The Hasse–Davenport Relations

Tools

3.7 The Hasse–Davenport Relations

3.7.2 The Hasse–Davenport Relations

We now have all the tools necessary to prove the Hasse–Davenport relations, at least whenp3.

Theorem 3.7.3 (Hasse–Davenport product relation). Letρbe a mul- tiplicative character of exact orderm|(q−1), and letχbe any multiplicative character on the finite field Fq. For any nontrivial additive character ψ we have

0a<m

τ(χρ^a, ψ) =−χ^−m(m)τ(χ^m, ψ)

0a<m

τ(ρ^a, ψ).

Proof.Since we know the valuations of Gauss sums for all primes and also their moduli, the proof of this theorem amounts to computing a specific root of unity, and forp3 this will follow from the distribution relation for the functiont proved above. There is a similar proof for p = 2, which will be given as Exercise 54 of Chapter 11.

First note that ifχis equal to a power ofρthenχ^m=ε, and the identity is trivial sinceτ(ε, ψ) = −1. We may therefore assume that this is not the case, hence that none of the characters occurring in the identity is the trivial character.

We use the notation of the proof of Stickelberger’s Theorem 3.6.6, and especially that of Section 3.6.2. Let P be a prime ideal of L =Q(ζq−1, ζp) abovep, and letω=ω_Pbe the corresponding character of orderq−1. Since ρ has exact order m we have ρ = ω^−dk with d = (q−1)/m for some k coprime tom, and replacingρbyω^−k (which does not change the set ofρ^a for 0a < m), we may assume thatk= 1. On the other hand, letbbe such thatχ=ω^−b. We thus have τ(χρ^a, ψ) =τ(ω^−(da+b), ψ). Set

ζ=−χ^m(m)

0a<mτ(χρ^a, ψ) τ(χ^m, ψ)

0a<mτ(ρ^a, ψ) =−ω^−mb(m)

0a<mτ(ω^−(da+b), ψ) τ(ω^−mb, ψ)

0a<mτ(ω^−da, ψ) . It follows from Stickelberger’s theorem (more precisely from Corollary 3.6.8) that

vP(ζ) =

0a<m

s(da+b)−

0a<m

s(da)−s(mb) = 0

by the distribution relation for the functions(a). Since this is true for every prime idealPabovepand since these are the only ones that can occur in the prime ideal decomposition of Gauss sums over the fieldFq, it follows thatζ is a unit inQ. Furthermore, from Proposition 2.5.14 we have

ζ=χ^m(m)Jm(χ, χρ, . . . , χρ^m−1)

Jm(χ^m, ρ, . . . , ρ^m−1) ∈Q(ζq−1).

Sinceχis not a power ofρ, each Gauss sum has modulus equal toq^1/2except τ(ρ⁰, ψ) =−1, soζhas modulus 1. Sinceσk ∈Gal(Q(ζq−1)/Q) sendsζq−1to ζ_q−1^k , henceχtoχ^k, it follows for the same reason that all the conjugates of ζhave modulus equal to 1. Thus by Kronecker’s theorem (Corollary 3.3.10) it follows that ζ is a root of unity belonging to Q(ζq−1); hence ζ =±ζ_q−1^k for some k ∈ Z. From the above expression of ζ in terms of Jacobi sums, it is clear that ζ does not depend on the choice of the nontrivial additive characterψ, so as usual we will chooseψ=ψ1.

Let again P be any prime ideal of L. By Stickelberger’s Theorem 3.6.8 we have

τ(ω^−r, ψ₁)≡ −(ζ_p−1)^s(r)

t(r) (modP^s(r)+1).

Since the powers ofζ_p−1 cancel by the distribution formula fors, it follows from the distribution formula fortthat

ζ≡ t(mb)

0a<mt(da) ω^mb(m)

0a<mt(da+b)≡(m/ω(m))^mb≡1 (modP), since by definitionω(m)≡m (modP). Now

1k<q−1(1−ζ_q^k₋₁) =q−1;

hence ifζ=ζ_q^k₋₁for 1k < q−1 we would have (1−ζ)|(q−1), so that 1−ζ would be prime top, in contradiction with the above congruence. Similarly, it is immediate that forp3 we have

0k<q−1, k =(q−1)/2(1 +ζ_q−1^k ) =q−1, giving again a contradiction if ζ =−ζ_q^k₋₁. Thusζ =ζ_q⁰₋₁ = 1, proving the

theorem forp3.

Remarks. (1) For p = 2 we have

0k<q−1, k =(q−1)/2(1 +ζ_q−1^k ) = 2, so that the possibility thatζ=−1 is not excluded by the last argument of the proof. To be able to exclude it we must computeζ moduloP², thus essentially modulo 4, and this will be done in Exercise 54 of Chapter 11.

(2) Since m^p−1 ≡ 1 (modp), it follows that as an element of Fq we have m^p−1= 1, henceχ(m^p−1) = 1, so thatχ(m) is a (p−1)st root of unity, and not simply a (q−1)st root of unity.

(3) By Proposition 2.5.9 the modulus of the product on the right-hand side is equal to q^(m−1)/2. We will see in Theorem 11.7.16 that it is in fact equal toq^(m−1)/2 times an explicit fourth root of unity.

Theorem 3.7.4 (Hasse–Davenport lifting relation). Let Fqⁿ/Fq be an extension of finite fields of degreen, letχ(respectivelyψ)be a nontrivial mul- tiplicative(respectively additive)character onFq. Define χ⁽ⁿ⁾=χ◦ N_Fq n/Fq

andψ⁽ⁿ⁾=ψ◦Tr_{Fq n/Fq}. Then

τ(χ⁽ⁿ⁾, ψ⁽ⁿ⁾) = (−1)ⁿ⁻¹τ(χ, ψ)ⁿ.

Despite its simplicity, this theorem is not so easy to prove, although in contrast to the HD product relation there does exist a relatively simple direct proof, which we give in Exercise 32.

Proof. Ifx∈Fqⁿ, the conjugates ofxare thex^qi for 0i < n. Thus N_Fq n/Fq(x) =

0i<n

x^qi =x^{(qn−1)/(q−1)}. On the other hand, by transitivity of the trace we have

Tr_Fq/Fp(Tr_{Fq n /Fq}(x)) = Tr_{Fq n}/Fp(x).

Keeping the notation of the preceding sections, we haveψ=ψbandχ=ω^−a for a uniqueb∈F^∗_q and integeramoduloq−1. It follows that

ψ⁽ⁿ⁾(x) =ψ◦Tr_{Fq n/Fq}(x) = Tr_{Fq n/Fp}(bx) =ψ⁽ⁿ⁾_b (x) χ⁽ⁿ⁾(x) =ω^−a(x^{(qn−1)/(q−1)}) = (ω⁽ⁿ⁾)^{−a(qn−1)/(q−1)},

using the additional upper index (n) on ψ_b and ω to indicate that they are with respect to the larger finite fieldFqⁿ. This implies in particular thatχ⁽ⁿ⁾ is also a nontrivial character since (qⁿ−1) |a(qⁿ−1)/(q−1) is equivalent to (q−1) | a, which is excluded since χ is nontrivial. Equivalently, this also immediately follows from the surjectivity of the norm from Fqⁿ to Fq

(Proposition 2.4.12), and the surjectivity of the trace (Proposition 2.4.11) implies thatψ⁽ⁿ⁾is also nontrivial. These immediate preliminaries out of the way, we can now mimic the proof of the HD product relation.

As above, setL=Q(ζq−1, ζp), and setζ= (−1)ⁿ⁻¹τ(χ⁽ⁿ⁾, ψ⁽ⁿ⁾)/τ(χ, ψ)ⁿ. Since the values of χ and ψ are in L, we have ζ ∈ L. Furthermore, Gal(L/Q(ζq−1)) is the group of mapsαkdefined byσk(ζp) =ζ_p^kforkcoprime top(see the footnote in the proof of Lemma 3.6.18). Clearly

αk(τ(χ, ψ)) =τ(χ, ψ^k) =χ⁻¹(k)τ(χ, ψ) by Lemma 2.5.6, and similarly

α_k(τ(χ⁽ⁿ⁾, ψ⁽ⁿ⁾)) = (χ⁽ⁿ⁾)⁻¹(k)τ(χ⁽ⁿ⁾, ψ⁽ⁿ⁾)

=χ⁻¹(N_Fq n/Fq(k))τ(χ⁽ⁿ⁾, ψ⁽ⁿ⁾) =χ⁻ⁿ(k)τ(χ⁽ⁿ⁾, ψ⁽ⁿ⁾). It follows thatα_k(ζ) =ζfor allkcoprime top, hence thatζ∈Q(ζ_q−1) (this part of the proof is the analogue of the use of Jacobi sums to prove the same result).

LetPbe a prime ideal ofLabove p. By Corollary 3.6.8 we have vP(τ(χ⁽ⁿ⁾, ψ⁽ⁿ⁾)) =vP

τ(ω^{−a(qn−1)/(q−1)}, ψ⁽ⁿ⁾)

=s(a(qⁿ−1)/(q−1)). Sincea(qⁿ−1)/(q−1) =

0i<naqⁱ andq is a power ofp, it follows from the definition ofsthat when 0a < q−1 we have

s(a(qⁿ−1)/(q−1)) =

0i<n

s(aqⁱ) =

0i<n

s(a) =ns(a) =nvP(τ(χ, ψ)), proving thatvP(ζ) = 0 for allPabovep. Since the moduli of all Gauss sums are powers of p, it follows that ζ is a unit. More precisely, since ψ⁽ⁿ⁾ and ψ⁽ⁿ⁾are nontrivial we have|τ(χ⁽ⁿ⁾, ψ⁽ⁿ⁾)|=q^n/2and|τ(χ, ψ)|=q^1/2, hence

|ζ| = 1. Changing ζq−1 into ζ_q−1^k for some k coprime toq−1 is equivalent to changingχ into χ^k henceχ⁽ⁿ⁾ into (χ⁽ⁿ⁾)^k. Thus all the conjugates of ζ also have modulus equal to 1. Since ζ is a unit it follows from Kronecker’s theorem thatζis a root of unity, and sinceζ∈Q(ζ_q−1), thatζis a (q−1)st root of unity. Finally, by Theorem 3.6.6 we have

τ(χ⁽ⁿ⁾, ψ⁽ⁿ⁾)≡ − (ζp−1)^ns(a)

t(a(qⁿ−1)/(q−1)) (mod P^ns(a)+1).

As we have seen above, the digits in basepofa(qⁿ−1)/(q−1) are the same as those ofarepeated ntimes; hence by Lemma 3.6.7 for 0a < q−1 we have t(a(qⁿ−1)/(q−1)) ≡ t(a)ⁿ (modp). Thus ζ ≡ 1 (modP), and we conclude as for the HD product relation thatζ= 1, finishing the proof.

Corollary 3.7.5. Ifχ1, . . . ,χk are multiplicative characters onFq that are not all trivial, andψ is a nontrivial additive character, then

Jk(χ⁽ⁿ⁾₁ , . . . , χ⁽ⁿ⁾_k ) = (−1)^{(n−1)(k−1)}(Jk(χ1, . . . , χk))ⁿ.

Proof. Immediate by inspection of the cases of Proposition 2.5.14.

Corollary 3.7.6. Assume thatpis odd, letq=p^f, and letχq be the unique multiplicative character of order2 of Fq. We have

τ(χ_q, ψ₁) =

(−1)^f−1q^1/2 if p≡1 (mod 4), (−1)^f−1i^fq^1/2 if p≡3 (mod 4).

Proof.Denote byχpthe Legendre symbol modulop. We haveN_Fq/Fp(x) =

0i<fx^pi =x^{(q−1)/(p−1)}, hence

χp◦ N_Fq/Fp(x)≡x^(q−1)/2≡χq(x) (modp),

since evidentlyχq(x) =x^(q−1)/2in the fieldFq. It follows from Theorem 3.7.4 that (with an evident abuse of notation)

τ(χq, ψ1) =τ(χ^(f)_p , ψ₁^(f)) = (−1)^f−1τ(χp, ψ1)^f .

This last Gauss sum over the finite field Fp is the same as the Gauss sum corresponding to the Dirichlet characterχ_p, so thanks to the determination of the sign of the quadratic Gauss sum (Theorem 2.2.19) we know its exact value: ifp≡1 (mod 4) it is equal top^1/2, so thatτ(χ_q, ψ₁) = (−1)^f−1q^1/2, and ifp≡3 (mod 4) it is equal top^1/2i, so thatτ(χq, ψ1) = (−1)^f−1i^fq^1/2.