REND. SEM. MAT. UNIVERS. POLITECN. TORINO

Vol. 44°,1 (1986)

J.C. Parnami, M.K. Agrawal, A.R. Rajwade

ON THE FOURTH POWER STUFE OF p-ADIC COMPLETIONS OF ALGEBRAIC NUMBER FIELDS

§ 1. Introduction and notation.

The problem of representation of numbers of a field (ring) by mth

powers of elements of the field (ring) has been extensively discussed by var­ious authors such as Siegel [9], Birch [1] and [2], Ramanujam [8], Tornheim [10], Niven [5] and others. Of special interest is the minimal such represen­tation of -1 (when possible) and leads, in the case m — 2, to the notion of the stufe (length) of the field (ring) in question. This has been investigated by many and recently Pfister [7] has proved some very striking results about it.

Following Pfister, we define the fourth power stufe s4 (K) of a field K to be the least value of the positive integer 5 for which the equation

-1 =*} + ... +a* (*)

is solvable with aj GK If (*) is not solvable in K, we put s4(K) = <». In contrast to s4(K), the Pfister stufe of K will be denoted by s2(K).

When K is a totally complex algebraic number field, it is easy to prove, using the Hasse principle, that s2 (K) < 4 and in a recent paper of ours [6], we have shown that s4(K)< 16. For a given algebraic number field K the exact de­termination of s4(K) is difficult; indeed even s2(K) is known only when K is an imaginary quadratic field or a cyclotomic field, apart from perhaps some stray isolated examples. In [6], we showed that s4(Q(V~ m)) = 1 5 if m — = 7 (mod 8) and worked out some inequalities for s4(K) for other ima­ginary quadratic fields K. With the idea of determining s4(K) exactly for at

Classificazione per soggetto AMS (MOS, 1980): 12B99

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least some class of fields, we discuss, in this paper, the determination of s4(Kn), where Kn is the p-adic completion of K at p . This leads to certain non-trivial lower bounds for s4(K) and in some cases determines s4(K) exactly (see §6).

We use the following notation:

k as algebraic number field. # the ring of integers of k. P a finite prime ideal of k. &p the p-adic completion of k at p . i9p the ring of integers of &« . e = ep the ramification index of p .

f — fn the residue class degree of p.

s = s4(kn) the fourth power stufe of kp .

We note to begin with that

(i) s4(k9)<s4(k),

(ii) for p | 2 , by Hensel's lemma, 5 •= s4(#/n ) = s4(Wpf)f where p is the unique rational prime lying below p , and this has been determined exactly in [6], viz

!

1 if p01 = 1 (mod 8) or if p = 2, 3 if p« = 29, 4 if p« = 5, 2 otherwise.

So in what follows, we shall assume that <P| 2. In this paper we shall prove that s < 15 and equality holds if and only if e = / = 1. We also prove that 5 < 11 if k is a cyclotomic field and we determine s exactly when k is either a quadratic field or a cubic field (see theorems 2 and 3). All the results have been compiled in a table at the end of § 5.

In § 3 and 4 we shall frequently use the following elementary result: if p is a prime ideal with norm 2 in an algebraic number field k and a, |3 E #

are such that pw exactly divides both a and |3 then pw + 1 divides a -I- /3. This result follows immediately by observing that P 7 p w + 1 — #/ P — IF2

(the field of 2 elements).

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§ 2. An analogue of Hensel's lemma.

By Hensel's lemma it follows immediately that s is the least positive integer such that

*? + ... + x4+1 = 0(modp4<?+1)

has a solution in # in which at least one Xj is not divisible by p. We shall show that the exponent 4e + 1 of p can be replaced by 3 £ 4 - 1 .

Suppose x\ 4- ... + # 4+ 1 = 0 (modp^) has a solution in # with 9\xl

and 0 > 3 * 4 - l . Let x 4 4- ... + a£+1 = «, p ^ | u . Let T T G P ^ ^ - P ^ 2 ^

and z G d . Then

( # ! + 7TZ)4 4" #2 4" ... 4" X 4+ 1 = ft 4" 47TZ#3 4- 6lT2Z2xl + 47T3Z3tf1 + 7T4Z4

= u + 4TTZX\ (modp^+ 1 )

(since 2j3~ 4e + <? > |3 + 1 , 2e 4- 3 0 - 6e >j3 4- 1 , 4 0 - 3e >$ 4-1).

We have (47T#i ,p ^+ 1) = P^ | u. Hence there exists a z such that 47r#3z = = — u (mod P^+1) and for this choice of z, (#j 4- 7rz)4 4-#4 -!- ... 4- #4

+ 1 = = 0 (mod pP+1) and p | (# t + 7rz). NOW proceeding as in the proof of Hensel's lemma, we obtain a solution of x\ 4- ... 4- #4

+ 1 = 0, with xl G #p and the remaining #y G #.

§ 3. We prove the following

THEOREM l. s < 15 and equality holds if and only if' e = / = 1. 16 _

Proof. 2 ay = 0 (mod p3 e + 1) on taking each Xj = 1 and so by §2, s < 15. 7 = 1

When e=f=l, 0,1,2, . . . ,15 is a complete residue system modulo p 4 . Hence for any p-adic integer a, ce4 = 0 or 1 (mpdp 4 ) . It follows that 5 > 15 (for a rational integer n = 0 (mod p 4 ) if and only if n = 0 (mod 16)).

Let now ef > 1 and we have to show that 5 < 15.

Case 1: / > 1. Take an a G # , a ^ O , 1 (mod p) (such an a exists since N(P) = 2f>2). Then a2 + a ^ 0 (mod p). Find yGi^ such that y4(a2 + a) == = l (p) (such a y exists since #/p is a field with 2^ elements and so every element of #/p is a fourth power). Put t = y ( 1 4 - 2 a), then £4 = = y 4 ( l + 8(a2 + ce))(mod 16) = y 4 4-8 (mod 8p), and so 3 4- 3y4 4- t* 4-4- 4(1 + P ) 4 + ( 1 4 - 2y 2 ) 4 = 8 4-1* 4- 16y 4- 32y 2 4- 16y3 + 31y4 4- 32y 6 + + 16y 8 = 0 ( m o d 8 P ) = 0 (modP 3 e + 1 ) . It follows that s < l l .

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Case 2: e > 1, e even. Let (2) = P<?p[1 ... p r ^ be the decomposition of 2 into prime ideals in #. Take a = 0 (mod P ^ P * 1 . . . p ^ )

^ 0 ( m o d p e / 2 + 1 p ^ . . . p ; r ) . Then a4 = 40 where p | j3 (0 G #). Find 7 such that 74/3 = 1 (mod P). Then (a7)4 = 4 (mod 4 p) and so 8 + 2(«7)4 = 0 (mod 8p) = 0 (modp3*+1). It follows that 5 < 9.

Case 3: e > 1, e odd. We need only look at the case when / = 1 since the case / > 1 has already been dealt with.

Choose 7 r G p - p 2 and let d = (e~l)/2. Then 4d = 2e-2>e + 1,

and 8 + 3ir*d 4- 7r4<^+1) 4- (TT^(1 4- TT))4 = 8 4- \n*d 4 4?r4rf+1 4 6TT4</+2 4-_

4- 4 T T 4 ^ 3 + 27r4(d+1) = 8 4- 2?r4rf+2 (mod p3<?+1) = 8 4 2TT2* (mod P 3 e + 1 ) =

s O ( m o d p 3 e + 1 ) since p 3 * ' | | 8and \\2ir2e and p 3 7p 3 * + 1 s #/p (afield

of 2 elements, since / = 1). It follows that 5 < 12.

Remark. When k — Q(e21Tl/rn) (m > 2) is a cyclotomic field, we have 5 < 11.

Proof. If 81 m then 5 = 1 (since (£27r ' /8)4 = - 1 in k and so in &«). If / 1 4- A 4

41 m then Q ( f ) c & and since - 1 = 4 I—~—) so 5 < 4 . If m is odd

then e = 1 and / > 1, hence 5 < 11.

§ 4. We consider the case ef = 2 and deduce the following

THEOREM 2. i>£ k = ( ? (vw) , m square-free # 1, Z?̂ a quadratic extension °f Q, real or imaginary; then (cf theorems 2,3,4 of[l])

2 if m = 5 (mod 8) , . 4 1/ w = 3 (mod 4)

5 4 ( * P ) - ) 6 1/ m = 2 ( m o d 4 ) 15 if m= 1 (mod 8)

for every prime ideal p in k, dividing 2.

LEMMA l. (i) If / is even then s < 2. (ii) 7 / 5 = 1 £/?£« 4 I £.

Proof. Find an integer a G i } such that a3 = 1 (mod p) and a ^ 1 (mod p) (this is possible since 3 | 2^ - 1 and #/p is a field with 2^ elements). Next

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find a 0 such that 03 = 1 (mod p 3 e + 1 ) , (S — a (mod p) (by Hensel's lemma). 1 - B3

Then 1 4- 0:4 + 08 = 1 4- 0 + 02 = _ fl = 0 (mod p 3 e + 1 ) . It follows that

Now suppose 5 = 1. Then x4 + y4 = 0 (mod p3<?+1) is solvable in #

with p | A?. Find z G i> such that xz = 1 (mod P3 e + 1) ; then 1 4- (yz)4 =

= 0(P3 e + 1)- Thus for t=yz we have

(t-l)(t + l)(t2 4-l) + 2 = 0(modp3< ? + 1) .

This gives pe\\t4~l and so p | * - l . Let pa\\t-lr then p a | | / : 4 - l , p | t 2 4 - l and therefore 2a + Ke and hence p 2 a || t2 - 1 4- 2. Conse­quently p 4 a | | t 4 - l or 4 a = e.

COROLLARY. If e = 1, / = 2 then s = 2.

LEMMA 2. L££ £ = 2, / = 1. l££ 7r £>£ d»y element of & such that p ||7r

/• <=« ^ ™ i 6 '/ ^2 = 2(m^p4) (i.e. 7 r E p - p ^ ) . 70£» s= {

t 4 otherwise. (Note that if the condition 7r2 = 2(p 4) holds for any ir E P ~ P 2 , then it holds for all 7r E p - p 2 ) .

/Voo/! First let 7r2 =- 2 (mod p4) hold for each TT E p - p 2 . Then 7r2 = 2 + 7 ( 7 E P 4 ) and TT4 = 4 4-7 2 4 -47 = 4 ( m o d p 8 ) and 4 - 1 4 4 - 3 - T T 4 =

s 0 (mod p8) = 0 (mod p 3 e + 1 ) . It follows that s < 6. Now suppose x\ 4- ... 4- x4

+l = 0 (mod p7) (*) where $\x1. We may suppose that p 2 | # ; for any i (otherwise p 8 | x\ and the s in (*) can be reduced). Now. observe that x = y (p2) implies x4 =y4

(P7) (for x4 -y4 = {x ~y){x +y)(x2 4-y2) = 0 (modp 6 ) , since each term = = 0 ( m o d p 2 ) . If p1)fx4—y4 then p2 | |#~~.y, x +y, x2+y2, which gives P | y and then P4 | x2 4-y2 which is a contradiction). So without loss of generality we may suppose that each xt; = 1,7r, 1 4-7T (since 0, l,7r, 14-7T is a complete residue system modulo P2). We have (14- 7r)4 = 1 4- 47r 4- 6ir2 4-4- 47T3 -f n4 = 5 4- 47r 4- 67r2 (mod p7) (since TT4 = 4 (mod p8)) = 1 (mod p4) , and 7r4 = 0 (mod p4). So the number of i in (*) with p\x{ must be a mul­tiple of 4 and hence equals 4. Without loss of generality suppose p | and #4 = 7r4 = 4 (mod p7) if i > 4. Then x] = 1 or 5 4- 47T 4- 67r2 (mod p7) ( K i < 4 ) and so x4 4- ... + #£ = 4 or 8 4- 47r 4- 67r2 (mod p7) (^ 0 (mod p7)) since 2(5 4- 4tf 4- 67r2)= 10 4- 12 ir2 (mod P 7 ) = 10 4- 24 (mod p7) (because IT2 = 2 (modp 4 ) ) = 2 (mod p7), and hence x4 4- ... 4- #5 = 8 or 12 4- 4TT +

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41 6ir2 (mod p7) (^ 0 (mod p7) and so finally x\ 4- ... 4- x\ = 12 or 4TT 4-4- 67r2 (mod p7) (^ 0 (mod p7). It follows that s > 5 and so 5 = 6.

Now suppose that ir2 ^ 2 (mod p4) for any rr £ p - p 2 . Then p3 17r2 - 2 (since / = 1) and so we have 4 4- 7r4 = (7r2 4- 2)2 - 47r2 = 0 (mod p7). It follows that s < 4. Finally let x\ 4- ... 4- x*+l = 0 (mod p7) with P'jff,-. As above we may suppose that Xj = l,7r, 1 + 7T for all i. Now (1 4- 7r)4 = 14-4- 47T 4- 67T2 4- 47T3 4- 7r4 =- 1 (mod p5). It follows that the number of i with p | a?,- is a multiple of 4 and hence equals 4. Again without loss of generality p\xlx2x3x4. Then x\ 4- ... 4- x\ = 4 (mod p5).

^ 0 (mod P7). It follows that s > 3 and hence equals 4.

Proof of theorem 2. First let ra = 1 (mod 8), then (2) = pp' and so e = = / = 1 and by theorem 1, s = 15.

Next let m = 5 (mod 8), then (2) = p and so e = 1, / = 2 and by the corollary to lemma 1, 5 = 2.

Again let m = 2 (mod 4), then (2) = p2 and so e = 2, / = 1. We take 7r = \/?w so that IT2 = m = 2 (mod p4) and so by lemma 2 ,5 = 6 (note that 7 r € p - p 2 ) .

Finally let m = 3 (mod 4), then (2) = p2 again and so e — 2, / = 1. We take 7T = 1 4- s/w, so that TT2 = m 4-1 4- 2y/m = 2 \/m (mod p4) =£ 2 (mod p4), (since 2 j l - y ^ ) (note that this 7r E p - p 2 ) and so by lemma 2, 5 = 4.

§ 5. We consider the case ef = 3 and deduce the following

THEOREM 3. Let k = Q(6) be a cubic extension of Q where 6 is a zero of the integral polynomial

X3 + CX + D

where without loss of generality for any prime p* either p2\C or p3\D. Write C = 2XCx, D = 2tlDl and A = - 27 D2 - 4C3 (the discriminant, of the cubic) —2vb!, where Cl,Dl,A

f are odd integers. Then

(i) / / D is odd and C is even, then (2) ~ px p2 , degPx = 1, deg$2 = 2

and 54(&fl )'•= 15 and s4(k~ ) = 2.

(ii) / / C,D are &o£/? odd, then (2) ~ P tfrcd 5 = 54(&p) = 5.

(iii) / / C is odd, D is even, v is even and A* = 1 (mod 8), then ( 2 ) ^ P1P2P3 arcd 5 = 54(&p.) = 15 (i = 1,2,3).

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(iv) / / C is odd, D is even, v is even and A' = 5 (mod 8), then (2) ~ P1P2, degPi = 1, degp2 = 2 and s^(kp{)-= 15, s4(£p2) = 2.

(v) / / C is odd, D is even, v is even and A' = 3 (mod 4), tfow (2) ~ P t P ^ arcd s 4 (*5 i ) '=15 , 54(^p2) = 4.

(vi) / / C is odd, D is even and v is odd, then ( 2 ) ~ p xP2 *w^ s^ik^) = = 15, s4(& 2) = 6.

(vii) / / C,D are both even and \>ix>0, then ( 2 ) ~ p 3 #w<i s4(&«) = 9.

(viii) / / C, D are both even and 0 < X < / i (*.£. X = 1, J U > 1 ) , t^^w

( 2 ) ~ P i P 2 arcd s*(kpl)= 15, s4(fy2) =

LEMMA 3. Let e = 3, / = 1, fltew 5 = 9.

Proof. Let 7r be any element of P - P 2 . Then TT3 = 2 (mod p4). We have 2(1 + TT)4 4- 6 + TT4 4- (?r(l 4- TT))4 = 8 + 4TT2 4- 2TT6 + 7r8(mod p10) = 4?r2 4-4- TT8 (mod p10) = - 4?r2 4- TT8 (mod p10) = ?r2 (?r3 - 2)(TT3 4- 2) (mod p10) = 0 (mod p10). It follows that s < 9. Now suppose

x\ 4- ... + x4s+1 = 0 (mod p10) (*)

We may suppose that p 3 | # * , p\xl...xr, r > l and p|#,- if i>r. Notice that if # = y ( m o d P 3 ) then #4 = y 4 (mod p10). So without loss of genera­lity we may suppose that

, ( 1, 1 4" TT, 1 + 7T2 , 1 4- TT + TT2 for 1 < I < f ,

(7T, 7T 4" 7T2, 7T2 for I > r .

Now for i < r, #,• = 1 (mod p) and therefore x\ = 1 (mod p4) , and for i > r, # 4 = 7r4, 7r4 4- 7r8 4- 8,7r8 (mod p10) = O(mod P4). Hence 4 | r giving r = 4 or 8. Now suppose r = 4. Then direct calculations show (using the fact that . 47r3 = 8(modp 1 0 ) ) tha t

*i + ... + x\ = {4, 4 4- 4TT2 4- 2TT4 , 12 4- 4?r 4- 6?r2 + TT4 , 4 + 4?r2 4- 2?r4.+ TT8 ,

12+47r + 27T24-37r44-7r8, 12 + 4?r + 2TT2 + 3TT4, 4 4-TT8,

12 4-47T4-67T2 4-7r44-7T8} (modp 1 0) ( t )

and a£+1 4-... 4- *4+ 1 =?TT4 4-/7T8 + &-S ( 0 < i < 3 , 0 < / , * < l ) (mod p1 0) .

For i odd, the ^cact power of p in x\ 4- ... 4- # 4 4- ... 4- #4+ 1 can be 4 or 5

while for i even, this exact power can be 4 or 6 and this contradicts (*) above and so r =£ 4.

4 if M = 2 ,

6 if fx>2.

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Next suppose r = 8. Then again a direct calculation shows that x\ 4-4- ... 4-#£ = 4 added to each element of the set (f) above (modp1 0) and none of these is = 0, -7r4 , - (irq 4- 7r8 4- 8), - 8 (mod P10) as a simple cal­culation shows. Hence s > 8. This completes the proof.

LEMMA 4. Let 3 | / , then s < 5.

Proof. #/p contains IF8, the field of 8 elements. Hence there exists an a e # such that oP = 1 (modp), a ^ l ( m o d p ) . Find j3 such that j3 = = a (modp), j37 = l (mod P3<?+1) (in fact x1 = 1 has a solution in &« with x = a (modp)). Then (1 4- j8 + ... 4- j36)(l -j3) = 0(mod p3<?+1) giving 14-4-0 4-.... 4-/36 = 0 (modp 3 e + 1 ) (since 1 - j3 £ P). We now have

3 - l 4 + ()3 4- j36)4 4- (02 4- 05)4 4- (03 4- /34)4 = 3 4- j34 4- 4/32 + 6 4- 4/35 4-

4- /33 4- 0 4- 4j34 4- 6 4- 403 4- j36 4- j32 4- 4j3 + 6 + 4]86 4- ]3S (mod p3*+1) =

= 21 4- 5(/3 4- jS2 4- ... 4- /36) (mod p3*+1) = 0(mod p3 e + 1) .

It follows that s < 5 .

LEMMA 5. Let e = 1 and let f be odd, then s > 4.

Proof Suppose to the contrary that 5 < 4. Then there exists X1,X2PX3,X4

£ # such that

x? 4- x\ 4- #4 4-#4 = 0 (modp 4 ) ( p { * i ) . (*)

We can then find ^E i} such that xxp= l ( m o d p 4 ) and multiplying (*) by t\ we get x,y,zG& such that

1 4 ^ 4 4 / 4 2 4 S 0 (mod p4) (1)

Hence (1 + x + y + z)4 = 0 (mod p) and therefore z = l + x+y($), giving z2 = 1+ x2 4-y2 + 2(x +y + xy) (mod p2) and z4 = 1 4- x4 + y* + 2x2 4-4- 2y2 4- 2#23/2 4- 4 [(# 4-y 4- ̂ y ) 2 4- (# 4-y 4- # y ) ( l 4- #2 4-y2)] (mod p) and by (1) this implies 2(1 4-#4 4-y4 4- x2 + y2 4- # 2 y 2 ) = 0 (mod p2), giving 1 4- x4 4- y 4 4- x2 4- y 2 4- x2y2 = 0 (mod p). This implies

(x2 4-1)2 + (j/2 4-1)2 + (*2 + l ) (y 2 4-l) = 0 ( m o d p ) (2)

Now since f is odd so 3 | 2^—1 and so 7 3 = l ( m o d p ) implies 7 = 1 (mod p). Thus for any 6 6 #, 52 4- 5 4-1 ? 0 (mod p). This gives a2 4- j32 + 4-aj3 = 0 (mod p) (a , /3G#) implies plot and pi 0. (2) now gives pi x2 4-1

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and P\y2 4- 1 = (y — l ) 2 4- 2y. Hence x = 1 (mod P), y = 1 (mod p) and then z = l(modp). It follows that 1 4- x4 + y4 4- z4 =4(mod p3) and this contradicts (1). That completes the proof.

LEMMA 6. Let e — 1, f ~ 3, then the congruence

2 + x4 4-y4 + z 4 = 0 ( m o d p 3 ) (1)

te wo solution in #.

Proof. If a solution exists then z = # + y (mod p) and vso z4 = (x 4-y)4 = = (#2 + y2 4- 2^y)2 = #4 + y 4 4- 4x2y2 4- 4x3/(x2 4-y2) 4- 2x2y2 (mod p3). Eliminating z, using (1) gives

2 (a:4 4-y4 4- x2y2 + 1) + 4*y(#2 4-y2 + xy) = 0 (mod p3) (2)

This implies x4 4-y4 4- x2y2 + 1 = 0 (mod p), i.e.

x2 +y2 + xy 4- 1 =0 (mod P) (3)

(2) and (3) give 2(*4 4-y4 4- x2y2 4-1) 4- \xy = 0 (mod p3) or

*4 + y 4 + ^2y2 + 1 + 2#y = 0 (mod p2) (4)

But by (3),

x4 + y4 4- x2y2 4- 2xy(x2 4-y2) = 1 (mod P2) (5)

(4) and (5) imply 2#y(#2 4-y2 - 1) = 2(mod p2) i.e. xy(x2 + y2 - 1) = = l(modp), i.e. xy(x2 + y2 4-1) = l(mod P). But by (3) now this gives (x2 +y2 + l)2 = l(modP) i.e. (x2 + y2 4-1)= l(mod p), i.e. ^24"3/2 = = 0(modp), i.e. x=y(modp). Hence z =x +y = 2* = 0 (mod p). By symmetry # = y = 0 (mod p) too, hence (1) gives 2=0(modp 3 ) which is a contradiction. This proves the lemma.

LEMMA 7. Let e - 1, / = 3, £&£ra 5 = 5.

Proof. Suppose s =£ 5, then by lemmas 4 and 5, 5 = 4. Let

*4 + ... + 4 = 0(modp4) (*)

have a solution, where jp|ay for any /. We may assume that the Xj are distinct modulo P, for otherwise, multiplying by a suitable fourth power, (*) will yield a solution of 2 4- x4 + y4 4- z4 = 0 (mod pa), contradicting lemma 6.

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Now observe that (*) implies xx 4- ... +xs = 0 (modp), and then it is easy to check that the elements 0, XifX<2fX^.XApXcf X t 4" X 2 , X i 4" X 3 ,

#! + # 4 , #! + #5 (and indeed 6 other similar ones) are distinct modulo p . But #/p is a field of 2-̂ = 8 elements and this gives a contradiction.

Proof of theorem 3. The decomposition of (2) in k, as written in the state­ment of theorem 3 has been given by Wahlin [11]. For any p | 2, we have the following possibilities:

1. e•= 1 = / . Then by theorem 1, s = 15. 2. e — 1, / = 2. Then by lemma 1, 5 = 2. 3. £ = 1, / = 3. Then by lemma 7 , 5 = 5. 4. * = 3, / = 1. Then by lemma 3, 5 = 9. 5. e = 2, / = 1. Then by lemma 2, 5 = 4 or 6.

It remains to determine 5 in the last case i.e. when e = 2, / = 1, i.e. to de­termine 54(&p2) in the cases (v), (vi) and (viii) of the theorem.

Proof of (viii) for p2 . We have Q3 =- Cd — D and so since C,D are both

even, px | 0, P2.| 0. In case p 2 | 0 we have 2 | 0 which is not possible since

2 2 | C and so one can take the ir of lemma 2 to be = 0. Now 0(02 + 2) =

= - ( C - 2 ) 0 - D , so p2* | | 0 (0 2 +2) if M = 2 (since p | | ( C ~ 2)0 and

i\\D) and P ^ | 0 ( 0 2 + 2 ) if n>2 (since p^ | (C - 2)0 and $62 | )) . For

M = 2, P2 || 02 4- 2 and hence 5 = 4 by lemma 2, while for JU > 2, P 2 | 02 4-

4- 2 and again by lemma 2, 5 = 6.

Proof of the cases (v) and (vi) /o r P 2 . Let K = k(\/~A), the splitting field of X3 4- CX 4- D. We have [K: Q] = 6. Let t> be the ring of integers of K. A simple observation reveals that P i $ = P i i , P2$— P21 P22 is t n e decom­position of PJ and P2 in t3\ Thus (2) =11^ p 2 1 p 2 2 . Here clearly / , the residue class degree of P2i is 1 and e, the ramification index of p2 J is 2. We have now the two cases v even and odd.

Case (v) (v even). Then A' = 3 (mod 4) and (1 + V A 7 ) 2 " = 1 4- A' 4- 2%/A7. Here 4 | 1 + A', so p ^ 11 4- A' and p2

21 || 2 X / A 7 . Hence p 2x || (1 + VA7)2

and so p 2 1 || (1 4- N / A 7 ) . NOW in K, for P = P 2 1 , we take ir = 1 4- VA ' (the 7T of lemma 2). Then TT2 4- 2 = 1 + A' 4- 2 ( 1 + x/A7) ? 0 (mod p ^ ) . It follows that ^4(Kjp21) = 4.

Now take any 7r' G # such that p2 || 7r'. Then P21 || IT' i n l3 \ Therefore

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IT'2 4- 2 f£ 0 (mod p ^ ) (since s4(#p^ ) = 4) and hence n'2 4- 2 ^ 0 (modp*). It follows that 54(^p2) = 4 (again by lemma 2).

Case (vi) (p odd). Here V^A7 G tf and p 221 || 2A\ i.e. p 21 || N/^A 7 . Further

2A' 4- 2 = 0 (mod 4) and so = 0 (modp^). By lemma 2 it follows that s4(/irp2l) = 6. But now kp2 CKn and so s 4 (£p 2 )>6. But by lemma 2 again s4(£p2) = 4 or 6. It follows that1 s4(£'p2) = 6. This completes the proof of theorem 3.

Table of results

Let P| 2 and e,f be the ramification index, residue class degree of p. We use 'arb' for arbitrary and 'mult' for multiple.

e 1 1 2 1 3 arb > 1 arb even arb arb 1

f 1 2 1 3 1 arb arb > 1 arb mult of 3

even odd

s 15 2 4 or 6 5 9 < 1 5 < 1 2 < 1 1 < 9 < 5 < 2 > \

§ 6. s4(&) in some special cases.

I. If k is an algebraic number field such that

(i) some \/-m Gk where m is a natural number of the form Sn + 7, (ii) there exists a prime divisor p of 2 in k with en = /p = 1,

then s4(&) = 15.

Proof. By theorem 1, s4(&p)=15; hence s4(k)>!5. But s4(&)< <s 4 ( ( ? ( \ / -m) ) = 15, by theorem 2 of [6],

As an explicit example of the above situation, we may take

k = Q(Vnh, ..., yfm~r)

where the mj are integers of the form Sn 4- 1 and at least one of them is negative. To see this we note using Hasse [3] or proposition 17 of Lang [4] (chapter 3, page 68) that for p | 2, en = 1. Further for

srf= (2, (1 + >/w7)/2, ... , (1 4- Vm~r)/2)

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routine direct calculations show that N(stf) = 2, and so J#=p (a prime ideal dividing 2) with /« = 1.

It may similarly be shown, using theorem 2 and theorems 3-6 of [6] that

II. s4(&) = 6 if \/^~2E& and there is a prime divisor p of 2 in k with

As an esplicit example of this situation we may take

k - Q(V^2, \frnl, ... , \fm~r) , mj = l(mod 8) .

III. s4(&) = 4 if %/-T or \/^lE:k and there is a prime divisor p. of 2 in k with en = 2, /„ = 1.

As an example we may take k = Q(\f-5, \frn[, ..., y/m^) or Q(y/^1,

Vm1, ... , Vra^), with my = 1 (mod 8).

REFERENCES

[1] Birch, B.J.: "Waring's problem for p-adic number fields", Acta Arith. 9 (1964), 169-176.

[2] — : "Waring's problem in algebraic number fields", Proc. Camb. Phil. Soc. 57 (1961), 449-459.

[3] Hasse, H.: "Zur Geschlecbtertheorie in quadratischen Zahlkorpern", J. Math. Soc. Japan, 3 (1951), 45-51.

[4] Lang, S.: "Algebraic number theory" (Addison-Wesley), 1970.

[5] Niven, I.: "Sums of nth powers of quadratic integers", Duke Math. J. 8 (1941), 441-451.

[6] Parnami, J.C., Agrawal, M.K. and Rajwade, A.R.: "On the 4-power stufe of a field", Rendiconti del Circolo Math, di Palermo, 30 (1981), 245-254.

[7] Pfister, A.: "Darstellung von -1 als summe von Quadraten in einem Korper",

J. London Math Soc. 40 (1965), 159-165.

[8] Ramanujam, C.P.: "Sums of in*^ powers in p-adic rings", Mathematika, 10 (1963), 137-146.

[9] Siegel, C.L.: "Sums of mtf? powers of algebraic integers", Ann. Math. 46 (1945), 313-339.

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[10] Tornheim, L.: "Sums of n1^ powers in fields of prime characteristic", Duke Math. J. 4 (1938), 359-362.

[11] Wahlin, G.E.: "The factorization of the rational prime in a cubic domain", Amer. J. Math. 44(1922), 191-203.

J.C. PARNAMI, M.K. AGRAWAL, A.R. RAJWADE - Department of Mathematics, Panjab University, Chandigarh, India.

Lavoro peruenuto in redazione in 26/IX/1984

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