On Shioda’s problem about Jacobi sums II by

LXXIII.4 (1995)

On Shioda’s problem about Jacobi sums II

by

Hiroo Miki (Kyoto)

In the present paper, we will give a complete affirmative answer to the l-part of Shioda’s problem ([5, Question 3.4]) on Jacobi sums J _l ^(a) (p), and to the conjecture (F. Gouvˆea and N. Yui [1, Conjecture (1.9)]) which comes from Shioda’s problem and my congruences for Jacobi sums (see [3, Theo- rem 2]) (see Theorem 1 and its Corollary of the present paper).

We retain the notation of [4], but l is any odd prime number here. Fur- thermore, let n be any positive integer and let ζ m be a primitive mth root of unity in C for any positive integer m. Let Q be the algebraic closure of Q in C. We fix an algebraic closure Q _l of Q l , and by a fixed imbedding Q ,→ Q _l we consider Q as a subfield of Q _l . Let M be any finite unramified extension of Q _l in Q _l , and put M _n = M (ζ _l

) and π _n = ζ _l

− 1. Then π _n is a prime element of M n . Let σ −1 ∈ G = Gal(M n /M ) (the Galois group of M _n over M ) be such that ζ _l ^σ

= ζ _l ⁻¹

. Let ord _M

denote the normalized additive valuation of M _n , and let U _n = U (M _n ) be the group of principal units in M n :

U _n = U (M _n ) = {x ∈ M _n | ord _M

(x − 1) ≥ 1}.

As is well known, U _n is a multiplicatively written Z _l -module. In particular, x ^1/2 ∈ U _n makes sense for x ∈ U _n .

Lemma 1. Let the notation and assumptions be as above. Furthermore, let J ∈ U n be such that J 6∈ M . Put q ⁰ = J ^1+σ

, and assume q ⁰ ∈ M . Then ord _M

(1 − Jq ^0−1/2 ) is odd. In particular , ord _M

(1 − J) is equal to ord M

(1 − q ⁰ ) or odd.

P r o o f. Put e − = (1 − σ −1 )/2 and e + = (1 + σ −1 )/2. Note that e − , e + ∈ Z _l [G] (the group ring of G over Z _l ), since l 6= 2 and 1/2 ∈ Z _l . Put A = J ^e

. Since e ₋ + e ₊ = 1, we have

(1) A = J ^1−e

= Jq ^0−1/2 .

[373]

On the other hand, the equality e − σ −1 = −e − implies

(2) A ^σ

= A ⁻¹ .

If A = 1, then by (1) we have J = q ^01/2 ∈ M ; this contradicts the assump- tion. Hence A 6= 1, so we can write

A ≡ 1 + λπ _n ⁱ (mod π ⁱ⁺¹ _n ) with some unit λ in M and an integer i ≥ 1. Since

π _n ^σ

= ζ _l ⁻¹

− 1 = (1 + π _n ) ⁻¹ − 1 ≡ −π _n (mod π ² _n ), we have

(π _n ⁱ ) ^σ

≡ (−1) ⁱ π _n ⁱ (mod π _n ⁱ⁺¹ ).

Hence

(3) A ^σ

≡ 1 + (−1) ⁱ λπ _n ⁱ (mod π ⁱ⁺¹ _n ).

On the other hand,

(4) A ⁻¹ ≡ 1 − λπ _n ⁱ (mod π _n ⁱ⁺¹ ).

Since λ is a unit, by (2)–(4) we have (−1) ⁱ = −1, so i is odd.

For any positive integer m and any a ∈ Z and for any prime ideal p of Q(ζ _m ) which is prime to m, let

g m (p, a) = g m (p, a; ζ p ) = − X

x∈F

χ ^a _p (x)ψ p (x) ∈ Z[ζ mp ]

be the Gauss sum, where F _q = Z[ζ _m ]/p, q = N p = #(F _q ), χ _p (x) = ^x _p

m is the mth power residue symbol in Q(ζ m ), i.e., χ p (x mod p) is a unique mth root of unity in C such that

χ _p (x mod p) ≡ x ^{(N p−1)/m} (mod p)

for x ∈ Z[ζ _m ], x 6∈ p, χ _p (0) = 0, and ψ _p (x) = ζ p ^{T (x)} (p is a prime number in p and T is the trace from F q to Z/pZ).

For arbitrary positive integers m, r and any a = (a ₁ , . . . , a _r ) ∈ Z ^r (the direct product of r copies of Z) and for any p as above, let

J _m ^(a) (p) = (−1) ^r+1 X

x

+...+x

=−1 x

,...,x

∈F

χ ^a _p

(x ₁ ) . . . χ ^a _p

(x _r ) ∈ Z[ζ _m ]

be the Jacobi sum.

Theorem 1. Let the above notation and assumptions hold. Then:

(i) Assume that a 6≡ 0 (mod l ⁿ ) and that

(∗) g _l

(p, a) 6= q ^1/2 .

Then ord M

(1 − g l

(p, a)q ^−1/2 ) is odd, where M = Q l (ζ p ). In particular , ord _M

(1 − g _l

(p, a)) is equal to ord _M

(1 − q) or odd.

(ii) Assume that a = (a 1 , . . . , a r ) 6≡ (0, . . . , 0) (mod l ⁿ ) and that (∗∗) J _l ^(a)

(p) 6= q ^(r

^−2)/2 ,

where r ⁰ = #{0 ≤ i ≤ r | a _i ≡ 0 (mod l ⁿ )} and a ₀ = − P _r

i=1 a _i . Then ord M

(1−J _l ^(a)

(p)q ^−(r

^−2)/2 ) is odd, where M = Q l . In particular , ord M

(1−

J _l ^(a)

(p)) is equal to ord _M

(1 − q ^r

⁻² ) or odd.

P r o o f. (i) Put J = g l

(p, a) and χ = χ ^a _p . Since a 6≡ 0 (mod l ⁿ ), we have χ 6= 1. Hence by [6, Lemma 6.1(b)], we have

(1) J ^1+σ

= χ(−1)q = q,

since (−1) ^l

= −1 and χ(−1) = χ(−1) ^l

= 1. If J ∈ M , then by (1) we have J ² = q, so J = ±q ^1/2 . Since J ≡ q ≡ 1 (mod π _n ) and l 6= 2, this implies J = q ^1/2 ; this contradicts our assumption (∗). Hence J 6∈ M . Thus the assertion follows from Lemma 1 for q ⁰ = q.

(ii) Put J = J _l ^(a)

(p). It is well known that J = q ⁻¹

Y r i=0

g _l

(p, a _i )

if a 6≡ (0, . . . , 0) (mod l ⁿ ). By this equality and (1), we have J ^1+σ

= q ^r

⁻² .

If J ∈ M , then J ² = q ^r

⁻² , so J = ±q ^(r

^−2)/2 , hence J = q ^(r

^−2)/2 , since J ≡ q ≡ 1 (mod π _n ) and l 6= 2. This contradicts the assumption (∗∗). Hence J 6∈ M . Using Lemma 1 for q ⁰ = q ^r

⁻² , we have directly the assertion.

If r ≥ 3 is odd (r is as in the definition of Jacobi sums) and if a _i 6≡ 0 (mod l) for all i (0 ≤ i ≤ r) (a 0 = − P _r

i=1 a i ), then by Shioda [5, Corol- lary 3.3], we can write

N _Q(ζ

_)/Q (1 − J _l ^(a) (p)q ^−(r−1)/2 ) = Bl ³ /q ^w ,

where B and w are non-negative integers, and w is defined by (2.8) of [5].

Shioda’s problem (see [5, Question 3.4]). Is B a square?

By (ii) of the above Theorem 1, we have directly the following affirmative answer to the l-part of Shioda’s problem.

Corollary. Let the notation and assumptions be as above. Assume that B 6= 0. Then ord Q

(B) is even.

In the following, we will show that the case where J 6= q ^01/2 and

ord _M

(1 − J) = ord _M

(1 − q ⁰ ) actually happens in the above Theorem 1

when n = 1, as an application of our congruences for Gauss sums and Jacobi sums previously obtained by the author ([3, Theorems 1 and 2]).

Assume l ≥ 5. For any odd m (3 ≤ m ≤ l − 2), put ε m =

l−1 Y

d=1

(1 − ζ _l ^d ) ^m

,

where m _d ∈ Z is such that m _d ≡ d ^m−1 (mod l) and P _l−1

d=1 m _d = 0. Put k = Q(ζ l ) and K = k( √

ε m | m odd, 3 ≤ m ≤ l − 2).

Theorem 2. Let l, k, and K be as above and put K ⁰ = K( √

l). Then:

(i) If a 6≡ 0 (mod l) and deg p = 1, then the following (a)–(c) are equiv- alent:

(a) g l (p, a; ζ p ) ≡ 1 (mod l) for a suitable choice of ζ p .

(b) g _l (p, a; ζ _p ) ≡ 1 + ^p−1 ₂ (mod π ₁ ^l ) for a suitable choice of ζ _p . (c) p is completely decomposed with respect to K ⁰ /k.

(ii) (cf. [4, Theorem 3]). The following (d)–(f) are equivalent:

(d) J _l ^(a) (p) ≡ 1 (mod l) for any a ∈ Z ^r .

(e) J _l ^(a) (p) ≡ 1 + ^r

⁻² ₂ (q − 1) (mod π ₁ ^l ) for any a ∈ Z ^r , where r ⁰ is as in (ii) of Theorem 1.

(f) p is completely decomposed with respect to K/k.

P r o o f. (i) If r 6≡ 0 (mod p), then

g _l (p, a; ζ _p ^r ) = χ ^−a _p (r)g _l (p, a; ζ _p ).

Note that χ ^−a _p (r) is a primitive lth root of unity if r 6∈ (F ^× _p ) ^l . Hence by [3, Theorem 1] we see that g _l (p, a) ≡ 1 (mod π ² ₁ ) for a suitable choice of ζ _p if and only if α 1 ∈ F l (α 1 is as in [3, Theorem 1]). By [3, Theorem 7], this is equivalent to χ _p (l) = 1, i.e., l mod p ∈ (F ^× _p ) ^l . Hence by [3, Theorem 1] we have the assertion.

(ii) See [4, Theorem 3].

Lemma 2. Let k and K ⁰ be as in Theorem 2. Then K ⁰ and k( √

ζ _l ) are linearly disjoint over k. In particular , there exist infinitely many prime ideals p of k of degree 1 satisfying the condition (c) in Theorem 2 and p − 1 6≡ 0 (mod l ² ).

P r o o f. The proof of the first part is similar to that of [4, Lemma 2].

The last part follows from the first part and Chebotarev’s density theorem.

Concerning condition J 6= q ^01/2 , the following theorem is known.

Theorem 3. (i) ([2, (10)]). Assume that deg p = 1 and a 6≡ 0 (mod l).

Then Q(ζ p )(g l (p, a)) = Q(ζ pl ). In particular , g l (p, a) 6∈ Q l (ζ p ) and g l (p, a) 6=

q ^1/2 .

(ii) ([2, Theorem]). Assume that l - r, r 6≡ 1 (mod p) and deg p = 1. Put a = (1, 1, . . . , 1) ∈ Z ^r . Then Q(J _l ^(a) (p)) = Q(ζ l ). In particular , J _l ^(a) (p) 6∈ Q l

and J _l ^(a) (p) 6= q ^(r−1)/2 .

(iii) ([5, Theorem 7.1]). Let a = (a 1 , a 2 , a 3 ) ∈ Z ³ be such that a i 6≡ 0 (mod l) for all i (0 ≤ i ≤ 3) and such that a i + a j 6≡ 0 (mod l) if i 6= j, where a ₀ = −(a ₁ + a ₂ + a ₃ ). Then J _l ^(a) (p) 6= q if deg p = 1.

By Theorems 2 and 3, Lemma 2, Lemma 2 of [4], and Chebotarev’s density theorem, there exist infinitely many prime ideals p of k of degree 1 satisfying both J 6= q ^01/2 and ord _M

(1 − J) = ord _M

(1 − q ⁰ ), where J = g _l (p, a) or J _l ^(a) (p) according to (i) or (ii) of Theorem 2.

I would like to thank Professors Don Zagier, Yuji Kida, and Masanobu Kaneko for supplying me further numerical data on Shioda’s problem.

References

[1] F. G o u vˆea and N. Y u i, Arithmetic of Diagonal Hypersurfaces over Finite Fields, London Math. Soc. Lecture Note Ser. 200, Cambridge Univ. Press, 1995.

[2] M. K i d a and T. O n o, A note on Jacobi sums, Proc. Japan Acad. 69 (1993), 32–34.

[3] H. M i k i, On the l-adic expansion of certain Gauss sums and its applications, Adv.

Stud. Pure Math. 12 (1987), 87–118.

[4] —, On Shioda’s problem about Jacobi sums, Acta Arith. 69 (1995), 107–112.

[5] T. S h i o d a, Some observations on Jacobi sums, Adv. Stud. Pure Math. 12 (1987), 119–135.

[6] L. W a s h i n g t o n, Introduction to Cyclotomic Fields, Graduate Texts in Math. 83, Springer, New York, 1982.

DEPARTMENT OF LIBERAL ARTS AND SCIENCES FACULTY OF ENGINEERING AND DESIGN KYOTO INSTITUTE OF TECHNOLOGY SAKYO-KU, KYOTO 606, JAPAN

Received on 24.10.1994 (2684)