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 s 4 (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 G K 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
142
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 •= s 4 (#/n ) = s4(Wpf)f where p is
the unique rational prime lying below p , and this has been determined
exactly in [6], viz
!
1
3
4
2
if p01 = 1 (mod 8) or if p = 2,
if p« = 29,
if p« = 5,
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 p w + 1 divides a -I- /3.
This result follows immediately by observing that P 7 p w + 1 — #/ P — IF2
(the field of 2 elements).
143
§ 2. An analogue of Hensel's lemma.
By Hensel's lemma it follows immediately that s is the least positive
integer such that
*? + ... + x 4 +1 = 0(modp 4<?+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
4
and 0 > 3 * 4 - l . Let x 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
+ 47T 3 Z 3 tf 1 + 7T4Z4
= u + 4TTZX\ ( m o d p ^ + 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# 3 z =
= — 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 p 3 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 a 2 + a ^ 0 (mod p). Find y G i ^ 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(a 2 + ce))(mod 16) = y 4 4-8 (mod 8p), and so 3 4- 3y 4 4- t* 44- 4(1 + P ) 4 + ( 1 4 - 2y 2 ) 4 = 8 4-1* 4- 16y 4- 32y 2 4- 16y 3 + 31y 4 4- 3 2 y 6 +
+ 16y 8 = 0 ( m o d 8 P ) = 0 ( m o d P 3 e + 1 ) . It follows that s < l l .
144
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(modpe/2+1p^...p;r).
Then a 4 = 40 where p | j 3 (0 G #). Find 7 such that 74/3 = 1 (mod P).
Then ( a 7 ) 4 = 4 (mod 4 p) and so 8 + 2(«7) 4 = 0 (mod 8p) = 0 (modp 3 * + 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,
d
4 +1)
d
4rf+1
and 8 + 3ir* 4- 7r <^ 4- (TT^(1 4- TT))4 = 8 4- \n* 4 4?r
4 6TT 4</+2 4-_
4- 4 T T 4 ^ 3 + 27r 4(d+1) = 8 4- 2?r4rf+2 (mod p 3<?+1 ) = 8 4 2TT2* (mod P 3 e + 1 ) =
s O ( m o d p 3 e + 1 ) since p 3 * ' | | 8 a n d \\2ir2e and p 3 7 p 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 (£ 2 7 r ' / 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 = ( ? ( v w ) , m square-free # 1, Z?^ a quadratic
°f Q, real or imaginary; then (cf theorems 2,3,4 of[l])
54(
,
*P)-
2
.4
) 6
15
if
1/
1/
if
m=
w=
m=
m=
extension
5 (mod 8)
3 (mod 4)
2(mod4)
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
145
find a 0 such that 0 3 = 1 (mod p 3 e + 1 ) , (S — a (mod p) (by Hensel's lemma).
1 - B3
:4
8
2
Then 1 4- 0 + 0 = 1 4- 0 + 0 =
_ fl = 0 (mod p 3 e + 1 ) . It follows that
Now suppose 5 = 1. Then x4 + y4 = 0 (mod p 3<?+1 ) is solvable in #
with p | A?. Find z G i> such that xz = 1 (mod P 3 e + 1 ); then 1 4- (yz)4 =
= 0(P 3 e + 1 )- Thus for t=yz
we have
(t-l)(t
+ l)(t2 4-l) + 2 = 0(modp 3 < ? + 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^p 4 )
(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 p 4 ) hold for each TT E p - p 2 . Then 7r2 = 2 + 7
2
8
(7EP4)
and
TT4 = 4 4 - 7 4 - 4 7 = 4 ( m o d p )
and
4-144-3-TT4 =
s 0 (mod p 8 ) = 0 (mod p 3 e + 1 ) . It follows that s < 6.
Now suppose x\ 4- ... 4- x4+l = 0 (mod p 7 )
(*)
2
8
where $\x1.
We may suppose that p | # ; for any i (otherwise p | x\ and
the s in (*) can be reduced). Now. observe that x = y (p 2 ) implies x4 =y4
(P 7 ) (for x4 -y4 = {x ~y){x +y)(x2 4-y 2 ) = 0 ( m o d p 6 ) , since each term =
= 0 ( m o d p 2 ) . If p1)fx4—y4
then p 2 ||#~~.y, x +y, x2+y2,
which gives
4
2
2
P | y and then P | x 4-y
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 P 2 ). We have (14- 7r)4 = 1 4- 47r 4- 6ir2 44- 47T3 -f n4 = 5 4- 47r 4- 67r2 (mod p 7 ) (since TT4 = 4 (mod p 8 )) = 1 (mod p 4 ),
and 7r4 = 0 (mod p 4 ). 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 p 7 ) 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 p 7 ) (^ 0 (mod p7))
since 2(5 4- 4tf 4- 67r 2 )= 10 4- 12 ir2 (mod P 7 ) = 10 4- 24 (mod p 7 ) (because
IT2 = 2 (modp 4 )) = 2 (mod p 7 ), and hence x 4 4- ... 4- #5 = 8 or 12 4- 4TT +
146
41 6ir2 (mod p 7 ) (^ 0 (mod p 7 ) and so finally x\ 4- ... 4- x\ = 12 or 4TT 44- 67r2 (mod p7) (^ 0 (mod p 7 ). It follows that s > 5 and so 5 = 6.
Now suppose that ir2 ^ 2 (mod p 4 ) for any rr £ p - p 2 . Then p 3 17r2 - 2
(since / = 1) and so we have 4 4- 7r4 = (7r2 4- 2) 2 - 47r2 = 0 (mod p 7 ). It
follows that s < 4. Finally let x\ 4- ... 4- x*+l = 0 (mod p 7 ) with P'jff,-. As
above we may suppose that Xj = l,7r, 1 + 7T for all i. Now (1 4- 7r)4 = 144- 47T 4- 67T2 4- 47T3 4- 7r4 =- 1 (mod p 5 ). 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 p 5 ).
^ 0 (mod P 7 ). 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) = p 2 and so e = 2, / = 1. We take
7r = \/?w so that IT2 = m = 2 (mod p 4 ) and so by lemma 2 , 5 = 6 (note that
7r€p-p2).
Finally let m = 3 (mod 4), then (2) = p 2 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 p 4 ) =£ 2
(mod p 4 ), (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 - 4C 3 (the discriminant, of
the cubic) —2vb!, where Cl,Dl,Af
are odd integers. Then
= 2
(i)
/ / D is odd and C is even, then (2) ~ px p 2 , degPx = 1, deg$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).
147
(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, s 4 (£p 2 ) = 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 ) ' = 1 5 , 5 4 (^p 2 ) = 4.
(vi)
/ / C is odd, D is even and v is odd, then ( 2 ) ~ p x P2 * w ^ s^ik^)
= 15, s4(& 2 ) = 6.
(vii)
/ / C,D are both even and \>ix>0,
(viii) / /
C, D
are both even and
( 2 ) ~ P i P 2 arcd s*(kpl)=
=
then ( 2 ) ~ p 3 #w<i s4(&«) = 9.
0<X</i
15, s 4 (fy 2 ) =
(*.£. X = 1, J U > 1 ) , t^^w
4 if M = 2 ,
6 if
fx>2.
LEMMA 3. Let e = 3, / = 1, fltew 5 = 9.
Proof. Let 7r be any element of P - P 2 . Then TT3 = 2 (mod p 4 ). We have
2(1 + TT)4 4- 6 + TT4 4- (?r(l 4- TT))4 = 8 + 4TT2 4- 2TT6 + 7r8(mod p 10 ) = 4?r2 44- TT8 (mod p 10 ) = - 4?r2 4- TT8 (mod p 10 ) = ?r2 (?r3 - 2)(TT3 4- 2) (mod p 10 ) = 0
(mod p 1 0 ). It follows that s < 9. Now suppose
x\ 4- ... + x4s+1 = 0 (mod p 1 0 )
(*)
We may suppose that p 3 | # * , p\xl...xr,
r > l and p|#,- if i>r. Notice
3
4
4
that if # = y ( m o d P ) then # = y (mod p 1 0 ). 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 p 4 ), and for i > r,
# 4 = 7r4, 7r4 4- 7r8 4- 8,7r8 (mod p 1 0 ) = O(mod P 4 ). Hence 4 | r giving r = 4
or 8. Now suppose r = 4. Then direct calculations show (using the fact that .
47r3 = 8 ( m o d p 1 0 ) ) t h a 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 =?TT 4 4-/7T8 + &-S ( 0 < i < 3 , 0 < / , * < l ) (mod p 1 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.
148
Next suppose r = 8. Then again a direct calculation shows that x\ 44- ... 4-#£ = 4 added to each element of the set (f) above (modp 1 0 ) and
none of these is = 0, -7r 4 , - (irq 4- 7r8 4- 8), - 8 (mod P 10 ) as a simple calculation shows. Hence s > 8. This completes the proof.
LEMMA 4. Let 3 | / , then s < 5.
Proof. #/p contains IF 8 , 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- j3 6 )(l -j3) = 0(mod p 3<?+1 ) giving 144-0 4-.... 4-/36 = 0 ( m o d p 3 e + 1 ) (since 1 - j3 £ P). We now have
3 - l 4 + ()3 4- j36)4 4- (0 2 4- 0 5 ) 4 4- (0 3 4- /34)4 = 3 4- j34 4- 4/32 + 6 4- 4/35 44- /33 4- 0 4- 4j34 4- 6 4- 40 3 4- j36 4- j32 4- 4j3 + 6 + 4]86 4- ]3S (mod p 3 * +1 ) =
= 21 4- 5(/3 4- jS2 4- ... 4- /36) (mod p 3 * +1 ) = 0(mod p 3 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
£ # such that
x? 4- x\ 4- # 4 4-#4 = 0 ( m o d p 4 )
(p{*i) .
X1,X2PX3,X4
(*)
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
14^
4
4/42
4
S 0 (mod p 4 )
(1)
Hence (1 + x + y + z) 4 = 0 (mod p) and therefore z = l + x+y($),
giving
2
2
2
2
4
4
z = 1+ x 4-y + 2(x +y + xy) (mod p ) and z = 1 4- x + y* + 2x2 44- 2y2 4- 2# 2 3/ 2 4- 4 [(# 4-y 4- ^ y ) 2 4- (# 4-y 4- # y ) ( l 4- # 2 4-y 2 )] (mod p)
and by (1) this implies 2(1 4-# 4 4-y 4 4- x2 + y2 4- # 2 y 2 ) = 0 (mod p 2 ), 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 #, 5 2 4- 5 4-1 ? 0 (mod p). This gives a2 4- j32 +
4-aj3 = 0 (mod p) ( a , / 3 G # ) implies plot and pi 0. (2) now gives pi x2 4-1
149
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- z 4 =4(mod p 3 ) and this
contradicts (1). That completes the proof.
LEMMA 6. Let e — 1, f ~ 3, then the congruence
2 + x4 4-y 4 + 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 z 4 = (x 4-y)4 =
= (#2 + y2 4- 2^y) 2 = # 4 + y 4 4- 4x2y2 4- 4x3/(x2 4-y 2 ) 4- 2x2y2 (mod p 3 ).
Eliminating z, using (1) gives
2 (a:4 4-y4 4- x2y2 + 1) + 4*y(# 2 4-y2 + xy) = 0 (mod p 3 )
(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-y 4 4- x2y2 4-1) 4- \xy = 0 (mod p 3 ) or
* 4 + y 4 + ^ 2 y 2 + 1 + 2#y = 0 (mod p 2 )
(4)
But by (3),
x4 + y4 4- x2y2 4- 2xy(x2
4-y 2 ) = 1 (mod P 2 )
(5)
(4) and (5) imply 2#y(# 2 4-y2 - 1) = 2(mod p 2 ) 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. ^ 2 4"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 ( m o d p 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- z 4 = 0 (mod p a ), contradicting
lemma 6.
150
Now observe that (*) implies xx 4- ... +xs = 0 (modp), and then it is
easy to check that the elements 0, XifX<2 X^.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.
f
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.
2.
3.
4.
5.
e•= 1 = / . Then by theorem 1, s = 15.
e — 1, / = 2. Then by lemma 1, 5 = 2.
£ = 1, / = 3. Then by lemma 7 , 5 = 5.
* = 3, / = 1. Then by lemma 3, 5 = 9.
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 p 2 . We have Q3 =- Cd — D and so since C,D are both
even, p x | 0, P 2 .| 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(0 2 + 2) =
= - ( C - 2 ) 0 - D , so p 2 * | | 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 || 0 2 4- 2 and hence 5 = 4 by lemma 2, while for JU > 2, P 2 | 0 2 44- 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 X 3 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 P 2 in t3\ Thus (2) = 1 1 ^ p 2 1 p 2 2 . Here clearly / , the
residue class degree of P2i is 1 and e, the ramification index of p 2 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 p221 || 2 X / A 7 . Hence p 2 x || (1 + VA 7 ) 2
and so p 2 1 || (1 4- N / A 7 ) . NOW in K, for P = P 2 1 , we take ir = 1 4- V A ' (the
7T of lemma 2). Then TT2 4- 2 = 1 + A' 4- 2 ( 1 + x/A 7 ) ? 0 (mod p ^ ) . It
follows that ^4(Kjp21) = 4.
Now take any 7r' G # such that p 2 || 7r'. Then P 21 || IT' i n l 3 \ Therefore
151
IT'2 4- 2 f£ 0 (mod p ^ ) (since s 4 (#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^A 7 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 s 4 (£p 2 ) = 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
<15
<12
<11
<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, s 4 (&p)=15; hence s4(k)>!5.
But s 4 (&)<
=
< 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)
152
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.:
169-176.
"Waring's problem for p-adic number fields",
[2]
— : "Waring's problem
(1961), 449-459.
[3]
Hasse, H.: "Zur Geschlecbtertheorie
Japan, 3 (1951), 45-51.
in quadratischen
[4]
Lang, S.: "Algebraic number theory"
(Addison-Wesley), 1970.
[5]
Niven, I.: "Sums of nth
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
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?
313-339.
in algebraic number fields",
Acta Arith. 9 (1964),
Proc. Camb. Phil. Soc. 57
Zahlkorpern",
powers of quadratic integers",
J. Math. Soc.
Duke Math. J. 8 (1941),
von Quadraten in einem
powers of algebraic integers",
Korper",
Ann. Math. 46 (1945),
153
[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
If
I