Determination of Some Properly Irregular Cyclotomic Fields

VoL. 16, 1930 MATHEMATICS: STAFFORD AND VANDIVER

139

4. It is probable that ammonia combines stoichiometrically with very few of the proteins. A case of stoichiometrical combination might perhaps be found in a protein consisting chiefly of dibasic a-mino-acids, because ammonia does combine stoichiometrically with one of the carboxyl groups in glutamic acid. 5. The fact that there are no solid ammonium caseinates or ammonium gelatinates at room temperature does not prove that there are no solid sodium caseinates or sodium gelatinates. 6. By using a solvent which will dissolve sodium hydroxide or acetic acid but which will not peptize or dissolve gelatin, sodium gelatinate, or gelatin acetate, it should be possible to tell whether solid sodium gelatinate and solid gelatin acetate are possible stable phases. This work is part of the programme now being carried out at Cornell University under a grant from the Heckscher Foundation for the Advancement of Research established by August Heckscher at Cornell University. *

DETERMINATION OF SOME PROPERLY IRREGULAR CYCLOTOMIC FIELDS By ELIZABUTH T. STAFFORD AND H. S. VANDIVER DZPARTMZNT OF MATHIRMATICS, UNIVZRSITY OF WISCONSIN DEPARTMENT OF PURE MATHEMATICS, UNIVERSITY or TExAs Communicated January 4, 1929

1. The integer h which represents the number of classes of ideals in a cyclotomic field defined by e2iT/l, where I is an odd prime, can be written in the form h1h2, where hi and h2 are both integers. If h _ 0 (mod 1) the field is called irregular. The necessary and sufficient condition that hi be divisible by I is that one of the first (1-3)/2 Bernoulli numbers be divisible by 1.1 A necessary, but not a sufficient, condition that h2 be divisible by I is that hi be divisible by 1.2 A cyclotomic field in which hi is divisible by 1, but h2 is prime to 1, is called a properly irregular cyclotomic field. By the use of the formulas derived in §2 it can be shown that, for I < 211, the only primes for which hi is divisible by I are I = 37, 59, 67, 101, 103, 131, 149, 157. By the methods outlined in §3 it can be demonstrated that h2 is not divisible by I for any of these primes. It follows that the cyclotomic field defined by e2iT/l will be a properly irregular cyclotomic field when I assumes any of these values. 2. The problem of determining irregular cyclotomic fields reduces to that of investigating the divisibility of the Bernoulli numbers. This problem is simplified by the use of the formulas which will now be derived.

140

MA THEMA TICS: STA FFORD A ND VANDI VER PROc. N. A. S.

The congruence3 (1 - ni)bi [u_n] 1 1=1i i: s'--l (mod 1) u=1 S=1 sfli where [ul/n] denotes the largest integer contained in ul/n, is satisfied if bi = -1/2, b2a+i = 0, b2a = (-)a-1 Ba, B1 = 1/6, B2 = 1/30, etc., being the Bernoulli numbers, 0 < i < I - 1. Define for I > 6

=i-

Ak

[kl/4] =

E

[(k-1)1/4] [kl/6]

s=

CR =

S2a-1

(2)

2al

(3)

1

S s= [(k-1)1/6] +1

where [x] is the greatest integer in x. Let n = 2 in (1) 22a) = [1/21 (1(1 - 2

b2a~ )__a

2a. 22a-1

Using the relation 2'1

(2"J2a

-

2a

2

S

'2

(mod 1).

s=1

(mod 1), this reduces

1

[1/4]

S2a-1

2a_

s=i

to

[1/2] E s2a-1 (modl) s=Vl/4]+1

+

or

A1 + A2 (mod 1).

(212G -2)b2a Let n =3 in (1) (1

-

(4)

t

[1/31 32a)bb Ja _E 2a1

2a.33

(31G - 3)b

S=1

_

[21/3]

S2a-1 + E S2a-1

(mod 1)

S=1

2(C1+ C2) + C3 + C4 (mod 1).

(5)

Since we may take l > 3, then I = 3k + 1 or 3k + 2. Consider

[I] + =

=

+

2([3

-

1= -1-[3] +

In- 1});n= 1, 2,..[]

Then

([L] + n)

([2]_ in _

-

1})2a-1

(-1)2a1 (mod 1),

VoL. 16, 1930 MA THEMA TICS: STA FFORD AND VANDI VER

_-_

([21]

in

1})2a1

-

141

(mod 1).

'rhe number of terms in the sum C3 + C4 is even, since if I = 3k + 1, (k even), I [3 ] 3] and if I 3k + 2, (k odd),

[2I A

k

[21I [1l = k + 1. Then

([s]

+ ([2!]

)

itn

)1

-

0 (mod 1).

It follows that

C3 + C4 Using this relation, (5) becomes

(31-2a -3)b2a

2(Cl + C2) (mod 1).

(5')

3A1 + 2A2 + A3 (mod 1).

(6)

2a

Let n = 4 in (1), (4-

we -

0 (mod l).

have

4)b2a

2a

But

A2 + A3 =

[31/4] s s=

[/4] -1

The number of terms in the sum A2 + A3 is even, since if I = 4k + 1, [31 -[fJ = 3k - k = 2k and if I = 4k + 3,

[31-[:1 = 3k + 2 - k = 2k + 2. Consider

[IJ + n = (3-1)-[3] + n; =1(31]{n1)

2,

1

MA THEMA TICS: STA FFORD AND VANDI VER PROC. N. A. S.

142

It follows that

+ ([3j

([i] + n)

- {n - 1)1

0 (mod 1)

and (6) becomes, since

A2 + A3 0 (mod 1) 2a-4)b2u 3A1 + A2 (mod 1). -

(6')

2a

Finally, let n = 6 in (1). The result may be written

(6l2a-6)b2_ 5C1 + 4C2 + 3C3 + 2C4 + C5

(mod 1).

(7)

In C2, the number of terms is given by [1/3] - [1/6]. If I = 6k + 1,

[~I-[~] =2k -k =k

and if l = 6k + 5,

[3I

[6]

= 2k + 1

-k = k + 1.

In C5 the number of terms is given by [51/6]- [21/3]. If I = 6k + 1, [51 [21] = k- 4k =-k

and if I = 6k + 5,

[5 2-1[] = 5k + 4 - 4k - 3 = k + 1. The number of terms in C2 equals the number of terms in C5, and ([6] +n)

(I

[6]+ {In

{n -

1})1

})

- ([6]-

(mod 1), n= 1,2, ... [h]

It follows that

C2 + C5

_

O (mod 1).

Then (7) becomes

(6'~- _ 6)b2a 2a

=

5C1 + C2 + C3 (mod 1).

(7)

VoL. 16, 1930 MA THEMA TICS: STAFFORD AND VANDIVER

143

From the definitions of Ak and Ck, it is evident that [1/21

(8)

.

Al+A2 = C+ C2+ C3= s-1S2 From (7')

6)b2a =2(C1 + C2) + 2C1 + A1 + A2 (mod 1) 4 41-2a 31-2a -_ 3 312a b2a + b2a + 2(C1-A1) (mod 1)

(61

-

2a

2a

or

2(Al

C1)

-

(31-2a

-

3)

+

(41-2a

-

4)

-

(61-2a

-

6) b

2

(mod

1).

2a In this congruence substitute for A1 - C1 and b2a the values given by (2)

and (3), we obtain for 2a < I - 1, for any odd 1, since it obviously holds for I = 5,

3'-2a -4-2a + 1-2a ( -52a-l _ 16" 1)aB

[1/4] s- [1/61 +1

(mod 1), (9)

4a

a result given without proof by Mirimanoff4 for the special case where I - 1 is divisible by 12. Also from (7')

2a

2 {2(C1 + C2)} 6) b2a= 2

+ A1 + A2

-2C2 (mod1)

2(31-2a - 3) + 21-2a - 2 2a or

C2 -

4a

(2-31- 6 +221

-

2-

6l1

+6) (modl).

Then (1/3]

s

E

1)aBa(2'-2a)B('4

1

2a-1 = 2a-

1)(31-2a )32

2a

[1/61+1

)(mod 0). (10)

From (6')

(3- 2- 3) b2a -2(C1 + C2) (mod 1) 2a 2(A1 + A2) - 2C3 (mod 1) 2)b2a - 2CW (mod 1) -2 (2 2 2a

or

2C24+1

4 4a

+

C-

1)a-' B.

(mod 1).

MA THEMA TICS: STA FFORD AND VA NDI VER PROC. N. A. S.

144

This gives [1/2]

21"'a - 31-26

E[/ 21-2a4a s11/31 +1 s2a-1

( -

la-B. (modl).

(11)

Finally I-1

s

S2a-1

=

12a1+ 32a-1 +

+

(I- 2)2a1 + 221

[1/2]

s1

Also

A1+A2+A3+A4 - 22-(A1 + A2) 12a-1 + 32a-1 +

+ (1

-

2)2a-l (mod 1).

But

A2 + A3 _ O(mod )and A, + A4 O (mod 1). Therefore from (4),

- 22a(1 1-122)b2,

+

(I -

2)2a-l

~ + 32a-1 + .... + (- 1)ailB )al,==2l

(I-

2)2al (mod 1)

2a. 22ai

2a-1 + 32a- +

(mod 1)

and

22a 1 222a

2-

(12) The formula (9) was applied for all values of I < 211. In case the left member of the congruence was congruent to zero modulo 1, the factor 1 - 41-2a - 31-2a + 6- 2a was investigated. If it was not congruent to zero, then B. is divisible by 1. As stated in §1, this condition was satisfied only for I = 37, 59, 67, 101, 103, 131, 149, 157. In case the factor 1 - 41-2a - 31-2a + 61-2a was congruent to zero, the divisibility of Ba was still indeterminate and one of the formulas (10), (11) or (12) was applied. For values of I between 100 and 210, (9) established the divisibility or non-divisibility of the Bernoulli numbers with the exception of the cases listed here. (a) I = 127 anda = 18, 25, 32, 36, 39, 46, 53, 60. Formula (11) was applied for these values of a and in each case the left member was not congruent to zero modulo 127. (b) I = 137 and a = 29. Formula (10), when applied for a = 29 was not congruent to zero modulo 137. (c) I = 151 and a = 23, 38, 53, 68. These cases were all settled by formula (11). (d) I = 167 and a = 42. This was settled by formula (12) which had its right member not congruent to zero modulo 167.

VOL. 16, 1930 MA THEMA TICS: STA FFORD AND VANDI VER

145

(e) I = 181 and a = 47. This was settled by formula (11). (f) I = 191 and a = 25, 48, 68. Formula (11) was used for a = 25, and 68, but formula (12) was necessary for a = 48. (g) I = 193 and a = 48. Here (9), (10), (11) and (12) all vanish modulo 1, but we find B48 O0 (mod 193) using actual value of B48 from Adams' tables. (h) I = 197 and a = 51. This was settled by (11). (i) I = 199 and a = 50. This was settled by formula (11). The computation involved in the application of these formulas was carried through by the use of a table of indices such as that in Cahen, Theorie des Nombres, vol. 2, p. 39. For 1 = 127 the part of the computation with a = 17, 18 and 19 is indicated below. S

1

22 23 24 25 26 27 28 29 30 31

70 11 77 96 38 69 35 79 26 50

2I

33!

N

36!

N

37I

N

14 22 28 66 76 12 70 32 52 100

42 111 21 18 120 9 21 87 102 12

19 5 20 2 47 111 20 54 87 94 459

56 7 49 84 70 21 91 119 28 112

52 105 90 107 22 20 59 75 68 37 635

70 29 77 24 20 33 35 25 80 86

22 46 24 73 13 102 28 83 113 124 628

Since [127/6] + 1 = 22 and [127/4] = 31, the first column contains the values of s to be used for each value of a. The second column (headed 1) contains the indices of the numbers as determined from the table of indices for I = 127. The next column (headed 21) is obtained by multiplying each of the numbers in the second column by 2 and then reducing modulo 126. If a = 17, 2a - 1 = 33 and the column headed 33I contains the residues modulo 126 of S33 for s = 22, 23, ..., 31. Add to each of these numbers the number of the 2I column which appears in the same row and reduce modulo 126. The resulting numbers will be those appearing in the column headed 35I. A repetition of this process gives the numbers appearing in the 37I column. In this manner the columns for a = 12, 13, ..., (1- 3)/2 (or for 2a - 1 = 23, 24, ..., I-4) were computed. The column headed N, which follows each kI column, was filled in by finding, in the table of indices and numbers, the number which had the index given in the kI column. These N columns were summed and reduced modulo 127. For a = 18 (or 2a - 1 = 35), the N column sums to 635- 0 (mod 127). Since the left member of (9) is congruent to zero, the factor 1 - 4 -2a - 3'-2a + 61-' must be investigated for I = 127 and a = 18. 1- 2a = 91 and 1 491 391 + 6"l = 1- 1-24 + 24 -0 (mod 127). Formula (9) does not decide the problem of the di-

-

MATHEMATICS: STAFFORD AND VANDIVER PROC. N. A. S.

146

visibility of Bi8 by 127. Using the method outlined above with formula (11) and a = 18, the left member of (11) is not congruent to zero modulo 127 so that B18 0 (mod 127). 3. In order to show that the cyclotomic fields defined by for I = 37, 59, 67, 101, 103, 131, 149, 157 are properly irregular cyclotomic fields, it must be proved that h2 is prime to I for each of these values. The divisibility of h2 by I can be investigated by means of certain properties of sets of independent units in the field defined by e21T/l. Kummer6 proves -

- e2 / ;i - 22 (ay-2); L'Y' primitive root of 1] a

,

that the set of units e(a), e(ay), . where

a -') =

e(a) = i(1 - a7)(1a

(1a)(a1-a')

a1-r/2

(13)

a- 1 a

-i

forms an independent set. Hence there exists a certain integral power of every unit in the field such that the unit raised to that power can be expressed as the product of integral powers of these independent units. Let ei(a), e2(a) . e,e..(a) form a fundamental set of real units.6 They satisfy the following regulations.

l(a!)"l

{ 61(a)'"

= =

e(a)rle(aT2

r2

e2(a)"' = e(a)'r1 e(aT) e,_1(a)"P-1 = e(a 1

....

2 e(a

-1 e(a"-2)r -2 e(a'y 2)r 1 )r2 .... e( A

(14) )r,

-1

where each nk has no factor in common with all of the integers rt, 4,r, rk..-1; the parentheses being omitted in the superscripts of the r's. Form

nlhlEi(a) = le(a) + rke(ay) + ... + rl_2le(a7 ) (15) { n2ke2(a) = r2le(aX) + rl22e(ae) + ... + r2,_le)a(7,A) n,-1 ks_.(a) = rF le(a) + r2 le(a') + + r,4 le(aY2 ) where lEk(a) stands for the real value of the logarithm of ek(a), etc. Now define D, A and R by the following determinants. -

D

=

e(a) le(a) le(a2) k(ae) le(a'2 le(ae3)

.

..(ay

...

le(ay"2) le(ae-') le(a'o) .

)

(eYPi) ke(a -4 )

(16)

VOL. 16, 1930 MA THEMATICS: STAFFORD AND VANDIVER

(a)

hl(a) 1E2(0i) .l.E.... .

A =

eIl(ae) le2(ad)

Ilf(aCz

..I

.-I(eY) (

....

(17)

-

r(18)

r1r2 1

... . . . . . . . .

le2(a""o-2)

)

147

2

K

If we form RD and actually carry out the multiplication of the determinants, using the multiplication theorem, and then substitute the values ., given by (15) where a is replaced successively by a, a'y, at'2 . a. relation we obtain the (19) RD = n1n2 .- n, A. But, R 0 0 is a necessary condition for an independent set of units,7 so that (19) may be replaced by

D/A

=

fnln2

n,

(20)

The expression D/A is the second factor of the class number.8 It is evident that if D/ A is to be divisible by 1, then at least one of the numbers nl, n2 ... n,,-1 must be divisible by 1. From this it follows that there exists a unit ea(a) such that (21) (ea(a))'g = e(a)T1e(a)" ... e(aY`2),i_,j where not all of the numbers r1, r2, ..., r..1 are divisible by 1. We have easily

e(z)-u

=

EJ 2'tY

...

E-y )

,a unit in k(a); i = 0, 1, 2, ..., ,- 1. Substituting in (21) gives Ea(a)lgP = E"E12 ... Et;,'# where

r2-Y2i

r;..l72i(I-2) i=1, 2, ...,,Now, not all the t's are divisible by I for if so we obtain tO0 (mod 1); i= 1, 2, ...,, - 1. Putting this in terms of the r's we get , - 1 congruences from which we X

=

rl +

may eliminate -y and obtain

+

...

+

MA THEMA TICS: STA FFORD A ND VA NDI VER PRoc. N. A. S.

148

r0-O (mod l); i = ly,2, ...,,u-1; since the determinant formed by the coefficients of the r's is prime to 1, but this contradicts the hypothesis that not all these r's have a common factor. Hence not all the t's have a common factor and by using a result that is established elsewhere9 it follows that we have Es = 51

for at least one value of i. As shown in another paper10 it follows that 0 (mod 1) and we can show that E. 76 el, B, _ 0 (mod 1). Hence if B where 51 is a unit in k(a), then it follows that k(a) is a properly irregular cyclotomic field for that value of 1. To effect this we select the smallest national prime q such that q - 1 (mod 1). Then q decomposes in k(a) into the product of I - 1 distinct ideal factors. Using Cahen's and Jacobi's tables of indices (and others constructed by the authors) for the prime q we select the rational integer d whose index is (q - 1)/i. Consider the ideal P = (a - d, q) which is a prime ideal factor of q. From this we obtain a- d (mod P). If we represent by En(d) the number obtained by substituting d for a in En we derive, from E. = 1,

En(d)

cl (mod q),

where c is a rational integer. We then reduce En(d) by the table of indices and determine if ind. En(d) = 0 (mod 1). If it is not, then k(a) is a properly irregular cyclotomic field. The table below gives the values of n, y, d, q for each irregular I < 211. The last column, headed ind. E"(d) gives the residue of ind. E"(d), modulo I in each case. d

I

n

37

16

2

17

59 67 101

22 29 34

2

IND. ES(d)

24

385

709

50

2 2

47 100

4 45

315 100 354

269 809 619

103

12

5

131

11

2

149

65

10

157

31 55

139 139

157

q

149 .

65

263

52

1093

127

1024

1571

1024

1571

150 39

Tables of indices for the primes 709, 269, 809, 619 and 263 are found in the "Canon Arithmeticus," C. G. J. Jacobi (1839), but tables of indices had to be constructed for the primes 1193 and 1571, the former to the base 10 and the latter to the base 2.

Vo. 16, 1930

MA THEMA TICS: STA FFORD AND VANDIVER

149

As the table shows, the computation established the fact that E, (a) is not the Ith power of a unit of the field for any of these cases. It follows that the cyclotomic fields defined by e2iT/l where I takes the values 37, 59, 67, 101, 103, 131, 149, 157 are properly irregular cyclotomic fields. All the calculations mentioned above were carried out by Miss Stafford with the exception of those relating to the proof that the second factor of the class number is prime to I for I = 157, the latter having been carried out by Mr. Vandiver. Miss Stafford made all her computations concerning the second factor of the class number for the various primes I in two different ways. The computations of Mr. Vandiver concerning the case I = 157 were all checked by Mr. S. S. Wilks, tutor in mathematics, University of Texas. For the case I = 37, Mr. Vandiver went through the computations concerning the second factor of the class number as a pattern for his work on the case I = 157. He thereby checked again the computations of Kummer and Miss Stafford for I = 37. The definition that was given at the beginning of this paper of a properly irregular cyclotomic field is equivalent to that which was given for such a field in another article,1' as it will be proved in another paper"2 that a necessary and sufficient condition that the second factor of the class number of a cyclotomic field defined by an odd prime I be divisible by 1 is that one of the units Ei(a) is the Ith power of a unit in a field; i = 1, 2, ..., (I- 3)/2. 4. In the determination of the irregular primes only the data was mentioned above concerning the computation for the primes 100 < I < 211. Using the process above described, Mrs. A. C. S. Williams checked all the primes < 100, and found only the irregular primes 37, 59 and 67 as had Kummer. Miss Stafford's computations as to the irregular primes check with Kummer's conclusion that the only irregular primes 100 < I < 167 are 101, 103, 131, 149 and 157. She found in each case the Bernoulli numbers divisible by I which Kummer had found. Her work concerning the divisibility of the second factor of the class number of k(a) by I for I = 37, 59 and 67 agrees with the results of Kummer's computations on the same.'3 The papers containing all the above-mentioned computations are now in the possession of Mr. Vandiver at the University of Texas; ultimately, they will probably be deposited in some University library. 1 Kummer, J. Math., 16, 1851, p. 479. A fraction in its lowest terms is said to be divisible by I if its numerator is divisible by 1. 2 Kumner, loc. cit., p. 486. 3Vandiver, Ann. Math., 18, 1917, p. 114. 4Crelle, 128, 1905, pp. 45-68. 5 J. Math., 16, 1851, p. 396. 6 Kummer, Did, 16, 1851, p. 399. 7 lbid, 16, 1851, p. 387.

MA THEMA TICS: E. W. BROWN

150

PROC. N. A. S.

8 Kummer, J. Math., 16, 1851, p. 471. 9 Vandiver, Bull. Amer. Math. Soc., 35, 1929, 333-35. 1"Transactions, Amer. Math. Soc., 31, 1929, pp. 628-9. The result is obtained by treating the relation E- = 8' in the same way that relation (15) p. 628 is treated. 11 Vandiver, Proc. Ndt. Acad. Sci., 15, 1929, p. 202. 12 Vandiver, Ibd., 16, 1930. 1I Berlin Abhandlungen, Math. Phys. Klasso, 1857, pp. 73-74.

ON AN EXTENSION OF THE FOURIER THEOREM GIVING RAPID METHODS FOR CALCULATING THE CONSTANT PART AND THE COEFFICIENT OF ANY PERIODIC TERM IN THE DISTURBING FUNCTION By

ERUms1 W. BROWN

DUPARTmNTS oF MATEMATICS AND ASTRONOmy, YALE UNIVRSrty Communicated December 27, 1929

1. Let F(X), G(X) be functions which may be expanded into cosines (or sines) of angles of the form iX + b, where i, b are constants. Let w be another variable related to X by means of the equation, (1.1) w = X + Gi(X). Then F(X) may be expressed in terms of w by means of Lagrange's theorem G*F1 + 2!dw-(G2GF) 3.!djW2 (G3-F1)+ *.., (1.2) where the notation, F = F(W), F1 = d F(w), G = G(w), Gi = d G(w), dw dw

F(X)

= F-

is used, and where it is supposed that G contains a small parameter as factor such that the series (1.2) can be used for calculation. Let

Hi

d H(w) Hi(w)_ dw

dH dw

be a function of the same nature as G and F, i.e., it is expressible as a sum of cosines (or sines) of angles of the form iw + b. In the applications, H1 contains no constant term, so that H contains no term of the form kw, but it will be shown that the result obtained is independent of the presence of such a term in H. We have, if D. d/dw,