divisor of h(x) over F by k(x). From (172.12), (172.15) it follows that for some i (= 1, ...,.r) gi(x) I k(x).
(172.16)
We can write
1(x) _ (x - iI) ... (x - 0.) so that N = F(t91,...,
(0 = 'b),
is the normal field of G I F. Because of the
separability of f(x), X91, ..., 0," are distinct. Now N(gi(x)) is the product of m factors, which result from g,(x) by applying the isomorphisms # --> 0, (s = 1, ..., m) of G I F (in N), and which, according to (172.12), are, in order,
divisors of g(x) - 01, ..., g(x) - 1,,, and consequently pairwise relatively prime. Since because of (172.16) they are all divisors of k(x), which follows
by applying the same isomorphisms, we find that N(gr(x)) I k(x).
Since the right-hand side is irreducible over F, it follows that every irreducible factor of h(x) over F is associated with one of the r factors of the righthand side of (172.15). This proves the Proposition. We shall now prove Theorem 428. For the assertion "if" this is simple, for if (172.10) holds, then
x"-a=x"-fld is divisible by x"" - f, but if (172.11) holds, then we have R
R
R
x" - a = x" + 4y4 = (x 2 - 2yx 4 + 2y2)( x 2 + 2yx 4 + 2y2).
CYCLIC FIELDS
677
Conversely, we suppose that (172.9) is reducible over F. We have to show that then (172.10) or (172.11) is true. If n is a prime number, then, according to the first part of Theorem 427, (172.10) is true. Now let n be composite. We suppose that the assertion is false for n but true for smaller n. Hence it follows that all the polynomials
xd - x
(d I n, < n)
(172.17)
are irreducible over F.
We denote by p (Z 0) the characteristic of F and first consider the case p,f'n. We begin with the case
n = q`
(q prime number, e >_ 2).
Since x4 - a, by (172.17), is irreducible over F and, because q # p, is even separable, a separable field G such that [G : F] = q is defined over F by G = F(O)
(VQ = 0C).
(172.18)
Since, moreover, we have
TV - a = (xV-')a - a, it follows from the reducibility of (172.9) and from the Proposition that
x°`-' is reducible over G.
-
19
(172.19)
If q # 2, it follows hence and from the induction hypothesis that there is an equation of the form
P_A4 On the other hand (since p
(AEG).
2), a = NGIF (0')
follows from (172.18). Since we now have a = (NGIF (A))4,
we have obtained the result that (172.10) is true for d = q.
If, however, q = 2, then it follows from the reducibility of (172.19) and from the induction hypothesis that
either 0 = (a# + v)2 or t = -4 (0 + v)4,
GALOIS THEORY
678
where µ, v denote suitable elements from F. In the first case 0 = µ2a + v2 + (21Av - 1)19,
thus µ2a + V2 = 0, 2µv - 1 = 0, and hence (because p
2)
a= -4v4. In the second case
0=0+ 4(µ2a+ v2+2µv+9)2= = 4(p2a + v2) + 16µ2 v2oc + (16p v(µ2a + v) +1)0,
thus (µ2a + v2)2 + 4µ2v2a = 0,
16µv(µ2a + v2) + 1 = 0,
and hence
j4.
1 8µv
So we see (in both cases) that (172.11) is t rue.
We now consider the case where n is not a prime power, i.e., it has a decomposition
n=uv
(u,vE7; (u,v)=1; u,vZ2).
Since, according to (172.17),
x" - a, x° - a are irreducible over F, and also
x"-a=(x°)"-a=(x')°-a, it follows from the Proposition that every irreducible factor of x" - a over F has a degree divisible by u and v, i.e., by n. This results in the contra-
diction that x" - a is irreducible over F. Lastly we consider the case p I n. Since, by hypothesis, a is not a pt power in F, the field G defined by
(P = a)
G = F ($)
(172.20)
is, according to Theorem 341, pure inseparable over F with [G : F] = p. Then
x"-a=(xp-," -#)°
(172.21)
An irreducible factor over F of this polynomial cannot be a power of .Cp-1n
- 0,
(172.22)
679
CYCLIC FIELDS
since 6, . . ., #°-' lie outside F. Consequently (172.22) is reducible over G. Therefore, because of the assumption, 4 1 n,
O= -4y4,
where fi, y are elements of G. (In the second case p must be odd, but we do not need this.) Since a = #P, it follows that a = (F'P)d
or
at
= -4 (yP)',
respectively. Since now G I F is pure inseparable and of degree p, it follows
from Theorem 343 that PP and yP lie in F. This contradiction completes the proof of Theorem 428. Cf. CAPELLI (1901), who considered merely a number field of finite degree.
Evidently the validity of Theorem 426 disappears if q = p. In this case, however, the following holds : TttEoi t 429. If G I F is a cyclic field of degree p and of characteristic p (p prime), then there is a A such that G = F(A) and AP - A E F.
Let s denote a primitive automorphism of G F and Y(s) the natural group ring of {s}. For an element of this and for an element of G we define the operator product n
w= E a,(s'w)
ais` 1
(ai E 7; w E G).
t=o
It is evident that G+ is transformed by this into an Y(s)-module (with the customary properties of an operator), furthermore F+ is an admissible submodule. In particular, (because sP = 1 and pw = 0)
(s- 1)P0.) =(SP- 1)w=0.
Now we take an co from G - F. For this (s - 1) w = sco - w 0 0, thus there is a k (1 _ 0) where p,?n and F contains the nth roots of unity, then we can obtain G from F by the adjunction of an nth radical.
§ 173. Solvable Equations
Let us take a field F. We take an integer t (>_ 0) and give, over F, polynomials of the form
9; = gi(x1, ..., x,_1)
g = 9(x1, .. , xr)
(1
(173.1)
and integers q1, . . ., qt (>_ 2). (In particular, g1 is an element of F.) If t91, ..., 0t are elements of an extension field of F, where
..., $;_1)
(1= 1, ..., t),
(173.2)
then we call
0 =g(131,...,$e)
(173.3)
SOLVABLE EQUATIONS
R
681
a radical expression over F belonging to the polynomials g1, ..., g1, g and the root exponents q1, . . ., q, The 01, . . ., 0, are themselves called generating radicals. Allowing for the obvious ambiguity, if necessary we say more precisely that 0 is a value of this radical expression. We can (173.3) an irreducible radical expression, if the generating radicals occurring in it are irreducible (i.e., every 13; is an irreducible radical over F (01, . . ., 0I-1)).
Radical expressions can be easily characterized. In order to do this, we put Gi = F(01,
. .
(i = 0, ..., t).
., 0I)
(173.4)
Then G; = Gr-1(01),
MP E G,_I
(i = 1, ..., t),
(173.5)
(173.6)
0 E Lr L.
Hence we see that 0 is a radical expression over F if, and only if, there exists a chain of fields
F=Go9G1c...9G,
(173.7)
such that every term arises from the preceding by the adjunction of a radical and 0 is an element of Gt. Incidentally we sometimes call (173.7) a chain of fields belonging to the radical expression 0.
As an example of a radical expression (with t = 2) we have:
x+ A V a+ (µ + v (.,/ a)3), J e+ or
+
(Ja)z
(a, K, ..., r E F).
We call the irreducible equation
f(x) = 0
(f(x) E F[x])
(173.8)
solvable (by radicals) if it has a root which is a radical expression over F. By the above this means that there is a chain of fields (173.7) whose last term contains a root of (173.8) and that moreover (173.5) holds. On the other hand, suppose that f(x) is the product of separable polynomials over F. Let N I F be a splitting field of f (x), which is thus a Galois field. We call it and its Galois group the Galois field and the Galois group of the equation f(x) = 0, respectively (but cf. § 174). Since N I F is, apart from equivalent extensions, uniquely determined, the Galois group of an equation is uniquely determined to within isomorphism. The solution of equations by radicals was the main problem in early
algebra which had no other (algebraic) methods for the solution of an algebraic equation. Although Theorem 299 entirely solves this problem by the help of the splitting field, nevertheless solvable equations still form
an important branch of algebra, this being explained by the fact that
GALOIS THEORY
682
at least one solution of them is easily determined, i.e. by reducing them to
the solutions of binomial equations. One of the best results of Galois theory will turn out to be the close connection between the solvability of an equation and its Galois group. Before beginning our considerations we take a simple example: every cyclotomic equation F (x) = 0 is solvable (over . o), since every root of it may be given as the
radical expression V1. Still this is not a satisfactory solution of the problem of determining the solutions of this equation, since not all the values of this radical are solutions, but only the primitive nth roots of unity among them. This remark is reasonable,
in particular, already for the case n = 3, where the equation x2 + x + 1 = 0 is involved;
now, besides the previous expression, the radical expression 2 (-1 + V=3) is also a solution of the equation, and that, "fortunately", for all its possible values, therefore it is far preferable to the other.
For a solvable equation (173.8) an irreducible radical expression is called
a solution formula for this equation, if all its values are solutions of the equation and, conversely, all the solutions are furnished by its values. Next we prove the following THEOREM 430. Every irreducible radical expression over a field F, which furnishes a solution of an irreducible (solvable) equation (173.8), is a solution formula for this equation.
For this we suppose that the radical expression 0 given by (173.3) is irreducible and a solution of (173.8). Here 0, is a solution of the equation xq` - g,(01, ..., oi_1) = 0
U= 1, ..., t)
(173.9)
whose left-hand side is irreducible over F(81, . . ., O'i-D We consider a further value 0' of the same radical expression. For this
0' = 01, ..., 07,
(173.10)
where 0l, ..., 6' are elements of an extension field of F such that 0 is a solution of xqt
- g.(1 , ..., O'_-D = 0.
(173.11)
We have to prove that 0' also satisfies the equation (173.8).
To do this, we show that the isomorphisms F(01,
.,,&j) I F x F(a9l, ... 0r) I F (01 -* #1, ..., 0, -+ 0j'),
(i = 1, . . ., t) (173.12)
hold. For i = 1 it is true since, according to (173.9) and (173.11), 01, O' are solutions of the equation xq' - g1 = 0, irreducible over F. We then consider an i (= 2, . . ., t) and make the induction hypothesis that the
683
SOLVABLE EQUATIONS
isomorphism (173.12) holds for i - 1 instead of i. By this the left-hand side of (173.9) is converted into that of (173.11). Since, of these two polynomials, the first is irreducible over F(01, . . ., #t_1), and since t; and 0; are roots of (173.9) and (173.11), respectively, it follows from Theorem 298 that (173.12) is true for the i concerned and therefore in general.
We now take into consideration the isomorphism (173.12) for i = t. Because of (173.3) and (173.10) this maps 0 onto 0'. Hence it follows that, together with 0, 0' is also a root of (173.8) as asserted. We have still to prove that conversely every root of (173.8) is a value
of the radical expression (173.3). With this end in view we consider a normal field N of F(01, ..., i9) 1 F. From f($) = 0 and (173.3) it follows that N contains a splitting field of f (x). Therefore it is sufficient to prove that all the zeros of f(x) lying in N occur among the values of the radical expression (173.3).
Let 0' be such a zero :
f(0') = 0, 0' E N. By Theorem 329 there exists an isomorphism e -> sLo of F(01, ..., 0) 1 F such that 0' -1- s O. Now from (173.3)
so = 9(sh'1, ..., so), and it follows from (173.9) that sli, is a root of x41 = 9,(s#1, ..., s$,_1)
(i = 1, ..., t).
Accordingly, 0' = sO is, in fact, a value of the radical expression (173.3). Consequently Theorem 430 is proved. As a further preparation we prove the following THEOREM 431. Every normal subfield G I F of a solvable field N I F is solvable.
Since the assertion is trivial for G = F and N, we may assume that F c c G c N. Let co denote the Galcis group of N I F, which is, by hypothesis, solvable. Further let cY, denote the invariance group of G. According to Supplement 3 of the fundamental theorem of GALOis theory (Theorem 417),
Y is a normal subgroup of cPi, furthermore the Galois group of G I F is . isomorphic with the factor group d I Now, according to the JORDAN-HOLDER theorem (Theorem 136), the
normal series c0 X D 1 may be refined to a composition series
(`j _)Xp aYl
...
Xr (_ t) D ... D 1.
According to Theorem 138 in this all the factors Xi/
r+l are of prime
GALOIS THEORY
684
order. According to the second isomorphy theorem (Theorem 126')
`4/X D t1/
D ... D Xr1J7(= 1)
is likewise a composition series whose factors are again of prime order. Accordingly 4g/P, on account of Theorem 138, is solvable, so that the proof of Theorem 431 is complete. THEOREM 432 (main theorem for solvable equations). Let
f (X) = 0 (f(x) E F[x])
(173.13)
be a separable equation over a field F of characteristic p (Z 0), further let cP be its Galois group. If (173.13) is solvable, then C4 is likewise solvable. If, conversely, 4 is solvable and in the case p > 0 all the prime factors of 0 (4) are smaller than p, then (173.13) is likewise solvable and has a solution formula in which all the root exponents are prime numbers, which do not exceed the greatest prime factor of 0 (4). In order to prove the first part of the theorem we suppose (173.13) to be solvable. Then there is a radical expression t over F with f(t9) = 0.
(173.14)
On account of the rule
we may suppose that in 0 all the root exponents are prime numbers. These, and the corresponding radicals, are denoted by q1, . . ., q, and' 191, ..., so that i97' E F(191, ..., X9,_1)
(i = 1, ..., t),
0EF(61,.. hold. We show that q1i . . ., qt (if p > 0) may be supposed to be different from p.
First let us consider the case, where for some i (>_ 0)
g1, ,qr#p; q,+1=...=q,=p By Theorem 341, of the fields F(61,
..., 0,) 1 F,
F(191, ..., 't) I F(01, ..., 0t)
the first is separable, the second pure inseparable. Since 19 is separable, it follows that 19 E F(01, ..., 19,). Accordingly, the (superfluous) radicals a91+1, , 0t may be omitted, so that the assertion is true for this case.
SOLVABLE EQUATIONS
685
Secondly consider the case, where there exists an i with
qi = p, q;+1 9 p It suffices to prove that 0 may be transformed into a radical expression, where the root exponents
q1, . , q;- l, q;+1,p,p,q,+2,-- ,qi occur in order of succession, since hence, and from what has been said above, the assertion will follow by induction. For our purpose we insert in the sequence of radicals 791, . . ., 09, between
19; the term
u*_0+1' from which we obtain the sequence $1,
..., $;-1, (97 (_ X91+1), 991, O'i+1, ..., 19'.
Since
0*Q'+' = ( +i')° E F(19i , . . ., 9f) c F(191, . . ., 0r-1), = 07' E Fo91, . . ., 0,-,) c F(61,
..., 49;-v $*),
091P+1 = 6* E F(01, ..., 09;-1, #
,
01)
hold, the truth of the assertion follows. We now put G = F(01, ..., 0,). The splitting field of the polynomials
x41 - 1,...,x4'over G is denoted by G'. Then G' contains a splitting field F of the san. polynomials over F, furthermore evidently
G'={F',G}. By Theorem 424, F may be obtained from F by successive cyclic field extensions. Since finite cyclic groups are solvable, it follows from Theorem 420 that here extensions of prime degree suffice.
Further let us denote the normal field of G' F by N. Together with this, N I F is separable and thus a Galois field. Lets denote an automorphism of it.
Because of the supposition we can put (Oct-, E
Hence sO,
=
°j sx; --I.
GALOIS THEORY
686
Evidently we obtain N from F if we first successively adjoin all the conjugates sh1 of #,, by which we obtain a normal field over F, then we successively adjoin to this all the conjugates she of 02, by which we again obtain a normal field over F, and continue in the same way. But since F contains all the q;th roots of unity (i = 1, . . ., t), it follows from Theorem 427 that the adjunction of each s1, effects a cyclic extension whose degree is q, or 1. Together with the preceding statement this results in the existence of a chain of fields
F=No cN1 c... cN,=N, in which all the N,+1 1 N, are Galois fields of prime degree. This means, on account of Theorem 420, that the Galois group of N I F, i.e. N I F itself, is solvable.
Since, on the other hand, because of (173.14), N I F contains a splitting
field of f(x) over F, i.e., the Galois field of equation (173.13) as a normal subfield, it follows from Theorem 431 that this field, and thus also the Galois group c0 of (173.13), is solvable.
In order to prove the second part of the theorem, let us suppose that it is true for smaller O(o2). We denote by G I F the Galois field of the equation (173.13) and consider a composition series of its Galois group: (`xd =) -7p D'71 D ... Z) -7r (= 1). To this belongs, according to Theorem 420, a chain of fields
(F=)Go cGxc
... cG,(=G),
in which G; I G,_1 is cyclic of a prime degree q, (i = 1, ..., r). Hence
O(d)=q1...q,. A splitting field of the polynomials
xqI - 1,...,xgr- 1 over G is denoted by N. This contains a splitting field No of the same polynomials over F, which is obtained from F by the adjunction of all the q,th roots of unity (i = 1, . . ., r). Because of the induction hypothesis it follows from Theorem 424 that we can obtain No from F by successive adjunctions of irreducible radicals, where only prime numbers occur as root exponents which are smaller than the greatest prime factor of 0(c4). Finally, we consider the fields
N,={No,G,}(9 N)
(i= 1,...,r; N,=N).
Since we have
N, = {Ni-,, Gi},
687
SOLVABLE EQUATIONS
Ni I N;_1 is, according to Theorem 419, a Galois field, and [Ni Ni-11191 Since, furthermore, N,_ 1 contains all the qih roots of unity, either we have
Ni = Ni_1 or we obtain Ni from Ni_1 according to Theorem 426 by the adjunction of an irreducible qih radical (i = 1, . . ., r). If these radicals are adjoined to No, we obtain N, = N. By consideration of the above, because G c N. the second part of Theorem 432 follows. EXAMPLE 1. In the second part of Theorem 432 the condition "in the case p > 0
all the prime factors of 0(4) are smaller than p" may not generally be omitted. In order to give examples for this, we remark in advance that in Theorem 432 for a finite field F (since then by (173.13) a finite field is again defined) there are very simple conditions, as in this case equation (173.13) is always solvable and its Galois group c1/j is cyclic. For, the solvability of (173.13) follows from the fact that every element of a
finite field is a radical, and is either 0 or a root of unity. Nevertheless it can happen
that (173.13) does not have a solution formula. This is trivial in the special case F = .`, ffor non-linear equations (173.13)] since E has only the two elements 0, 1, therefore over it there exist no irreducible radicals at all. (There is no contradiction to Theorem 432, since now the above condition is not fulfilled.) EXAMPLE 2. Let f(x) be an irreducible polynomial of degree p over .2",. According to the above observation the equation f(x) = 0 is solvable, yet it has no solution formula. Otherwise it must also have (cf. the beginning of the proof of Theorem
432) a solution formula wherein only prime numbers other than p occur as root exponents, but this is impossible, since, by the equation f(x) = 0, a field of degree p is defined. EXAMPLE 3. Over the rational function field F = .z-$ (z) let us consider the separable equation
x2+x+z=0.
Its Galois group is of second order, thus solvable, although the equation itself is not solvable. (Of course, the condition quoted in Example I is not fulfilled.) Every root of unity over F is algebraic over 9 g. Thus if we adjoin to F arbitrary roots of unity, then the equation (because of the transcendence of z) remains irreducible. After adjoining suitable roots of unity we may always restrict ourselves to the adjunction
of irreducible radicals. When, for such an adjunction, the root exponent is odd, then the equation again remains irreducible. Finally, the adjunction of square roots leads to a pure inseparable extension, consequently it does not furnish a solution of the equation.
§ 174. The General Algebraic Equation First of all let us consider over a field F an equation (174.1) f(x) = 0, whose left-hand side is the product of separable principal polynomials
which we suppose without loss of generality to be different. Let ce, denote the Galois group of this equation. We put f (x) = (x - oc1) ... (x - xn) 23 R.- A.
(%I,. .., an
G),
(174.2)
GALOIS THEORY
688
where G = F(a1, . . ., an) denotes the Galois field of (174.1) over F, so that 4 is simultaneously the Galois group of G I F. If now t' -* n' is an arbitrary automorphism of G I F, then, because of (174.2),
J (X) = (x - aI) ... (x - an). Hence it follows that al ... and
a1...an
(174.3)
is a permutation of the roots of (174.1). Since furthermore al, ..., an are generators of G (over F), so the above-considered automorphism N -> t)' is already determined by the permutation (174.3); consequently the permutations
(174.3) must constitute a group isomorphic with c4. Generally we tacitly identify the Galois group of the equation (174.1) with this permutation group of the roots al, . . ., an. The index of this group, in the full permutation
group of the set _ 0). Here we always suppose that p # 2
EQUATIONS OF SECOND, THIRD AND FOURTH DEGREE
693
and when n = 3, 4 we also assume that p 3. In both the last cases we also assume for the sake of convenience that al = 0, by which, according to the inference of the preceding paragraph, we lose no generality. Since the full permutation group of at most, fourth degree is solvable, by Theorem 432, equation (176.1) is solvable under the convention adopted,
and it also has a solution formula. We now intend to deduce this formula.
The procedure consists in assuming from the outset a splitting field ..., x,J, where xI, . . ., x denote all the roots of (176.1), then to
F(x1,
construct this splitting field by the help of adjunctions of irreducible radicals
and to express the roots xI, . . ., xn by the same radicals. In this way we have not only set up a solution formula, but also simultaneously effected the actual decomposition of the left-hand side of (176.1) into the form
x"+alx"-I+...+a,, =(x-xl)...(x-xn), where the x1, . . ., x,, are given by radical expressions over F. In connection with this we shall show that this decomposition also remains valid in the
reducible case, by which we obtain the complete solution of (176.1). We denote by D the discriminant of (176.1). We may put
JD = II
(xi - xk),
(176.2)
since, according to Theorem 277, the square of the right-hand side of (176.2) is equal to D. It should be noted that, by Theorem 434, the Galois
group of (176.1) over F(JD) is equal to a subgroup of the alternating group vf,, . CASE n = 2: (176.1) now reads
x2+alx+a_-=0.
(176.3)
On account of (113.4) we have
D = ai - 4a2,
(176.4)
JD = xl - x2 .
(176.5)
further because of (176.2)
2 = 1, so, according to the above statements, F(xl, x2) = F(-,/Y)). Accordingly, x1, x2 may be expressed as elements of F(.,/5). This can be done using equation (176.5) and the equation As t
xl+x2=--a1
GALOIS THEORY
694
arising from (176.3). Then we obtain
x1=
2
(-al+ / ), x2= 2 (-a1-.J),
(176.6)
where we have taken into consideration that p # 2. Since, conversely,
x2+a1x+a2=(x-x) (x-x2) always follows from (176.4) and (176.6), the formula (176.6) gives all the solutions in the present case. CASE n = 3: We now write (176.1) (with a1 = 0) in a simpler notation in the form
x-3+ax+b=0.
(176.7)
In order to avoid the necessity to distinguish between different cases, we assume that the coefficients a, b are indeterminates over F. This means that (176.7) is regarded as being over the rational function field F (a, b) as a fundamental field, which we still denote, for the sake of simplicity, by F. According to the final observation of the preceding paragraph, (176.7) is then like the general equation of third degree, by which we mean that Theorem 433 can be applied to equation (176.7), i.e., this is affectless. The result for this case can be applied to every other case, if we replace the indeterminates a, b by elements of the fundamental field originally given. Since equation (176.7) is affectless, it follows, as above, that its Galois group over F(,FD) is equal to f3. On account of (113.5) we now have
D = - 4a3 - 27b2.
(176.8)
Further, according to (176.2),
VD = (x] - x2) (x1 - x3) (x2 - x3) =
xix., - Y- x32Lx3 , 3
(176.9)
3
where the symbol > is defined for arbitrary functions f(xl, x2, x3) by 3
E J (XI, x2, x3) = J
x2, x3) + fl X21 x3, x1) + f(x3, x1, x3]
a
As we have O(vE3) = 3, the Galois field F(x1, x2, x3) I F(N46) is cyclic of degree three. We adjoin both third roots of unity
_=
2 (- 1 + J- 3), 02= 2
(176.10)
EQUATIONS OF SECOND, THIRD AND FOURTH DEGREE
which is equivalent to the adjunction of the radical ,
695
. Of course,
also the field F(x1, x2, x3, J-3) 1 F(JD ,) is cyclic of degree three. We take the Lagrange resolvent [(172.2)]: y = x1 + Cx3 + e2x3 .
(176.11)
(The following will be a similar construction to that in the proof of Theorem 426.) From (176.11) and p3 = 1 we have
y3 = E xi + 3e E xix2 + 3e2 E xix3 + 6x1x2x3 . 3
3
3
After the substitution of (176.10) and then (176.9) we obtain y3 =
xi 3
2
6
xix2 + 6x1x2x3 + 2 V - 3 JD ,
(176.12)
where E denotes the summation over all the permutations of <x1, x2, x3>. 6
From the formula p3 = - ai + 3a1a2 - 3a3 (§ 115, Example 1) we get Z Al = - 3b. 3
Further,
Y, xl 3 After substituting
E x1x2 = E xix2 + 3x1x2 x3. 3
6
Y_ x1 = x1 + x2 + x3 = 0,
x1x2x3 = - b
3
we obtain
E xix2 = 3b. 6
Thus, from (176.12),
y= 3I- 2 b+ 2 y-3D , where
- 3D =
(176.13)
JD (therefore the occurrence of J - 3D does
not imply that any new radical has been adjoined). Since y is obviously an irreducible radical of degree three over the field F(%D , /----3), it follows that
F(x1, x2, x3, J- 3) = F(/D, J- 3 , y).
(176.14)
Hence, it is possible to compute x1, x2, x3 as elements of the right-hand side of (176.14). 23/a R.-A.
696
GALOIS THEORY
For this purpose we have at our disposal the equations
x1+x2+x3=0
(176.15)
and (176.11). Similarly to (176.11) we put (176.16) y' = x1 + e2x2 + ex3. For this a formula similar to (176.13) holds (see below), but it can also be computed as follows: because e3 = I and (176.10),
yy'
_ 3
+ (e + e2) E xlx2 =
3
3
xi -
3
x1x2 = (Z x1)2 3
- 31 x1x2 3
follows from (176.11) and (176.16). According to (176.15) and
E x1x2 = x1x2 + xlx3 + x2x3 = a 0
we have
(176.17)
YY' = -3a, therefore y' is determined in terms of y.
We now compute the roots x1, x2, x3 from (176.15), (176.11), (176.16),
taking into consideration e2 + e + 1 = 0 and p # 3,
(Y+y'),
x1=
x2 = 3 (e2Y + ey') ,
(176.18)
1
X3 = 3 W + e2Y')
The equations (176.8), (176.10), (176.13), (176.17), (176.18) together furnish the required result. Since with the radical expressions x1, x2, x3 so computed, the decomposition
x3+ax+b=(x-x1)(x-x2)(x-x3) holds, this also remains true in the case where the indeterminates a, b are replaced by arbitrary elements of the original fundamental field, so that our result, apart from the case when a = 0, is generally valid.
We can express the result in another form too. If /- 3 is replaced by
- J- 3, e becomes e2, thus (176.11) becomes (176.16). Accordingly, from (176.13) we obtain
y' = 3 -
22
b-3
- 3D
(176.19)
697
EQUATIONS OF SECOND, THIRD AND FOURTH DEGREE
with respect to which it should be noted, however, that the radicals (176.13), (176.19) are not arbitrary inasmuch as (176.17) must bold for them. After substituting (176.8), (176.10), (176.13), (176.19) in (176.18) we now obtain all three roots of the equation (176.7) in the form
x=
a
-
b
+ JI 212 + 13 I3
+ s - 2-
X2)2
+ Tr (176.20)
where the radicals occurring here are always to be chosen so that the
product of the two terms of the right-hand side should be -
. This
formula (176.20) is called the Cardan formula (for the equation of3 degree of course, the case a = 0 is no longer an exception. Finally, denote the two radicals in (176.20) by u and v, so that (176.20)
then assumes the form x = u+ v uv = six pairs of values are possible: u, v; 0u, 02v;
02U,
I. Then for u, v the following 3
Ova v, u; 0v,
e2U;
e2v, eu ,
of which the first three already furnish the three solutions of (176.7).
CASE n = 4: We can now take the given equation (176.1) (because a1= 0) in the form
x4+ax2+bx+c=0,
(176.21)
Again it will suffice to consider the case where a, b, c are indeterminates, i.e., that a rational function field F (a, b, c) is the fundamental field, which we denote, for the sake of simplicity, by F. The solution of (176.21) will be obtained by its reduction to the previous case n = 3. Because of the supposition, the Galois group of (176.21) is the full permutation group '94. All its composition series are,, according to Theorems 84, 85, the
.94D f4JG4z) HD 1,
(176.22)
where G4 denotes the four-group consisting of the permutations 1, (1 2) (3 4), (1 3) (2 4), (1 4) (2 3) ,
(176.23)
and H is an arbitrary subgroup of order two of G4. The invariance fields belonging to the subgroups (176.22) form a chain
F c F(,JD) c F. c F12 c F(x1, . . ., x4)
(176.24)
OAI.OIS THEORY
698
with [Fe : F] = 6, [F12 : F] = 12. Our next intention is to construct the field F6.
With this end in view we consider the elements
Y1=(x1+x2)(x3+x4), (176.25)
Y2 = (X1 + x3) (x2 + x4) ,
Y3 = (x1 + x4) (x2 + x3)
and assert that
F6
(176.26)
= F(Y1, y2, y3).
On the one hand (176.25) furnishes fixed elements of the permutations (176.23), whence F(Y1, Y2, Y3) c Fa
(176.27)
.
On the other hand, we determine the equation of degree three with the roots yl, y2, y3. For this we must determine the elementary symmetric polynomials of yl, y2, y3. These are in shortened notation [Yll = 2[x1x2] = 2a [YIY21 = [xix221 + 3 [xix2x3] + 6 x1x2x3x4
YLY2y3 = [x3jx2x3] + 2 WX2x3x4] + 2 [x2jx2x3] + 4 [xlx2xsx4l
where, e.g., [x2jx2] denotes that symmteric polynomial whose terms all have the coefficient 1 and result from xix22 by permutations of <x1, ..., x4). Since [xix2]2 _ [x2jx22] + 2 [xlx2x3] + 6 x1x2x3X4 = a2,
[x1] [XIX2x31 =
[x2jx.2x3] + 4 xlx2x3x4 = 0
,
X1x.2x3x4 = C
[xlx2x3] = - 4c,
[xix2]=a2+2c, [YLY2l =a2-4c. Furthermore [x1] [x1x21 [xlx2x3l = [xix2x3] + 3 [xlx2x3x4] + 3 [x1 2xg] + 8 [x2jx2x3x4] = 0 ,
[JGIX2X3X4] + 2 [ 'IX2x3X4] = 0 , [X1X2x3]2 =
[X1X21X1X2x3X1 =
[x2jx2x321 + 2 [x2Ix2X3x41 = b2,
[xix2x3x4] = ac
EQUATIONS OF SECOND, THIRD AND FOURTH DEGREE
699
hold, so that we have
[xixaxs] _ - 2ac + bz , 2ac, [-41x2x3x4] [xix22x3] = 4ac - 3b2 ,
Y1Y2Ys = - Y.
Consequently yl, y2, y3 are the three roots of the third degree equation
y3-2ay2+(a2-4c)y+b2=0,
(176.28)
which (or, at least its left-hand side) is called the cubic resolvent of equation
(176.21). [Often y3 + 2ay2 + (a2 - 4c) y - b2 = 0 is also called so.] We shall show that the discriminant of (176.28) agrees with that of (176.21). We obtain from (176.25)
Y1 - Y2 = (xl - x4) (x3 - x2) ,
Y1 -/Y3 = (x1 - x3) (x4 - x2) ,
Y2 - Y3 = (X1 - x2) (x4 - x3)
whence the assertion follows. Since D is not now a square in F, it follows from what has been proved,
and from Theorem 434, that the equation (176.28) is affectiess, i.e., [F (3'1, )'21 Y3) : F] = 6. Hence, and from (176.27), (176.26) follows. From (176.28), y1, y2, y3 may be computed by the Cardan formula. Accordingly, it still remains to construct the extension field F(xl, ..., x4) of F(y1, y2, y3) by the adjunction of radicals and to express XI, ..., x4 in terms of these radicals. This is easy since this field, according to (176.22), (176.24), has the four-group G 4 for Galois group, thus the adjunction of two suitable square roots will be sufficient. In order to carry through this construction, we take into consideration XI + x2 + x3 + X4 = O ,
(176.29)
therefore (176.25) may be written as y1 = - (X1 + x0)2,
y2 = - (XI + x3)2,
y3 = - (X1 + x4)2.
Accordingly,
x1 + x2 =I1 X1 + X3 =
-Y21
X1 -+4
Y3,
(176.30)
whence evidently F(xI,
. .
., x4) =
(176.31)
GALOIS THEORY
700
According to what was said in (176.31) a square root is dispensable. We also prove this directly by showing that Yi -,I-Y2 %-Y3
= -b .
(176.32)
Since, for the product of the left-hand sides of (176.30),
(x1 + x2) (xi + x3) (xl + x4) = xi [x11 + [xlx2x3l = -b, so
(176.32) is true. Lastly we obtain from (176.29), (176.30) 1
x1= 2 (,/-Y1+
2+
3),
/
G,
X2
(176.33)
1
x4 = 2 (- N
/
Yl -
/
-Y2 + V -Y3)
These formulae thus furnish all the roots of the biquadratic equation (176.21), where yl, y2, y3 denote the three roots of the cubic resolvent (176.28) and the three square roots in (176.33) satisfy condition (176.32). The solution formula (176.20) is wrongly attributed to CARDAN as he obtained it from TARTAGLIA, but it is not known whether the latter was the discoverer of this formula. Formulae (176.33) are essentially due to FERRARI. EXAMPLE 1. We obtain the Cardano formula by a simple but clever method as follows: put x = u + v, with two new unknowns u, v, in (176.7). We then have
(u + v)3 + a(u + v) + b = 0, i.e.,
u3+v3+(3uv+a)(u+v)+b=0.
if we subject u, v subsequently to the condition
3uv+a=0, then the former equation reads
u3 + v3 + b = 0. The solution of the system consisting of the last two equations entails no difficulties
and leads at once to (176.20). Of course, this elementary process gives no real insight into the cubic equation. (See also the following paragraph.)
EQUATIONS OF SECOND, THIRD AND FOURTH DEGREE
701
EXAMPLE 2. Also the biquadratic equation (176.21) admits an elementary solution, if we seek a factor decomposition
x°+axs+bx+c=(x2+rx+s)(x2-rx+t) of its left-hand side. For this purpose one must determine a solution r, s, t of the system of equations
s+t-r$=a, r(-s+t)=b, st=c,
which can be easily done. EXAMPLE 3. Over a field with characteristic other than 2 the polynomial
y2 + bx +
JOY -
2 ay - 4 (a$ - 4c)
4
n the two indeterminates x, y has the property that its discriminants, with respect to y and x, are the biquadratic polynomial x4 + ax2 + bx - c and its cubic resolvent y3 - 2ay2 + (a2 - 4c)y + b2, respectively. [Cf. R DEI (1959c)].
§ 177. The Irreducible Case
First of all let us consider an arbitrary polynomial over Col without multiple factors which we decompose into the product of irreducible factors : ,f (x) = 11(x) ...1,(x) q1(x) ... qs(x)
The lk shall be assumed to be linear and the qk quadratic. Since the discriminant of an 1k is equal to 1 and that of a qk is negative, it follows from the multiplication theorem for discriminants (§ 113, Example 2) that the discriminant of J (x) is positive if, and only if, s is even, i.e., the number of
the non-real zeros of f(x) [in 501(i)] is divisible by 4. We now examine a cubic equation over 9'tO) without multiple roots, which we assume without restriction of generality to be in the form
x3+ax+b=0,
(177.1)
further we denote the discriminant of (177.1) by D. According to the above, the number of the real roots of (177.1) is equal to 3 or 1 according as D is positive or negative.
On the other hand, we know that for computing the roots of (177.1) according to the Cardan formula it suffices to adjoin to the fundamental field for a fixed value of _-3D the three values of 22
b+2
(177.2)
702
GALOIS THEORY
[cf. (176.13)]. If there is only one real root, i.e., if D < 0, then among the values of the radical (177.2) a real one always occurs, so that this single real
root of (177.1), on account of the Cardan formula, may be computed within .7(o. It is different if all three roots are real, i.e., if D > 0, since then the radical (177.2) has no real value at all. This case of the cubic equation, in which it is has three real roots, is called the irreducible case. The terminology refers to
the fact that in this case equation (177.1), applying the Cardan formula, is reducible only after the adjunction of non-real radicals. [From (176.18) it
is clearly seen how the three real roots then arise as a sum of pairwise conjugate complex numbers.] One might think that in the irreducible case the difficulty could be avoided; namely, the Cardan formula might be replaced by another, which
in the case of three real roots would only require the adjunction of real radicals. The centuries old efforts regarding this problem were brought to an end with the Galois theory in the following theorem. THEOREM 436. If f(x) is an irreducible polynomial of third degree with three real zeros over a subfield F of then none of these is expressible by real radicals (over F). For, let D (> 0) be the discriminant of f(x). After the adjunction of .JD
we obtain a field F(.JD) of at most second degree over F, so that f(x) remains irreducible. Hence and from Theorem 434 it follows that the Galois
is the alternating group 'e, group of f(x) over We suppose that f(x), after the adjunction of certain further real radicals 01, ...,14',,, will be reducible. We may assume that all the root exponents therein are prime numbers. In the considered sequence of radicals there is a term #k= /w (q a prime number) such that, of the fields G = F(.[, t91, ..., Ok-1),
G1
= G (;/w)
f(x) is irreducible over the first and reducible over the second. If the radical Jco is reducible, then co is, according to ABEL'S theorem (Theorem 427), a qt' power in G, whence, because .Jw E 9o), we obviously have .,/w E G, G1= G. But this is absurd, thus ,A.) must be irreducible. Hence [Gl : G ] = q. Since, on the other hand, G I contains a zero of f(x), we have 3 1 q, i.e., q = 3. According to what has been said above, the Galois field of J (x) over G is of degree 3. Since G1 contains a zero of f(x) and is likewise of degree 3 over G, so G1 I G itself must be a Galois field. Thus the irreducible polynomial
f(x) over G decomposes into linear factors over G.I. Likewise, according to Theorem 427, G1 must contain all third roots of unity which is impossible,
because G1 9
o). Consequently the above theorem has been proved.
EQUATIONS OF THIRD AND FOURTH DEGREE
703
§ 178. Equations of Third and Fourth Degree over Finite Fields
If f(x) = 0 is an equation over a field F, then the question arises as to how many roots lie in the fundamental field F itself. If, e.g., F = .off, and f(x) is a polynomial of third degree (over Y(0)) without multiple zeros, then the number of these roots of f(x) = 0 is 1 or 3. On the other hand, for other fundamental fields this number can have the three values 0, 1, 3, e.g., if F is a finite field. Here, and from now on, we shall deal only with this case and we shall examine the question raised with regard to equations of third and fourth degree. The KONIG-RADOS theorem (Theorem 313) answers this question generally for finite fields, for equations of arbitrary degree, but much simpler results of a quite different kind hold in the special cases mentioned here. THEOREM 437. Let F denote the finite field of characteristic p (> 3) such that O(F) = q, further let n =
q-1
q+1 or
6
(178.1)
6
denote the integer nearest to 6 . Let the cubic equation
x3+ax+b,0
(178.2)
over F be given with the discriminant
D = -4a3+27b2#0.
(178.3)
If D is not a square in F, then (178.2) has exactly one root in F. If D is a square in F, and
t, (-27b2D'1) = 0 .
(178.4)
where
V(X)=l
2n
Jx2...+2n
(178.5)
then (178.2) has three roots in F. In the remaining cases (178.2) has no roots in F. In order to prove the theorem, let us denote by nz (= 0, 1, 3) the number
of roots of equation (178.2) lying in F. Then we have m = 1 if, and only if, the left-hand side of (178.2) splits into the product of two irreducible factors over F. This is the case, according to STICKELBERGER'S theorem
(Theorem 421), if, and only if, D is not a square in F. Hence we have shown that m = I if, and only if, D is not a square in F.
704
GALOIS THEORY
Henceforth we suppose that D is a square in F, whence m = 3 or 0. It suffices to prove that (178.4) is true if, and only if, m = 3.
Let Fe denote the field of 6" degree over F, for which we then have O(F6) = q6. We denote by F2 and F3 the subfields of F of degrees 2 and 3, respectively, over F. Let x1, x2, x3 be the roots of (178.2) in F6. But since
m = 0 or 3, so (178.6)
xI, x2, x3 E F3.
Since a -+ aq is a primitive automorphism of F6 I F, if m = 0 we can choose
the notation so that (178.7)
X1= x2, X2= X3, x3= x1. Because of the definition of n,
q=6n+e
(e=±1).
(178.8)
Certainly F2 (possibly also F) contains the radical ,/-3, therefore, also, the
third roots of unity
o2(-1+
,
022(-1-%J-3).
(178.9)
Furthermore, because of the equivalence of the propositions
OEF, x2+x+lIx°-1-1, x3-1Ixq-1-1, 31 q-1, the rule
EFae=1
holds.
(178.10)
We now consider the Lagrange resolvents y = x1 + 0X2 + 02x3 , y' = x1 + 0`'x2 + 0x3.
(178.11)
These lie in Fe. Just as in (176.13), (176.19) now
y3=2 (- 27b +
-27D), y'3=2 (- 27b -
27D). (178.12)
Here also y, y' are different from 0, for otherwise we have (27 b)2 = -27 D, a = 0, although we have restricted ourselves to the case a 0 0. We compute yq, where we distinguish, according to the different
values of m (= 0, 3) and e (= 1, - 1), four cases. We prove that
yq=o2y (m = 0, e = 1); yq = oy' (m = 0, e = - 1) ; yq=
(178.13)
y(m=3,e= 1) ; yq= y'(m=3,e= -1). (178.14)
EQUATIONS OF THIRD AND FOURTH DEGREE
705
If m = 0, according to (178.7) and (178.111), we have yq = x2 + ogxs + o2gxl = o2q(x1 + ogx2 + o2gx3)
But, according to (178.10), we have oq = o or oq = 02, according as e = 1 or e = -1. Hence, and from (178.11), the assertion (178.13) follows. If m = 3, x1i x2, x3 E F, then yq = x1 + ogx2 + 02gx3 .
If the cases are similarly distinguished as formerly, then (178.14) follows. From (178.13), (178.14) we obtain proper equations if we exchange o for o2 and y for y'. By division of the corresponding equations we obtain (yy'_1)q-I=o(m= 0,e= 1)
;
(yy'_I)q_1= 1(m=3,e= 1);
(yy'-I)q+1= o-1(m=0,e= -1); (yy'-1)q+1= 1(m=3,e= -1).
Accordingly, m = 3 if, and only if,
(yy'-1)q_e
= 1. This condition, by
(178.8) and (178.12), may be represented in the form
27b +
- 27D 2n=
27b-
-27D)
1
and then in the form (1 +
227b2 D- I)2n - (1 -
- 27b2D_ 1)2" = 0
.
Hence, after applying the binomial theorem and dividing by 2,/ - 27 b 2 D-1 we obtain equation (178.4). Consequently Theorem 437 is proved. THEOREM 438. Let a biquadratic equation
x4 + axe + bx + c = 0
(b & 0)
(178.15)
without multiple roots be given over a finite fielal F of characteristic p (> 3).
Let m and m' (< m) denote the number of the roots of the cubic resolvent
ys + 2ay2 + (a2 - 4c)y - b2 = 0
(178.16)
lying in F, and the number of those roots of (178.16) which are also squares in F, respectively. If m > m', then (178.15) has no roots in F; if, on the other hand, m = m', then (178.15) has exactly m + 1 roots in F. (It should be noted that for the pair m, m' only the following five possibilities may be taken into consideration: 0, 0; 1, 0; 1, 1; 3, 1; 3, 3.)
GALOIS THEORY
706
For the purpose of the proof, let n denote the number of roots of (178.15) lying in F. The theorem can be divided into the following part-assertions:
Assertion 1. n = 1 if, and only if, m = 0.
Assertion 2. n = 2 if, and only if, m = m' = 1. Assertion 3. n = 4 if, and only if, m' = 3. We denote the roots of (178.15) by x1, . . ., x4 in a suitable overfield of F. Because x1 + ... + x4 = 0, the three elements [cf. (176.25), (176.28)] Y1 = - (x1 + x2) (x3 + x4) = (x1 + x2)2, (178.17)
Y2 = - (x1 + x3) (x2 + x4) = (xj + x3)2, Y3 = - (x1 + X4) (x2 + X3) = (x1 + x4)2
are then the roots of (178.16). Hence it follows [cf. (176.31)] that F(x1i
.
. ., x4) = F(Jyl,
Y2,
Thus x1, ..., x4 E F if, and only if, we have Jy1,
.,/Y3)
Y2,
(178.18)
.
Ys E F. Thus Asser-
tion 3 is proved. We denote the left-hand side of (178.18) by G. Because of (178.18), the degree of G I F(Y1, Y21 Y3) is a power of 2. Thus the degree of G I F is divisible by 3 if, and only if, the same holds for the degree of F(y1, Y2, y3)1 F. Thus we have established Assertion 1. In order to prove Assertion 2, we first suppose that n = 2. We may even suppose that x1, x2 lie in F. Then 1
x3=- 2(x1+x2)+Jw, x4=- 2(x1+x2)-Jw 1
where w is an element of F, which is not a square in F. Consequently, accord-
ing to (178.17), we have z
Y1 = (xl + xz)2, Y2 = ( 2
(xl - x2) + ',/;) , Y3 = l2 (xl -
7e2)
-
2
Therefore m = m' = 1. Conversely, we suppose that m = m' = 1. Then among the yl, y2, Ys exactly one is a square in F. We may suppose that it is yl. According to (178.17) we then have x1 + x2 = r,
y1 = r2
(178.19)
707
EQUATIONS OF THIRD AND FOURTH DEGREE
where r is an element of F. After substitution in (178.16) we get
r6 + 2ar4 + (a2 - 4c)r2 - b2 = 0
(r # 0).
(178.20)
Also 4
i
(x - xI) = x2 - rx +
r3+ar+b IIx2 + rx + r3+ar- b ) 2r
2r
(178.21)
holds, since the right-hand side, because of (178.20), agrees with the lefthand side of (178.15). Because b 0 0 it follows immediately from (178.15) that the six sums x1 + xk (1 ajA,wi /=I
(i = 1, ..., n).
F) .
F.
734
GALOIS THEORY
We introduce the polynomial
g(Yi, ...,
(E N [Yi, ...,
=.f(E [y1A¢Ai , ..., E yiAnwi) i i
(182.2)
where we have to sum over the j = 1, ..., n. (The coefficients A,ai are deliberately placed on the right side.) Because of (182.1),
(a,,...,a,, E F).
g(al,...,a.) = 0 From this, because of Theorem 206,
0
g(Y1, ...,
arises from f(xl, . . ., follows. But g(yl, . . ., the linear transformation
xi -
(182.3)
because of (182.2), by
(i= 1, ..., n)
yyA,pii
(182.4)
whose matrix is (A,i). On account of (145.10) the square of the determinant I A,wi I is equal to the discriminant of the basis elements w1, ..., w,,. This discriminant is, according to Theorem 362, different from 0, therefore the same also holds for A,wi I. This means that the matrix (Ap) is invertible. also arises by a linear transformation Since, according to this, f(xi, ..., from g(yl, . . ., it follows from (182.3) that the proposition is true. We can now prove the theorem. Keeping the former notations we write A;A; = A(i.i)
(i, j = 1, . . ., n) ,
(182.5)
where (1, j) denotes the function of i, j with the range just defined by (182.5). Using this function we form the determinant
f(xl, ...,
I xy, i) I
(E F [x1, . . ., x )) .
(182.6)
For every a (E N), according to (182.5) and (182.6),
.f(Ala, ..., AA) = I A(i, )a I = I AiA,a 1.
(182.7)
But it follows from (182.5) that the condition (i, j) = 1 between i (= 1, . , n) and j (= 1, ..., n) produces a one-to-one relation. This means, because of (182.6), that f(l, 0, ..., 0) is a determinant which contains in each row and column exactly one element other than 0 and this element is equal to 1. Such a determinant is different from 0 (and equal to 1 or -1). 0 0, it follows from the Since accordingly we have, a fortiori, f(xl, . . ., proposition, and from (182.7), that an a (E N) exists such that I AiA,,x 1
0.
(182.8)
NORMAL BASES
735
The square of the left-hand side is, according to (145.10) the discriminant of the elements Ala, ..., Ana. (182.9) Since this discriminant, because of (182.8), is different from 0, the elements (182.9) constitute, because of Theorem 362, a basis, and this is evidently a normal basis of N I F. For the above proof cf. BOURBAICI (1939). For another proof, which also holds for finite fields, cf. PICKERT (1951).
EXAMPLE 1. Let N I F be a Galois field of degree 2 and of characteristic p (Z 0).
If p 96 2, then N = F (,) for a suitable element d from F. Now .1
+ ,Fd ,
1 - Jd
is a normal basis of N I F. If, on the other hand, we have p = 2, then, according to Theorem 429, N = F(8), 02 + 0 + a = 0 for a suitable element a of F. Now
0,1+0 are conjugate; consequently they constitute a normal basis of N I F. EXAMPLE 2. In the n`h cyclotomic field F (of characteristic 0) the primitive nth roots of unity 91, ..., e9,(")
(182.10)
constitute a basis, thus a normal basis if, and only if, n has no multiple prime factors.
We first prove the "if" part. This is true if n is either 1 or a prime number. If n = = pl . . . pk (k > 2) is the product of different prime numbers pl, . . ., pk, then it follows from the preceding statements and from Theorem 292 that the products
al...ak
(182.11)
constitute a basis of F, if a; runs through the primitive PC roots of unity (i = 1, . . ., k). But these products (182.11) now coincide with the elements (182.10), so that the assertion is also true in this case. In order to prove the "only if" part, we denote the n1h cyclotomic polynomial by F"(x). If d is the product of the different prime factors
of n, then F"(x) = Fd(x° '").
If n > d it follows that the sum of the elements (182.10) vanishes, therefore these elements now constitute a basis. This completes the proof. For an application cf. REDEI (1959, 1960).
EXAMPLE 3. The solution of the Exercise in § 172 is obtained as follows. Let us take an element 0 of G whose conjugates constitute a normal basis of G I F, together with a primitive automorphism s of G I F and a primitive n`h root of unity a (E F). The Lagrange resolvent [cf. (172.2)]
A = 0 + e(s6) + ... + e"-'(s"-'0) is then different from 0, moreover (as quoted above) sA = e''7. and A" E F. Accordingly
G=F(.jx)for rc=A" (E F).
CHAPTER XII
FINITE ONE-STEP NON-COMMUTATIVE STRUCTURES A non-commutative structure is called one-step non-commutative when its proper substructures (of the same kind) are all commutative. The determination of these structures is a difficult task, and has only been solved for groups, rings and semigroups; furthermore, we have to note WEDDER-
theorem (Theorem 318) which implies that no finite one-step non-commutative skew fields exist. Finite one-step non-commutative structures are of great importance because every finite non-commutative BURN'S
structure contains at least one one-step non-commutative substructure. § 183.* Finite One-step Non-commutative Groups In compliance with the above general definition a non-commutative group
with only commutative proper subgroups is called a one-step
non-
commutative group. All the finite groups of this kind are fully determined by the following two theorems. THEOREM 444. The finite one-step non-commutative p-groups are: the quaternion group (of order 8), every group (of order pn,+n+l) for given m, n (E .f') defined by the equations Apm = 1, BO = 1, C° = 1, AC = CA, BC = CB, BAB-' = AC (m>_n>_ 1)
(183.1)
and every group (of order p,n+n) defined by the equations
Ae = 1, B°" = 1, BAB-' =
Al+vm-'
(m >_ 2, n > 1; pm+n > 8). (183.2)
These groups are not isomorphic. THEOREM 445. The remaining finite one-step non-commutative groups are obtained as follows: Given two different prime numbers p, q and a natural number n, put (183.3) m = o(p (mod q)) , 736
GROUPS
737
take the finite field F such that
O(F) = p"
(183.4)
and from the group F* an arbitrary fixed element co such that
o(w) = q.
(183.5)
Then a group (of order pmq") exists such that by the help of certain special elements
Px(a E F), Q (Pi, the unity element; o(Pj = p for at
0; o(Q) = q") (183.6)
all its distinct elements may be written uniquely in the form P,.Qi
(aE F; i=0,...,q"- 1)
(183.7)
and the rule for multiplication of elements is
(a, P E F; i, k = 0, ..., q"- 1).
P«Qi PpQk = PQ+.r# Qi+k
(183.8)
Taking for each of the above triples p, q, n such a group, we just obtain the mutually non-isomorphic finite one-step non-commutative groups which are distinct from the p-groups.
It has to be proved that every finite one-step non-commutative group is isomorphic with one of the groups given in Theorems 444, 445 and that, conversely, these groups are one-step non-commutative and not isomorphic with each other. We begin the proof with the first of these assertions. We denote by 4 a finite one-step non-commutative group and show first that it is not simple. For the proof we assume c.P to be simple. We first remark that for two proper subgroups a°, . Y of c4, for which 2c° is maximal and .2' not contained in :72, "'
1 (15Y= 1. For, because of the supposition,
'Xj
C4
Since moreover
, W' are Abelian, ' (1 r is normal in JT and 5Y
and therefore also in 4. Because of the simplicity of 4 the truth of the remark follows. Since c.P is not commutative, 0(4) is a composite number. Let a maximal subgroup of 4 be denoted by A' and its index by h = O(C
:
Let Al, .., Ah
(183.9)
738
FINITE ONE-STEP NON-COMMUTATIVE STRUCTURES
be a left representative system of 4 mod .Y. We show that all the conjugates
1i = A;,7,4; I of fT.
(i= 1, ..., h)
(183.10)
are pairwise distinct. We assume that Xi = X, for two different
i, j. Itsfollows that Aj 1A,YoAr 1 Aj
so that
=
°,
is normal in the subgroup
_{
,A.'A,}
of 4. But since X is maximal, we have 4.l = 4. Since, according to this, Y is normal in 4 and 4 is simple, it follows that ' = 1. This contradicts the maximal property of Y, so that we have proved that the conjugates (183.10) are pairwise distinct.
Hence, and from the above remark, it even follows that
tf210;= 1
(1 _ 2. Symmetry allows us to assume that
mznZ2. We determine a solution z of the congruence yz
x (mod p)
(183.36)
and put B' = AJ "n-"zB .
As AB'
(183.37)
B'A, (183.27) implies the necessity of o(B') z o(B). Since,
on the other hand, because of (183.22), (183.25) and n >_ 2 ,
B'P"= 1
,
it follows from (183.252) that o(B') = o(B). Accordingly we may interchange B and B'. Because of (183.15), (183.16), (183.37),C remains unaltered.
Moreover, according to (183.22), (183.34), (183.36), (183.37), -IPM -'y
B'P"-'y = CC2
-Iy-1)
If the second factor on the right is 1, i.e., C = B"'"-'y, then the assertion is true. If the above-mentioned factor is not equal to 1, then the exponent in it is not divisible by p. This means that
p = 2, m=n=2. Then we have
o(A) = 4, o(B) = 4.
(183.38)
GROUPS
743
Further, because of (183.34), we have C = A2B2 .
Hence and from (183.15) A2B2 = A-1BAB-1. Thus, because of (183.38), (AB)2 = 1. We have found an element of order two, non-interchangeable with A. But this contradicts (183.27) because of (183.382). This contradiction proves that it may be assumed in (183.34) that p I x or p I y. Then C belongs to {B} or to {A}. If A, B are interchanged, then according to (183.15), (183.16), (183.20) C changes into C-1, so that we can put C E JA}.
If a suitable primitive element of {B} is substituted for B, we can show, according to (183.21), (183.24), that even
C = A""'-'.
(183.39)
Moreover, because of (183.14) and (183.20) we must have m Z 2. We see from (183.15), (183.25), (183.39) that the equations (183.2) are satisfied. But as the group defined by (183.2) is evidently of order at most pm+", so (183.332) implies that & is exactly this group. Consequently we have
proved our assertion for the case (183.19), and have shown that in this case c4 occurs among the groups enumerated in Theorem 444. Secondly we consider the case where c4' is not contained in Z. We denote, as above, a normal divisor of cP with prime index by A. In the above proof of the assertions concerning (183.14), (183.15), (183.16) we have seen that we may also require that
A E r° a (In the previous case keeping to this additional requirement would have been superfluous and even prejudicial.) Furthermore (183.23) may also be assumed, since the proof was independent of assumption (183.19). We show that in addition to (183.231) we may also assume that A° = 1
.
For since according to (183.15) we have BAB-1 = AC, it follows from (183.16) that
BA°B-1 = ARC". This results, because of (183.23,), in
C°= 1 On the other hand, we have
.
BC0CB,
since from BC = CB and (183.15), (183.16) it would follow that cj' _
744
FINITE ONE-STEP NON-COMMUTATIVE STRUCTURES
_ {C} c ;x which contradicts our assertion. Since, accordingly, we have
4={C,B} and also C E ', we only need to take C instead of A in order to ensure that the requirements for A shall be satisfied. If we write Q for B we can express what has been said so far as follows : In c0 there are two elements A, Q such that AE
(183.40)
o(A) = p .
(183.41)
O(Q) = q",
(183.42)
f
AQ 96 QA ,
cP={A,Q} AQQ = QQA .
(183.43) (183.44)
(183.45)
[These are not independent, (183.43) and (183.44) being equivalent, while (183.45) follows from (183.40) and (183.42), since T is an Abelian normal divisor of index q in c &..l
We put
(i=0,1,...; A0=A).
Ai=Q`AQ-I
(183.46)
According to this we have (i, k = 0, 1, ...) .
QkAiQ-k = Al+k
(183.47)
From (183.45), (183.47) it follows that
A,=AI
(i=j(mod q)).
(183.48)
From (183.41), (183.46),
o(A) = p
(i = 0, 1, ...) .
(183.49)
By (183.40) and (183.46), A. E
'
(I= 0, 1, ...) .
(183.50)
Because of (183.50) the Al generate a subgroup Iwo
= {A0, AI, ... }
(183.51)
GROUPS
745
of which is thus likewise Abelian, moreover because of (183.49) it is an elementary p-group. From (183.44), (183.46) it follows that o is a normal
divisor of 4. In the sequence A0, Al, ... there is a first term Am, for which we have Am E {Ao, ..., Am_i} , and, because of (183.48), certainly
1 - 2 and assume the assertion for smaller r. According to (183.60), QB1-1 Q-1Bi l 1 = BI
(i = 1 , 2, ...) .
Since, because of the supposition, B,_1 1, we have B,_2 Q that d4 _{B,_2,Q).
QB,_2, SO
Since the commutator B,_1 of these generators Q, Br_2 is interchangeable furthermore since dj' does not lie with B,_2 (because B,_1, Br_2 lie in in Z, it follows that Br_i, Q cannot be interchangeable. This implies that Br 0 1, so proving (183.65). Since, in FP[x], xP - 1 = (x - 1)P, it follows from (183.61), (183.65) that
p# q.
(183.66)
GROUPS
747
We prove that F(x) is irreducible. For this purpose we consider a factor decomposition (f(x), g(x) E FP[x]) F(x) = f(x)g(x) where x*_1
f(x) = xs + ds_l
+ ... + do
(s > 0) .
Because of (183.64), (183.65) we may assume that
f(x)#x- 1. It is sufficient to prove that necessarily s = m, since from this the irreducibility of F(x) follows because of (183.63). We therefore write
B = g(x),
Bi = Q'BQ-`
(i=0,1,... ; Bo = B) .
Because of the assumption we have
F(x) 0 (x - 1) g(x) .
From this, because of (183.60) and the definition of F(x) it follows that
QBQ-'B-' + 1, and so
di ={B,Q}. Since, furthermore, B E A o and so o(B) = p, B satisfies the requirements for A in (183.40) to (183.45). But we also have s
F(x) =
=0dkX'AX)
whence it follows, because of (183.56), (183.58), (183.59), (183.64), that
Bs Bds'l s-1
Bd. 0 =1
therefore, because of (183.53), s>-m, i.e., s = m. This proves the.irreducibility of F(x). Now, because of (183.56), we have a homomorphism Fp[xl+ ,.,
(f(x) -)-f (x))
,
(183.67)
the kernel of which we denote by a. Then
f(x) E a a f(x) = 1
.
(183.68)
We show that a is an ideal of FP[x]. By definition a is a submodule and so because of (183.59), (183.68) an 'ideal of Fp[x].
748
FINITE ONE-STEP NON-COMMUTATIVE STRUCTURES
0 1, a o Fp[x] follows from (183.67). FurtherSince we have more F(x) is, according to (183.64), (183.68), an element of a. From both, because of the irreducibility of F(x): a = (F(x))
(183.69)
follows.
Because of (183.61), (183.65) and (183.68),
x4-IEa, x-11a. Hence and from (183.69) owing to the irreducibility of F(x), it follows that x9 - 1 F(x)
X- 1
'
(183.70)
so that for the degree m of F(x), by § 133, Example 3,
m = o(p(mod q)).
(183.71)
We make use of the finite field F over FP such that
O(F) = p'
.
(183.72)
Since F(x) is irreducible and of degree m, we may put F = FP(w)
(183.73)
F(w) = 0 .
(183.74)
where
Because of (183.70), wQ = 1, co 0 1 also holds, thus o(w) = q .
(183.75)
is normal in d d, the elements of ci, because of (183.44), may be Since written in the form .f(x) Q' . From (183.68), (183.69),
f(x) = g(x) «f(x) = g(x) (mod F,(x)) . But, according to (183.73), (183.74),
(x) = g(x) (mod F(x)) a f(w) = g(w) .
GROUPS
749
Since, accordingly, the arbitrary element (,-r) of 2l° is completely deter. mined by the element f(co) of F, we may write the elements of X0 in the form f(x) = Pn.)-
(183.76)
Then the PP
(183.77)
(a E F)
are all the different elements of Ta. Because of (183.42), according to the above, the (a E F; i =0,...,q" - 1) (183.78) P. Q' are all the different elements of 4. Since from (183.59), (183.56) -f(X)Q' . g(x)Q" = f(x) + x'g(x)Q' +k
,
after writing the elements in the form (183.76) we get P,Q i
k - Pa+w'fQi+k PeQ-
(183.79)
Thus the elements of 44 have to be multiplied according to this rule. With regard to (183.71), (183.78), (183.79) we see that (4 is completely characterized by the determining elements p, q, n, co. The elemdht co of F according to (183.75), (183.74) is there subjected to the conditions o(co) = q, F(co) = 0. The second of these conditions may be neglected, i.e.,
we may take for co in (183.79) a quite arbitrary element of F such that o(w) = q, because from this, as will be shown immediately, we obtain groups isomorphic with 4. All the co such that o(w) = q may be given by any arbitrary one of them in the form wr (r q - 1) Now according to (183.79), P«Qri
PpQri =
Pa+wrrJ5Q('+k)
.
The comparison with (183.79) shows that the substitution of Qr for Q has the consequence that in the multiplication rule (183.79) the element to is exchanged for co". This establishes the required isomorphy. From this last proof and from (183.66), (183.71), (183.72), (183.78), (183.79) we see that c4 is in fact one of the groups given in Theorem 445. It still remains to be proved that, conversely, the groups given in Theorems
444, 445 are one step non-commutative and are not isomorphic with each other.
First of all it is evident that the quaternion group is one step noncommutative. Since, furthermore, condition (183.28) is not satisfied here, it is not isomorphic with any of the remaining groups given in Theorem 444.
750
FINITE ONE-STEP NON-COMMUTATIVE STRUCTURES
The groups defined by (183.1) and (183.2) are denoted by co, and 4 2 respectively. First of all we consider coil. It is evident that its elements may
be written as
(i==0,...,p' - 1;k=0,...,p"- 1;1=0,...,p- 1),
A'BkCI
(183.80)
If we show that equality holds here, it follows so that O(d4j) that all the different elements of coil are given by (183.80). For this purpose we consider the set G of triples i, k, I
of integers i mod p', k mod p", 1 mod p. From the full permutation group .9(S) we take the following three permutations
_
i
kl
B'
_ ik / C'-lik1+1)' = ik+1i+1)' r
k
l
(183.81)
which are evidently of orders p', p" and p. According to (1 83.81)
B'A'B" _
i
k+1 i+l
_
i
k1
i+1k+11+l+1,i+1 k1+1,
A'C'
'
and also A'C' = C'A', B'C' = C'B'. Accordingly, the equations (183.1) are satisfied by (183.81), from which by Theorem 101 the homomorphy
dil x {A', B', C) follows. Therefore it is sufficient to show that the group on the right-hand side is of order at least p"'+"''. It is not commutative and because
B'P =
I
k l)
ik+pl,
it contains the direct product {A'} ® (B'P} ® {C'}. This is of order p'+" and its index in (A', B', C) is at least p. Hence the assertion is proved. Since, according to this, the elements (183.80) are distinct, it follows also that cPil is not commutative. In order to show that c4JI is one-step noncommutative we have to demonstrate that any two non-interchangeable elements X = A`BkCI , Y = A"Bk'('c'
GROUPS
751
of 4, are already generators of this group. Here Ymay be replaced by any X'Y which, because of (183.1), allows us to restrict ourselves to the case where i
or i' is equal to 0. We can put i' = 0. The assumption and Theorem 446 implies that p,f' V. If we then replace X by a suitable XY', we also obtain k = 0. Since two elements of the form X = A'C' ,
Y = B"C" ,
where p,j' i, k', are evidently generators of d41, so the assertion has been proved.
The corresponding results may be proved for dg, as follows: What we have proved concerning co, did not necessitate the assumption m >_ n, and so remains true for all m >_ 2, n >_ 1. As APm-', C are central elements of dul, so {APm-' C-1} is a normal divisor of order p of 4g,. A glance at (183.1) and (183.2) shows that `42
zt; 41 / {APm-` C-1} ,
implying that co, is in fact of order p'"+" and one-step non-commutative. (This proof holds even for the case p°'+" = 8, although we have not yet made use of this fact.) We see from Theorem 446 that in c& the commutator group {C} is prop-
erly contained in a cyclic subgroup only when p = 2 and m = n = 1, i.e., O(c.Pg,) = 8. Since for 42, on the other hand, O(c42) > 8 and its commutator group {APm-'} is properly contained in {A}, it follows that no d%g, can be isomorphic with any c42. Both in 4,, if 0(ddl) > 8, and in 42 because of Theorem 446 the maximum of the orders of the elements is equal to p,a,(m,") Hence it is obvious that neither the d41 nor the 442 are isomorphic with each other. Finally, we consider the group defined in Theorem 445 which we denote
by cog. As opposed to the previous cases we have here to prove first the existence of cog, for cog is not defined by equations like cogs and 42, but by the
multiplication rule (183.8). We make the observation that all integers i, k may be admitted in (183.8) because o(w) = q and o(Q) = q". For the product of any three elements of c4 we have, according to (183.8), (PPQ1'PPQk) PyQI =
p.'d'3Q'-1k pYQl
-
Pn+w"/3+uat+k'Q1+k+1
P«Q'PP"u,'Qk+1 _ P.+w,(P+ruky)Qi+k+l
Hence the multiplication (183.8) is associative. Evidently P0 Q° is the unity element. The left inverse of PgQk is P_,0_k8Q-k. Thus we have proved the existence of the group jog. 25 R. - A.
752
FINITE ONE-STEP NON-COMMUTATIVE STRUCTURES
It is obvious that cQ,l is not commutative. We still have to show that c4 is generated by any two non-permutable elements
X=P«Q', Y=p'Qk. Since we may replace X and Y by XY' and X'Y, respectively, we may assume, e.g., that k = 0. Equation (183.5) and the assumption imply that q,j'i. Since X may be replaced by every primitive element of {X}, we may assume that i = 1. Since now
X=P,Q, Y=P,j, 0, we have fro ni (183.8)
where
(i=0,...,m - 1).
X'YX-l = Since
cof,
...,
O)m-I Q constitute th(e'
a basis of F, it follows that the group
P. (o E F). It then also contains P_,X = Q, con{X, Y} contains all sequently it is equal to L. Accordingly, dj is one-step non-commutative. Because of the diversity of the orders, df cannot be isomorphic to either ddI or c1762.
Since, according to (183.8),
Q'PIQ-` Pl I = P.,-1
(i=0, ..,m- 1),
the P. (a E F) make up the commutator group of c0,1, which is then of order
p'. From this it evidently follows that no two of the d3 are isomorphic. Consequently we have proved Theorems 444, 445. For finite one-step non-commutative groups cf. MILLER- MORENO (1903), SCHMIDT (1924), REDEI (1950a, 1958b). Further literature and applications see in GOLFAND (1948), IT6 (1955), REDEI (1951a, 1955-56), SUZUKI (1957).
EXAMPLE 1. From Theorem 445 it follows that there is at least one one-step non-commutative group of order pq" (p, q different prime numbers, m, n > 1) if, and only if, one of the conditions in = o(p (mod q)), n = o(q (mod p)) is satisfied. There are two such groups if, and only if, both these conditions are satisfied.
There are never more than two such groups. EXAMPLE 2. The centre of the group given in Theorem 445 is the cyclic subgroup (Q°} of order q"-1. The only centre free case is that with n = 1. In this case the commu-
tator group is the one and only one proper normal divisor. EXERCISE 1. A one-step non-commutative group (of order 8) is defined in the (excluded) case p'+" = 8 by the equations (183.2), which is, however, isomorphic with one of the groups given by (183.1). EXERCISE 2. Every finite one-step non-commutative p-group is the homomorphic image of a group defined by (183.1).
GROUPS
753
EXERCISE 3. Every finite one-step non-commutative group is a homomorphic image of the group defined by the equations ABTA-1 = Bt+1,
B;B, = Bi$,
(i,.% E J).
PROBLEM. Are there infinite one-step non-commutative groups?
§ 184.* Finite One-step Non-commutative Rings
In virtue of our general definition a one-step non-commutative ring means a non-commutative ring with only commutative proper subrings. The finite rings of such a kind are determined by the following two theorems.
As a preliminary, let us consider elements e, a from a ring R and a polynomial from 7[x]:
f(x) = c + g(x)
(c E J ; 9(x) E Y [x] = xxY [x]).
Though f(e) is then in general meaningless, the "product" f(e)a = Ca + A00, may always be defined. (The reader will easily see that this is an operator product and that this is essentially a special case of the construction in § 62, Exercise 5.) Then f(e)e is the substitution value of f(x)x for x = e. THEOREM 447. In order to determine the non-nilpotent finite one-step non-commutative rings, we define for every prime number p and every natural number m a ring
RI = {e, a}
(184.1)
p"', = 0, pa = 0, e2 = e, ea = or, ae = 0, a2 = 0;
(184.2)
by the equations
further, for any (not necessarily distinct) prime numbers p, q and natural numbers m, e, n (n < q) we define the ring
R11={e,a}
(184.3)
by the equations
e =0, pa=0, F(e)e=0, F(e)a=0, are = epa,
a2 = 0
(P = p" °e-,),
(184.4)
where F(x) (E J [x]) denotes a fixed arbitrarily chosen irreducible principal polynomial mod p of degree qe the particular choice of which is of no importance since the R11 belonging to fixed p, q, m, e, n and different F(x) are
754
FINITE ONE-STEP NON-COMMUTATIVE STRUCTURES
isomorphic. The rings RI (belonging to all the p, m), the rings opposed to these and the rings RII (belonging to all the p, q, m, e, n) then constitute a complete system of non-isomorphic non-nilpotent finite one-step noncommutative rings. There, RI is of order O(RI) = pm+l,
(184.5)
it has the basis elements e, a, where
o+(e) = p', o+(a) = p,
(184.6)
and
(ae + ba) (ce + do) = ace + ado,
(a, b, c, d E -.7),
(184.7)
for the product of two elements. RII is of order O(R1I) = p(m+1)WW ;
(184.8)
its elements may be uniquely given as
a(e) + b(e)a
(a(x) E 7[xlo, b(x) E 7[xl),
(184.9)
where a(x), b(x) run through a representative system of J [xb mod (pmx, F(x)x) and of J' [x] mod (p, F(x)), respectively. The rule
(a(e) + b(e)a) (c(e) + d(e)a) = a(e)c(e) + (a(e)d(e) + b(e)c(e)0a (a(x), c(x) E J' [x b, b(x), d(x) E J [x]) (184.10)
holds for the product of two elements. THEOREM 448. The nilpotent finite one-step non-commutative rings are the rings (184.11)
Rn1
defined by the equations
pme=0, p"o=0, er=0, a,=0, o2a = eoe = ae2,
(184.12)
eat = aoa = ate, pea = pae
and their non-commutative homomorphic images; here p is a prime numbe
and m, n, r, s are natural numbers such that m S n and r, s
2. R111 is of
order O(R1I1) =
p1+m (r- 1) s+ n (s- 1)
(184.13)
755
RINGS
with basis elementsoa - o, e'a'(01) for which
o+(Pa-aO)=p; o+(ea')=p' (1 s also disappear, i.e., the terms for which j > u may be omitted. After this only the terms for which j = u remain : a, x'ys-1 E r .
RINGS
785
(According as u = 0 or u > 0, one has to sum here over i = 1, ...,p - 1 or i = 0, . . ., p - 1, respectively.) The terms of the left-hand side are of different degree. But t has, according to (184.155), only homogeneous generators, so that the mentioned terms must each lie in t. Among these terms
there is, according to the definition of u, one which is different from 0. Suppose that ar" 0. According to what has been said we have armxrys-I E t .
Because of (184.155), army 'YS -1 E (pmx, p"y, xr, .3's, xy - Yx) .
since (after adding xy - yx) the cancelled generators of t lie in the principal
ideal (xy - yx) . Because i < r and s - i < s it follows that a;" xry"-1 E (pmx, p"Y, xy - yx)
(184.159)
.
The substitution y = x results in a,,,x`+s'' E (P-x, P"x) , where we now have an ideal of .7 [xb on the right. Because m _:!9 n it follows that Pm
a:"
But since for i > 0 a;,, only takes the values 1, ..., p-1, because of (184.157) and since a1 # 0, i = 0 (and u > 0). In the remaining case i = 0, (184.159) reads as follows: aouys-1 E (pmx, p"y, xy - yx)
After the substitution x = 0, y = x we obtain
.
E (p"x), thus
p' I a... But since, because of (184.157) and because ao, 0 0, aou only takes the values
1, ...,p"-1, we. have proved by this contradiction that the elements (184.157) represent all the different residue classes of . mod r. The number of these representatives (184.157) is 1
,
whence, because of (184.156) the validity of (184.13) follows. From (184.11), (184.123, 4, s.6) it follows at once that the elements 'Oo' -
eY, mentioned before (184.14), are generators of the module Ri1I+. On the other hand, from (184.1 21, 2, 7) and m < n it follows that for these elements equations (184.14) hold with " a > 1) and for a prime number p we define the semigroup (185.3)
S11 = {Lo, a}
by the equations
or = e, ape+I _ a, ape = eap, LOaiI = e, L02ak = eakQ = ake2
(k = 1, 2, ...) ,
(185.4)
thirdly for the natural numbers r, a, s, b (r > a > 1, s > b > 1) we define the semigroup Sin = {Lo, a}
(185.5)
by the equations Qr = 9a, or'
oure = aLo2 ,
Dr
(185.6)
a2LO = aqa = LOa2.
In addition, we denote by S1V a semigroup consisting of two right units. These semigroups S1, . . ., S1V, and their non-commutative homomorphic and
anti-homomorphic images and, finally, the finite one-step non-commutative
groups defined in Theorems 444, 445 are, apart from isomorphy, all the finite one-step non-commutative semigroups. S1 has order
O(S1) = r + 1
(185.7)
ee'(i= 1,...,r - 1),a,ago,
(185.8)
O(Sn) = pe(r + p - 1)
(185.9)
and consists of the elements
S11
has order
and consists of the elements ahoak, oiak, a1+k
(i=2,...,r- 1; h=0,...,p- 1; k=0,...,pe- 1),
(185.10)
finally, 5111 has order 0(S111) = rs
(185.11)
and consists of the elements
Loiak(i+k>-,1; i=0,...,r- 1; k=0,...,s- 1),aLo. (185.12) 26/a R.- A.
788
FINITE ONE-STEP NON-COMMUTATIVE STRUCTURES
NoTE. In order to give a complete system of non-isomorphic finite one-
step non-commutative semigroups, we still have to solve the homomorphy problem for the semigroups S, S11, 5111. This is simple for S1 and Sn, but for S111 it is an extremely fatiguing task. In order to prove the theorem, first of all we have to take into considera. tion the fact that, according to Theorem 33, a finite group has only groups
as subsemigroups. It follows that the only groups occurring among the finite one-step non-commutative semigroups are the finite one-step noncommutative groups. Consequently our proof should consider only those semigroups which are not groups. We now wish to prove the assertions listed in (185.7) to (185.12). From the defining equations (185.2), (185.4), (185.6) for S1i 511, S111 it follows that all the elements of these semigroups occur among the elements quoted in (185.8) (185.10) and (185.12) respectively. Therefore it is sufficient to
prove that these elements are all distinct in the three cases, for (185.7), (185.9) and (185.11) then follow by counting the elements. The fact that the mentioned elements are different, will be proved by Theorem 101, when we consider three semigroups belonging to the three equations (185.2), (185.4) and (185.6), respectively, for which we retain the notations (185.1), (185.3), (185.5), and show that the relations (185.7), (185.9), (185.11) hold for them
with Z instead of =. In all three cases we shall define the generators 'o, a as certain mappings of a set S1, S11 or S111, respectively, into itself.
For non-negative integers t, in, n (0 < m < n) we denote by (t);, that integer, for which
(t)M' = t (t < n) or
(t)om,-I (mod n - m), m 5(t)7 - pe (r + p - 1), as was required. For the definition of SIIl we denote by S1II the set of elements
(x,y)
(x=0,...,r- 1; s=0,...,s- 1), 9,
which are again to be regarded as pure symbols, and denote the mappings
a of this set into itself by 60(x, y) = ((x + 1)p, y),
eQ = (2, 1)
and
a(x, y) = (x, (y + 1)b) (except for x = 1, y = 0)
,
a(l,0)=d2, crQ=(1,2). Then
(i = 2, 3, ...) .
60'(x, Y ) = ((x + i)Q, y), o'Q = ((1 + i)a, 1)
and
(k=2, 3, );
ak(x,Y)=(x,(Y+k)b). akSa=(1,(1 +k)%) so that 60' = 60° , as = b Moreover we have the following relations 60 2(x, y) _ ((x + 2)a, y) ,
602Q
((3)a, 1)
,
a2(x, y) _ (x, (y + 2)b) , a2Q _ (1, (3)b) ,
ea(x, y) = ((x + 1)a , (y + 1)b) , eaQ = (2, 2) ,
so that we obtain easily e2a(x, y) _ ((x + 2)0 , (y + 1),) = eae(x, y) = ae2(x, y) , e2a.Q = ((3)a ,
2) = eaeQ =
ae2.Q
and
a2e(x, y) = ((x + l )y , (y + 2)b) = aea(x, y) = ea2(x, y) , a2eQ = (2,(3)b) = aeaQ = ea2Q .
792
FINITE ONE-STEP NON-COMMUTATIVE STRUCTURES
Hence
e2d'=eae-cie2, atLo =uea=Lo a2. Consequently SI1I belongs to the equations (195.6). Since the images
e'ak(0, 0) = e'(0, k) = (1, k)
(i+kZ 1; i=0,...,r- 1; k=0,...,s- 1), and
ae(0,0)=cr(1,0)=Q
of the element (0, 0) of SIII are different, 0 (SI11) >_ rs. Consequently the assertions in (185.7) to (185.12) follow. We denote the semigroups defined in (185.1) to (185.6) again by SI, SII, SIR, and now prove that they are one-step non-commutative. (The same holds for SIv.) It is sufficient to prove that the semigroups mentioned have non-interchangeable elements and are generated by any two such elements. Of all the elements (185.8) of SI only a and a, according to (185.2), consitute an (unordered) pair of non-interchangeable elements. From (185.1) it follows that they generate SI. Of all the elements (185.10) of S11, according to (185.4), only the
K=a"eo°, a=aw
(u=0,...,p- 1; v=0,...,pe- 1; w= 1,. ,pe;
P,f'w)
constitute non-interchangeable pairs of elements. If we determine a natural number t by wt = 1 (mod pe), then according to (185.4)
a=0.W1 =R'
0=
ap,Loap'=O,Pe_"KO'P'
=
= )(P'_u)1K;i.(Pe_v)t.
Hence, and from (185.3) it follows that K, ? are generators of SII. Among the elements (185.12) of SIII, according to (185.6), only a and Cr
constitute a non-interchangeable pair of elements and, by (185.5), they generate S111. Since, by the above, Si, . . ., S,v are finite and onestep non-commutative, the first part of the Theorem is proved. In order to prove the second part, we consider a finite one-step non-commutative semigroup S which is not a group. We may assume this to be of the form S
(185.13)
SEMIGROUPS
793
We have to show that S is a homomorphic or an anti-homomorphic image of one of the semigroups S1,. .., SIV. First of all we prove some propositions about S. PROPOSITION 1. If pS c S and
a, b, c, d, e,...
(185.14)
is a finite non-empty sequence of natural numbers with then
a+c+e+...?4,
(185.15)
eaabQcad ... _ pa+c+... ob+d{...
(185.16)
Before giving the proof we make the observation, important also for what follows, that oS is a subsemigroup of S and so by virtue of the assumption, commutative. We now denote by k and K the number of terms in the sequence (185.14) and the left-hand side of (185.16), respectively.
For k = 1, 2, the assertion (185.16) is trivial. For k = 3 we distinguish two cases : if c 2, then paab and p` lie in pS, thus we have K=
paab
pe = p'paab ,
so that (185.16) holds. In the other case, when c = 1, according to (185.15), we have a >_ 3. Hence, by an analogous reasoning, we obtain K = p2 . pa
tab p1
= pa
tab pe+Q,
so that we have reduced this case to the preceding one. The case k = 4, since
K= may be reduced to the case k
paob pc . Off,
3. Finally the case k >_ 5, since
paab . p°Qd oe = eeadpe+aab
may be reduced to a case with smaller k. Hence Proposition 1 is proved. PRoPosmoN 2. If eS e S, then OiOkp!_e'+Jok
(185.17)
For jk = 0 this is trivial. We suppose that (185.17) does not hold for some triple i, j, k of natural numbers. This means that the elements peak, pi are not interchangeable, i.e., S = (p'ak, ell .
794
FINITE ONE-STEP NON-COMMUTATIVE STRUCTURES
In particular, a can then be expressed as a finite product or = e"a°eW ...
(u, v, w.... Z 1).
(185.18)
Since e, a are not permutable, the case a = e" is excluded in (185.18), i.e., at least v must exist in addition to u. Let us substitute for a factor a of the right-hand side of (185.18) this right-hand side itself. Then we again obtain
an equation of the form (185.18), where u + w + ... >_ 2. After a further such substitution we even have u + w + ... >_ 3. Then Proposition 1 may be applied to both the products (ea =) a"+Ia°ew ..., (ae =)
a"a"e"
... e,
by which their equality is established, i.e., ea = ae. This contradiction proves Proposition 2. PROPOSITION 3. If Se = S, then e°(e) is a right identity for S.
From the assumption it follows that S = So = See = ... Since S is finite, so the mapping at -* aek ((x E S) is a. permutation of S for every natural number k. Consequently there are two natural numbers k, 1(k < 1) such that ao' = aek for all at. If we write this equation in the form aek er-k = = aek and consider that Sek = S, then we see that eI-k is a right unit for S. Hence, in particular, e'-k+I = Q. Let n denote the least natural number
(therefore existing) with a"+' = e. Then e" is a right unit for S since Se = S but clearly n = o(e). Consequently Proposition 3 holds. PROPOSITION 4. If eS c S, Se = S, then all the powers e, e2, ... lie outside the centre of S.
This is obvious from eS, e2S,... c S and Se = See = ... = S. PROPOSITION 5. If there exists a natural number n with a"+' = or, then
ea`'s aye
((k,n)= 1; k= 1,2,...).
(185.19)
For the proof take some 1(>_ 1) with kl =_ 1 (mod n). From ea ae and a = (ak)I the assertion follows. PROPOSITION 6. If, for a fixed natural number n and for all the natural numbers k,
eak = ake a (k, n) > 1
(185.20)
holds, then n is a prime power (>_ 1).
If the assertion were false, we could write n = kl, where k z 2, 1 > 2, (k, 1) = 1. From (185.20) e would then be interchangeable with ak and a1, and so also with ak+I. But according to (185.20) and (k + 1, n) = 1, this is impossible, so Proposition 6 is true.
SEMIGROUPS
795
We now want to prove the assertion stated after (185.13). Not all the
eS, Se, cS, Sa can be equal to S, for then from (185.13) S = aS = Sa would follow for all elements a of S, which is contrary to the assumption that S is not a group. Since the order of succession of e, a, and an anti-automorphism of S are immaterial we may assume that (185.21)
OS C S.
In the following proof we distinguish four cases.
1. Let Se = Sa = S. According to Proposition 3 (applied also with a instead of e) S contains two right units of the form (m, n ? 1),
9m, an
from which we obtain em+1
= e
a+1 = a.
If we have eman = an
em.
i.e. em = a", then [partly from (185.21) and Proposition 21 we have
ae = an+ie = em . ae = emae = em+l a = U. But since this is false, we must have ?n
a tae. n
IT m
Hence S contains two non-permutable right units, which are therefore necessarily generators. Thus S contains no further elements whatever and is, consequently, isomorphic with Siv.
2. Let Se = S, Sa c S. According to the Propositions 3, 4, S contains a right identity em which is not a central element. Hence from (185.13) ema # aem.
Because eS c S, and Se = S, we also have eS c S; Sem = S, so that we may take em instead of e. In other words this means that a may be assumed to be a right unit. Then every element of S is of the form ak
(k = 1, 2, ...) or ea! (1= 0, 1, ...).
Next we suppose that there is a k (> 2) for which eak
A ake.
From this it follows that S = {e, ak} so that or E {e, ak}, and a is of the form
a=ak' (x
1)
or a =eak-v (p
0).
FINITE ONE-STEP NON-COMMUTATIVE STRUCTURES
796
But the second case is impossible because p is a right unit for which ea 0 a, consequently the preceding case remains the only possibility. Because kx Z 2 there are thus natural numbers n for which
a"+1=a. We assume n minimal. Then kx 1 (mod n), thus (k, n) = 1. This result, together with Proposition 5, implies that the rule (185.20) holds. According to Proposition 6 we thus have n = pe (e z 0) for some prime number p.
If e z 1 then (p, n) > 1. Since (185.20) now holds and a is a right unit, we have
eaP+I =
ooP
, a = aPe , or = aP , a = aP+1 = aP+ie.
But this contradicts (185.20), so only the case e = 0, n = 1, a2 = a remains. By what has been said above all the elements of S occur among
a, e, ea. Then
aea
e
e = ea,
whence it follows that Sa = S. But as this contradicts the supposition, all the equations
(k = 2,3,...),
eak=ake must hold, in particular,
eat=ate=a2. From the above we must also have e2 = e,
ae = a.
The finiteness of S implies that
a' = a"
(for some r > a ? 1).
Here necessarily a > 1, since otherwise ea = ea' = a'e = ae would follow, which is false. We have found that all the equations hold which result from (185.2) after interchanging a and a. Consequently S is now a homomorphic image of S1.
3. Let Se c S, Sa = S. If aS c S, then after interchanging a and a [with regard to (185.21)] the assumptions of the preceding case are satis-
797
SEMIGROUPS
fled; therefore it is sufficient to consider the case aS = S. For the sake of convenience we summarize the suppositions:
eS,SocS; aS=Sa=S. These are self-dual, a fact which we have to consider repeatedly in the sequel. According to Proposition 2 we have
(i,j= 1, 2, .. .; k=0, 1,...),
of+Jak = Qiakef = ak`O1+J
and, in particular, 02ak = oako =
ako2
(k = 1, 2, ...).
From Proposition 3 (applied with a instead of o) ao(, is the unity element of S, thus a"+1 = a,
a"o = oa" = o
(n = o(a)).
We consider a k (>_ 2) for which oak # ako. Then it follows that S = {L0, ak}
and a E {o, ak}, so that we have an equation of the form a = ak-o''akZ, where x + y + z > 0. Since oa ao, necessarily y = 0. Consequently we can write a = a' (t > 0). Since here we have kt Z 2, it evidently follows that kt = I (mod n), thus (k, n) = 1. This result and Proposition 5 imply that rule (185.20) holds and so, according to Proposition 6, we have n = p° (e _>_ O, p prime). But because oa ao, n = 1 is not possible, so it necessarily follows that e >_ 1. Further, because of (185.20), we have gap = apLo-
On account of the finiteness of S an equation of the form Q
r = Qa
holds for r > a >_ 1. But here we must have a > 1, since the assumption a = 1, because r Z 2, gives the contradiction oa= o'a = ao' = ao The result is that all the equations (185.4) are satisfied, so that S is a homo-
morphic image of S. 4. Let So, Sa c S. If we have a S = S, then [with regard to (185.21)] after dualization and interchanging of o and a we obtain the case before last. Accordingly, we have only to consider the case aS c S. We now have the (self-dual) suppositions eS, So, aS, Sa c S.
798
FINITE ONE-STEP NON-COMMUTATIVE STRUCTURES
According-to Proposition 2 it follows that 92a
= o e = ae2 and
°.2e = QLoa = Nat.
From the finiteness of S we must have equations of the form Lo"=N
o'=o'b,
where r > a z I and s > b ? 1. But here a > 1, b > 1 must hold, since e.g., if a = 1 we obtain the contradiction Loa = 'a = aLor = a Lo. Since according to this all the equations (185.6) are satisfied, S is a homomorphic image of Sn1. We have now confirmed the assertion in all four cases. Consequently Theorem 449 is proved. EXAMPLE. The (infinite) semigroup defined by the equations
eEa = eae = aez, ate = aea = ea: consisting of the (distinct) elements
e'ak(i+k>0; i,k=0, 1,...), ae, has e, a as its one and only one non-permutable pair of elements. Consequently it is one-step non-commutative. EXERCISE. Solve the homomorphy problem for the semigroups SI, SE, Sm of Theorem 449.
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