THE HASSE DAVENPORT RELATION. F = F q. is an automorphism of F, and the group of automorphisms of F is the cyclic group of order r generated by σ p,

Transcription

1 THE HASSE DAVENPORT RELATION 1. Environment: Field, Traces, Norms Let p be prime and let our ground field be F o = F p. Let q = p r for some r 1, and let the smaller of our two main fields be The map F = F q. σ p : F F, σ p (t) = t p is an automorphism of F, and the group of automorphisms of F is the cyclic group of order r generated by σ p, Aut(F ) = σ p = {1, σ p, σ 2 p,, σ r 1 p }. All such automorphisms fix F o pointwise, and conversely any element of F that is fixed by the automorphisms lies in F o. It suffices to check whether an element of F is fixed by the generator σ p. The trace function from F to F o symmetrizes each element additively by summing it and all of its automorphisim-conjugates, tr F/Fo : F F o, tr F/Fo (t) = σ Aut(F ) Note that indeed tr(t) lies in F o because it is fixed by automorphisms. The trace is an additive homomorphism, i.e., tr F/Fo (t + t ) = tr F/Fo (t) + tr F/Fo (t ), t, t F. Similarly, the norm function from F to F o symmetrizes each element multiplicatively, N F/Fo : F F o, N F/Fo (t) = The norm is a multiplicative homomorphism, σ Aut(F ) N F/Fo (tt ) = N F/Fo (t)n F/Fo (t ), t, t F. Fix some s 1 and let the larger of our two main fields be K = F q s. Note that K contains F as a subfield. Since also K = F p rs, the previous discussion of trace and norm applies verbatim with rs in place of r to give tr K/Fo : K F o, tr K/Fo (t) = σ(t) 1 σ Aut(K)

2 2 THE HASSE DAVENPORT RELATION and N K/Fo : K F o, N K/Fo (t) = σ Aut(K) But also, we now have a relative trace and norm. The map σ q : K K, σ q (t) = t q is an automorphism of K that fixes F, and the group of such automorphisms of F is the cyclic group of order s generated by σ q, Aut F (K) = σ q = {1, σ q, σq, 2, σq s 1 }. All such automorphisms fix F pointwise and any element of K that is fixed by the automorphisms lies in F, and it suffices to check whether an element of K is fixed by σ q. The relative trace function from K to F is tr K/F : K F, tr K/F (t) = σ(t), and the relative norm function from K to F is N K/F : K F, N K/F (t) = σ Aut F (K) σ Aut F (K) The relative trace is again additive and the relative norm is again multiplicative, and the traces and norms compose as nicely as they possibly could, tr K/Fo = tr F/Fo tr K/F and N K/Fo = N F/Fo N K/F. 2. Additive Characters, Multiplicative Characters, Gauss Sums Recall that F o = F p. Let ζ p = e 2πi/p C. An additive character of F o is ψ o : F o C, ψ o (t) = ζ t p. The corresponding additive character of F is ψ F : F C, ψ F = ψ o tr F/Fo, and the corresponding additive character of K is ψ K : K C, ψ K = ψ F tr K/F, Given also a nontrivial multiplicative character of F, χ F : F C, the corresponding multiplicative character of K is χ K : K C, χ K = χ F N K/F. Definition 2.1. The Gauss sum of χ F is and the Gauss sum of χ K is τ(χ F ) = t F χ F (t)ψ F (t), τ(χ K ) = t K χ K (t)ψ K (t). Here we are tacitly defining χ(0) = 0. Alternatively, we could sum over t F for the first Gauss sum and similarly for the second.

3 THE HASSE DAVENPORT RELATION 3 3. Gauss Sum Terms and Minimal Polynomials Let t be a nonzero element of K. Let H be the subgroup of Aut F (K) that fixes t. Then H takes the form H = σ d q for some d s. Thus t has d distinct conjugates in K, including itself. Denote these conjugates t 1 through t d where t 1 = t. Then tr K/F (t) = (s/d)(t t d ) and N K/F (t) = (t 1 t d ) s/d. Also, consider the polynomial f(x) = d (X t i ) = X d (t t d )X n ( 1) d (t 1 t d ). i=1 Certainly f(t) = 0 since t = t 1. Also, because any automorphism σ of K over F permutes the conjugates of t, the product form of f(x) shows that it is invariant when its coefficients are passed through any such σ. Thus the coefficients of f lie in the smaller field F. In fact f(x) is the smallest monic polynomial in F [X] satisfied by t, making it irreducible. The polynomial f(x) is the minimal polynomial of t over F. Rewrite the minimal polynomial of t as Then (s/d)c 1 = tr K/F (t) and c s/d n f(x) = X d c 1 X d ( 1) d c d = N K/F (t), and so (ψ F (c 1 )χ F (c d )) s/d = ψ F ((s/d)c 1 )χ F (c d ) s/d = ψ F (tr K/F (t))χ F (N K/F (t)) = ψ K (t)χ K (t), giving a term of the Gauss sum τ(χ K ). And furthermore, since t and its conjugates all have the same trace and norm and hence all have the same ψ K - and χ K -values, d(ψ F (c 1 )χ F (c d )) s/d = d ψ K (t i )χ K (t i ). Let MI denote the set of monic irreducible polynomials in F [X]. Each t K satisfies some f MI with deg(f) s, and conversely each such f MI divides X qs X so that its roots lie in K = spl F (X qs X). If f MI is specified, let d = deg(f) and let c 1 and c d be the coefficients of f as displayed in the previous paragraph. Then the previous display and the reasoning of this paragraph combine to give the following formula. i=1 Proposition 3.1. The Gauss sum for χ K where K = F q s τ(χ K ) = d s d(ψ F (c 1 )χ F (c d )) s/d. is

4 4 THE HASSE DAVENPORT RELATION 4. An Euler Factorization for Polynomials The calculations of the previous section suggest a general definition. Definition 4.1. Let M denote the set of monic polynomials in F [X], not necessarily irreducible. Define a function λ : M C as follows: For any f(x) = X d c 1 X d ( 1) d c d M, λ(f) = ψ F (c 1 )χ F (c d ). Note that in particular, λ(1) = ψ F (0)χ F (1) = 1. A little algebra shows that λ is multiplicative, λ(fg) = λ(f)λ(g) for all monic f, g M. That is, λ gathers the additive character ψ F and the multiplicative character χ F into a single multiplicative character on the monoid M. (A monoid is like a group but without inverses.) Proposition 4.2. The following Euler factorization identity holds for any multplicative function λ : M C, λ(f)t deg f = (1 λ(f)t deg f ) 1. Furthermore, for the particular λ of the previous definition, the left side of the previous display simplifies to λ(f)t deg f = 1 + τ(χ F )T. Proof. The fact that every monic polynomial factors uniquely into monic irreducibles gives the crucial third equality (in which the symbol f changes its meaning from a general monic irreducible polynomial on the left side of the equality to a general monic polynomial on the right side) in the calculation (1 λ(f)t deg f ) 1 = (λ(f)) n T n deg f n 0 = n 0 λ(f n )T deg(f n ) = λ(f)t deg f. This gives the Euler factorization. For the second part, we have λ(f)t deg f. λ(f)t deg f = n 0 d=n For n = 0 the inner sum is 1. For n = 1, the monic irreducible polynomials are f(x) = X t for all t F, with c 1 = c d = t, and so the inner sum is λ(f)t = λ(x t)t = ψ F (t)χ F (t)t = τ(χ F )T. t F t F d=1

5 THE HASSE DAVENPORT RELATION 5 For n 2, note that for each choice of c 1 and c n in F there are q n 2 monic polynomials with those coefficients. Thus λ(f)t = q n 2 ψ F (c 1 )χ F (c n ) = q n 2 ψ F (c 1 ) χ F (c n ). d=n c 1,c n F c 1 F c n F But both character sums are zero (for the second sum it is relevant that χ F is nontrivial), and so the entire expression vanishes. 5. The Hasse Davenport Relation Theorem 5.1 (Hasse Davenport Relation). The relation between the Gauss sums τ(χ K ) and τ(χ F ) is τ(χ K ) = ( τ(χ F )) s. Proof. From the previous proposition we have the relation 1 + τ(χ F )T = (1 λ(f)t deg(f) ) 1. Take logarithmic derivatives and multiply by T, τ(χ F )T 1 + τ(χ F )T = deg(f)λ(f)t deg(f) (1 λ(f)t deg(f) ), n 1 and then expand the geometric series, ( 1) n 1 τ(χ F ) n T n = Equate the coefficients of T s, ( τ(χ F ) s ) = d 1 deg(f)λ(f) d T d deg(f). d s dλ(f) s/d. The right side is τ(χ K ) by Proposition 3.1, so the proof is complete.