Purity of exponential sums on An

Purity of exponential sums on An Antonio Rojas-Le´on Abstract We give a purity result for two kinds of exponential sums of the type Px∈knψ(f(x)), where kis a finite field of characteristic pand ψ:k→C?is a non-trivial additive character. In the first case f∈k[x1, . . . , xn] is a polynomial of degree divisible by pwhose highest degree homogeneous form defines a non-singular projective hypersurface, and in the second one fis a polynomial of degree prime to pwhose highest degree homogeneous form defines a projective hypersurface with isolated singularities. 1. Introduction Let kbe a finite field of characteristic pand cardinality q, and let f∈k[x1, . . . , xn] be a polynomial of degree d. Pick a non-trivial additive character ψ:k→C?, and consider the sum Px∈knψ(f(x)). In [Del74] Deligne proved, as a corollary to his proof of the Riemann hypothesis for projective varieties over finite fields, the following estimate: Theorem 1.([Del74], Th´eor`eme 8.4) Suppose that i) The highest degree homogeneous form fdof fdefines a nonsingular hypersurface in Pn−1 ¯ k. ii) dis prime to p. Then we have the estimate ¯ ¯ ¯ ¯ ¯ X x∈kn ψ(f(x))¯ ¯ ¯ ¯ ¯ ⩽(d−1)n·qn/2. Moreover, he showed that the sum is pure of weight nand rank (d−1)n. In particular, there are (d−1)ncomplex algebraic numbers α1, . . . , α(d−1)n, all pure of weight n(meaning that all their conjugates over Qhave absolute value qn/2) such that, for every integer m⩾1, if kmdenotes the degree mextension of kin a fixed algebraic closure ¯ k, we have (−1)nX x∈kn m ψ(Tracekm/k(f(x))) = (d−1)n X i=1 αm i. What can we say in the case where pdivides d? By perversity arguments (cf. [KL85], [Kat93], [Kat04]) we know that the sum is pure for almost all f∈k[x1, . . . , xn]. More precisely, if we add a sufficiently general linear form to f(one that is contained in a suitable Zariski dense open subset Uof the dual affine space ˆ An kdepending on ψand q), the sum becomes pure of weight n. However, these results do not give us any information about the sum associated to a particular f. On the other hand, in [AS00b] Adolphson and Sperber show, using p-adic methods, that if fsatisfies certain regularity hypotheses the L-function associated to the exponential sum (or its inverse) is 2000 Mathematics Subject Classification 11L03,11L07 Keywords: exponential sums, l-adic cohomology Partially supported by MTM2004-07203-C02-01 and FEDER Antonio Rojas-Le´ on a polynomial. In this article we will use these results to give a version of Theorem 1 for the case where pdivides d. Fix a prime `6=pand an isomorphism ι:¯ Q`→Cso that we can speak about absolute values of elements of ¯ Q`and weights without ambiguity. From now on we will assume that such an isomorphism has been chosen, without making any further reference to it. Thus, for every α∈¯ Q`, |α|will always mean |ι(α)|. We will also use this isomorphism to identify the sets of C?-valued characters and of ¯ Q? `-valued characters of any finite group. Consider the lisse Artin-Schreier ¯ Q`- sheaf Lψon A1 kassociated to the non-trivial additive character ψ:k→C?(cf. [Del77], 1.7). For every finite extension k0/k and every t∈A1(k0) = k0, the trace of the geometric Frobenius element in Gal(¯ k/k0) acting on the stalk of Lψat a geometric point ¯ tover tis ψ(Tracek0/k(t)). In particular, since ψtakes its values among the roots of unity, Lψis pure of weight 0. Let Lψ(f)denote the pull-back f?Lψon An k. The cohomology groups with compact support Hi c(An ¯ k,Lψ(f)) are endowed with an action of the absolute Galois group Gal(¯ k/k) and, in particular, of the geometric Frobenius element F∈Gal(¯ k/k). By the Grothendieck trace formula we have X x∈kn ψ(f(x)) = 2n X i=0 (−1)iTrace(F|Hi c(An ¯ k,Lψ(f))). Our first result is the following Theorem 2.Let dbe divisible by p. Write f=fd+fd0+f0, where fdis the degree dhomogeneous component of f,d0is the degree of f−fdand fd0is the degree d0homogeneous component of f. Suppose that a) d0/d > p/(p+ (p−1)2)and d0is prime to p. b) The equation fd= 0 defines a non-singular hypersurface in Pn−1 ¯ k. c) The hypersurface defined in Pn−1 ¯ kby fd0= 0 does not contain any of the common zeroes of ∂fd ∂x1, . . . , ∂fd ∂xnin Pn−1 ¯ k. Then 1. Hi c(An ¯ k,Lψ(f)) = 0 for i6=n. 2. Hn c(An ¯ k,Lψ(f))has dimension (d0(d−1)n+ (−1)n(d−d0))/d and is pure of weight n. 3. We have the estimate ¯ ¯ ¯ ¯ ¯ X x∈kn ψ(f(x))¯ ¯ ¯ ¯ ¯ ⩽d0(d−1)n+ (−1)n(d−d0) d·qn/2. For d0=d−1 (the generic case) the inequality in (a) holds as long as d⩾3, and we get Corollary 3.Assume d⩾3is divisible by p. Let f=fd+fd−1+f0be as above. Suppose that a) The equation fd= 0 defines a non-singular hypersurface in Pn−1 ¯ k. b) The equation fd−1= 0 defines a hypersurface in Pn−1 ¯ kwhich does not contain any of the common zeroes of ∂fd ∂x1, . . . , ∂fd ∂xnin Pn−1 ¯ k. Then 1. Hi c(An ¯ k,Lψ(f)) = 0 for i6=n. 2. Hn c(An ¯ k,Lψ(f))has dimension ((d−1)n+1 −(−1)n+1)/d and is pure of weight n. 2 Purity of exponential sums on An 3. We have the estimate ¯ ¯ ¯ ¯ ¯ X x∈kn ψ(f(x))¯ ¯ ¯ ¯ ¯ ⩽(d−1)n+1 −(−1)n+1 d·qn/2. As usual, (3) is a consequence of the vanishing of the cohomology together with Deligne’s theorem on weights (cf. [Del80], Corollaire 3.3.4). The second result deals with another kind of sum studied by Adolphson and Sperber in [AS00b] and is a generalization of ([Gar98], Theorem 0.4). Let f∈k[x1, . . . , xn] be a polynomial of degree d, which we will now assume to be prime to p. We will show Theorem 4.Write f=fd+fd0+f0as in Theorem 2. Suppose that a) d0/d > p/(p+ (p−1)2)and d0is prime to p. b) The hypersurface defined by fd= 0 in Pn−1 ¯ khas at worst weighted homogeneous isolated singularities of total degrees d1, . . . , dsprime to p(cf. [AS00b], Section 2 or [Gar98], 0.3 for the definitions). c) The hypersurface defined by fd0= 0 in Pn−1 ¯ kdoes not contain any of these singularities. Let µ1, . . . , µsbe the Milnor numbers corresponding to the singularities of fd= 0. Then 1. Hi c(An ¯ k,Lψ(f)) = 0 for i6=n. 2. Hn c(An ¯ k,Lψ(f))has dimension (d−1)n−(d−d0)Ps i=1 µiand is pure of weight n. 3. We have the estimate ¯ ¯ ¯ ¯ ¯ X x∈kn ψ(f(x))¯ ¯ ¯ ¯ ¯ ⩽((d−1)n−(d−d0) s X i=1 µi)·qn/2. 2. A cohomological vanishing result In this section we will begin the proof of Theorem 2. We will first use the method of pencils to show the vanishing of Hi c(An ¯ k,Lψ(f)) for i > n + 1. This requires studying the fibers of the map f, so the first thing we need to do is find a suitable compactification of f. Unfortunately, the compactification defined in [Kat99] by embedding Anas a dense open subset of the subscheme of Pn×A1given by the vanishing of F−λXd 0no longer works in this case. The reason is that we are compactifying a map of degree divisible by p, and this may introduce some wild ramification at infinity in the higher direct images of the constant sheaf with respect to the compactified map. Therefore, instead of directly compactifying f, the idea is to first write fas the composition of a closed embedding of Anin An×A1(given by the graph of f) followed by the projection, and then compactify the projection restricted to the image of An. Since we are compactifying a map of degree 1, we do not run into any problems caused by wild ramification. However, one disadvantage of this compactification is that the fiber at infinity will always have a singular point, so we will only be able to deduce the vanishing of the cohomology groups for i > n + 1. Proposition 5.Suppose that the equation fd= 0 defines a non-singular hypersurface in Pn−1 ¯ k. Then Hi c(An ¯ k,Lψ(f)) = 0 for i > n + 1. Proof. Define Zto be the hypersurface in Pn+1 k(where we take coordinates X0, . . . , Xn, T) defined by the vanishing of F−TXd−1 0, where Fis the homogenization of fwith respect to the variable X0(i.e. F(X0, . . . , Xn) = Xd 0·f(X1/X0, . . . , Xn/X0)). The affine space An kis naturally an open subscheme of Z(just by embedding it in An+1 kusing the graph of f, and then identifying An+1 kwith Pn+1 kminus the hyperplane X0= 0). 3 Antonio Rojas-Le´ on Next, we define the incidence variety ˜ Zas a divisor of Z×P1 k, given (with coordinates X0, . . . , Xn, T for the first factor and λ0, λ1for the second one) by the zero locus of λ0T−λ1X0. Thus ˜ Z(¯ k) = {((x0, . . . , xn, t),(λ0, λ1)) ∈Z(¯ k)×P1(¯ k) : λ0t=λ1x0}. Let ˜ f:˜ Z→P1 kbe the restriction to ˜ Zof the canonical projection π2:Z×P1 k→P1 k. It is a proper map, being the composite of a closed immersion and a proper projection (since Zis projective). The open subset An k,→Zcan be embedded as an open subscheme of ˜ Zin the obvious way. Namely, we identify the point x∈An(¯ k) with (x, f(x)) ∈˜ Z(¯ k). In this way we get a commutative diagram An k−−−−→ ˜ Z f  y  y ˜ f A1 k−−−−→ P1 k where the horizontal arrows are open embeddings. The image of An kin ˜ Zcan be described as the set of (x, λ)∈˜ Zsuch that x6∈ Z∩ {X0= 0}. Before going any further we need to show that ˜ fis a flat map. Lemma 6.The map ˜ f:˜ Z→P1 kis flat. Proof. By ([Har77], Proposition III.9.9) it suffices to show that all geometric fibers of ˜ fhave the same Hilbert polynomial. The fiber over a finite point λ∈A1(¯ k) is easily seen to be the complete intersection of the degree dhypersurface F−λXd 0= 0 and the hyperplane T−λX0= 0. Similarly, the fiber over infinity is the complete intersection of the hypersurface F= 0 and the hyperplane X0= 0. Since the Hilbert polynomial of a complete intersection only depends on its multidegree, we conclude that it is the same for all geometric fibers of ˜ f. We extend by zero the sheaf Lψto the whole P1 k, and take its pull-back by ˜ fto ˜ Z, which we will also denote by Lψ(f). This is compatible with the previous notation, since its restriction to An k is just the pull-back of Lψby f. Lemma 7.There is a quasi-isomorphism RΓc(An ¯ k,Lψ(f))∼ →RΓc(˜ Z⊗¯ k, Lψ(f)). Proof. To simplify the notation, we will identify each homogeneous form with the projective hypersurface defined by its vanishing. It is clear that ˜ Z1:= (Z∩T∩X0)×P1 kis contained in ˜ Zas a closed subscheme. Let ˜ Z0be its complement. The restriction of ˜ fto ˜ Z1is just the second projection. From the decomposition ˜ Z0 j ,→˜ Zi ←-˜ Z1 we get an exact sequence of sheaves 0→j!j?Lψ(f)→ Lψ(f)→i?i?Lψ(f)→0 from which we get a distinguished triangle in Db(¯ Q`−vector spaces) RΓc(˜ Z0⊗¯ k, Lψ(f))→RΓc(˜ Z⊗¯ k, Lψ(f))→RΓc(˜ Z1⊗¯ k, Lψ(f))→ Now in ˜ Z1∼ =(Z∩T∩X0)×P1 kthe sheaf Lψ(f)is just the external tensor product ¯ Q`£Lψ. Therefore by the K¨unneth formula we have RΓc(˜ Z1⊗¯ k, Lψ(f)) = RΓc((Z∩T∩X0)⊗¯ k, ¯ Q`)⊗RΓc(P1 ¯ k,Lψ) = 0 4 Purity of exponential sums on An since RΓc(P1 ¯ k,Lψ) = RΓc(A1 ¯ k,Lψ) = 0 (cf. [Del77], Th´eor`eme 2.7*). Hence we get a quasi-isomorphism RΓc(˜ Z0⊗¯ k, Lψ(f))∼ →RΓc(˜ Z⊗¯ k, Lψ(f)). The image of the open immersion h:An k,→˜ Z0is the set of (x, λ)∈˜ Zsuch that x6∈ Z∩X0. Its complement in ˜ Z0is the set of (x, λ)∈˜ Zsuch that x∈Z∩X0and x6∈ Z∩T, so it maps to the point at infinity under ˜ f. Since the stalk of Lψat infinity is zero, we have an equality h!h?Lψ(f)=Lψ(f), and therefore a quasi-isomorphism RΓc(An ¯ k,Lψ(f))∼ →RΓc(˜ Z0⊗¯ k, Lψ(f))∼ →RΓc(˜ Z⊗¯ k, Lψ(f)). We will also denote by ˜ f:˜ Z⊗¯ k→P1 ¯ kthe map deduced from ˜ f:˜ Z→P1 kby extension of scalars to ¯ k. Since ˜ fis proper, we have (by composition of derived functors) RΓc(˜ Z⊗¯ k, Lψ(f)) = RΓc(P1 ¯ k,R˜ f?Lψ(f)). On the other hand, by the projection formula we have R˜ f?Lψ(f)= R ˜ f?(¯ Q`⊗˜ f?Lψ) = R ˜ f?¯ Q`⊗ Lψ so Proposition 5 is equivalent to Proposition 8.Under the previous hypotheses the cohomology group Hi c(P1 ¯ k,R˜ f?¯ Q`⊗Lψ)vanishes for i > n + 1. Therefore we will prove Proposition 8 instead. Proposition 9.The sheaves Ri˜ f?¯ Q`on P1 ¯ kare lisse for i⩾n+ 1. For i=nit is the extension of a lisse sheaf by a punctual sheaf. Proof. The fiber of ˜ fat a point λ∈A1(¯ k) is defined in Pn+1 ¯ k(with the usual coordinates X0, . . . , Xn, T) by the homogeneous ideal (F−TXd−1 0, T −λX0) = (F−λXd 0, T −λX0). Its intersection with the hyperplane X0= 0 is then defined by the ideal (F, X0, T ), and is therefore isomorphic to the hypersurface defined in Pn−1 ¯ kby fd= 0, which is non-singular by hypothesis. Therefore, the fiber itself has at worst isolated singularities. On the other hand, the fiber at λ=∞is defined in Pn+1 ¯ kby the ideal (F, X0). This is the projective cone over the hypersurface defined in Pn−1 ¯ kby fd= 0, so it has only one singular point (the vertex). By ([SGA7I], Expos´e I, Cor. 4.3) we deduce that for every λ∈P1(¯ k) the Iλ-invariant specialization map (Ri˜ f?¯ Q`)λ→(Ri˜ f?¯ Q`)¯η(where ¯ηis a geometric generic point of P1 ¯ kand Iλthe inertia group at λ) is an isomorphism for i > n and surjective for i=n. As a consequence, Ri˜ f?¯ Q`is lisse at λfor i > n. For i=nwe have an exact sequence (cf. [Kat99], Theorem 13) 0→(punctual sheaf) →Rn˜ f?¯ Q`→j?j?Rn˜ f?¯ Q`→0 where jis the inclusion of an open subset of P1 ¯ kon which Rn˜ f?¯ Q`is lisse. But since the specialization map (Rn˜ f?¯ Q`)λ→(Rn˜ f?¯ Q`)¯ηis surjective and Iλ-equivariant, the action of Iλon (Rn˜ f?¯ Q`)¯ηis trivial. As a consequence, the sheaf j?j?Rn˜ f?¯ Q`is lisse at λ. Proposition 10.The cohomology group Ha c(P1 ¯ k,Rb˜ f?¯ Q`⊗ Lψ)vanishes for: i) a > 2, all b ii) b > n, all a iii) b=n,a > 0. 5 Antonio Rojas-Le´ on Proof. Part (1) is clear for cohomological dimension reasons. For b>n, the sheaf Rb˜ f?¯ Q`is lisse on P1 ¯ kby Proposition 9. Since P1 ¯ kis simply connected, it must be constant. Then, if ¯ηis a geometric generic point of P1 ¯ k, we get RΓc(P1 ¯ k,Rb˜ f?¯ Q`⊗ Lψ) = (Rb˜ f?¯ Q`)¯η⊗RΓc(P1 ¯ k,Lψ) = 0 since RΓc(P1 ¯ k,Lψ) = 0. This proves (2). To prove (3), let j:V ,→P1 ¯ kbe as in Proposition 9, where Vis a dense open set on which Rn˜ f?¯ Q`is lisse, and let H=j?j?Rn˜ f?¯ Q`. Then His lisse on P1 ¯ kby Proposition 9, so exactly as above we get RΓc(P1 ¯ k,H ⊗ Lψ) = 0. From the exact sequence 0→ I (= punctual sheaf) →Rn˜ f?¯ Q`→ H → 0 we get, after tensoring with Lψ, 0→ I ⊗ Lψ→Rn˜ f?¯ Q`⊗ Lψ→ H ⊗ Lψ→0. Now I⊗Lψis punctual, so Hi c(P1 ¯ k,I⊗Lψ) = 0 for i > 0. From the long exact sequence of cohomology associated to the exact sequence above we get isomorphisms Ha c(P1 ¯ k,Rn˜ f?¯ Q`⊗ Lψ)∼ →Ha c(P1 ¯ k,H ⊗ Lψ) = 0 for a > 0. This proves (3). We can now complete the proof of Proposition 8. We have a spectral sequence Ha c(P1 ¯ k,Rb˜ f?¯ Q`⊗ Lψ)⇒Ha+b c(P1 ¯ k,R˜ f?¯ Q`⊗ Lψ). Suppose a+b > n + 1. Then either -a > 2, so Ha c(P1 ¯ k,Rb˜ f?¯ Q`⊗ Lψ) = 0 by part (1) of Proposition 10, -b > n, so Ha c(P1 ¯ k,Rb˜ f?¯ Q`⊗ Lψ) = 0 by part (2) of Proposition 10 or -a= 2 and b=n, so Ha c(P1 ¯ k,Rb˜ f?¯ Q`⊗ Lψ) = 0 by part (3) of Proposition 10. Therefore, the spectral sequence implies that Hi c(P1 ¯ k,R˜ f?¯ Q`⊗ Lψ) vanishes for i > n + 1. 3. A sum of Milnor numbers computation Consider the L-function associated to the sheaf Lψ(f)on An k: L(T, Lψ(f)) = exp ∞ X m=1 Sm mTm where Sm=X x∈kn m ψ(Tracekm/k(f(x))) and kmis the extension of degree mof kin ¯ k. By the Grothendieck trace formula, we have L(T, Lψ(f)) = 2n Y i=0 det(1 −T·F|Hi c(An ¯ k,Lψ(f)))(−1)i+1 where F∈Gal(¯ k/k) is the geometric Frobenius element. The following result of Adolphson and Sperber ([AS00b], Theorem 1.11 and Proposition 6.5) gives an important restriction on the shape of this L-function: 6 Purity of exponential sums on An Theorem 11.Write f=fd+fd0+f0, where fdis the degree dhomogeneous component of f, d0is the degree of f−fdand fd0is the degree d0homogeneous component of f. Suppose that d0/d > p/(p+ (p−1)2)and d0is prime to p. Suppose also that ∂fd ∂x1, . . . , ∂fd ∂xnhave a finite number of common zeroes in Pn−1 ¯ k(which is automatic if the hypersurface fd= 0 in Pn−1 ¯ kis non-singular) and the hypersurface defined in Pn−1 ¯ kby fd0= 0 does not contain any of them. Then L(T, Lψ(f))(−1)n+1 is a polynomial of degree (d−1)n−(d−d0)Ps i=1 µi, where the sum is taken over the set {P1, . . . , Ps} of common zeroes of ∂fd ∂x1,..., ∂fd ∂xnin Pn−1 ¯ kand µidenotes the corresponding Milnor number µi= dim¯ kOS,Pi. Here Sis the zero-dimensional subscheme of Pn−1 ¯ kdefined by the ideal (∂fd ∂x1, . . . , ∂fd ∂xn), and OS,Piits local ring at Pi, which is a finite ¯ k-algebra. We will now compute this sum of Milnor numbers explicitly in the following more general setting Lemma 12.Let F1, . . . , Fn∈¯ k[x1, . . . , xn]be (possibly zero) homogeneous polynomials of degree d−1. Suppose that i) F1, . . . , Fnhave a finite number of common zeroes in Pn−1 ¯ k. ii) We have the relation n X i=1 xi·Fi= 0. Let {P1, . . . , Ps}be the set of common zeroes of F1, . . . , Fnin Pn−1 ¯ k, and for every i= 1, . . . , s let µibe the corresponding Milnor number. Then we have s X i=1 µi=(d−1)n−(−1)n d. Proof. By induction on n, we first prove it for n= 2. In this case, both F1and F2must be non-zero (otherwise, by (2) they would both be zero, and (1) would not hold). The relation x1F1+x2F2= 0 implies that x1divides F2and x2divides F1. Let F1=x2G1and F2=x1G2. Then x1x2(G1+G2) = 0, so G2=−G1. Therefore the subscheme defined by F1and F2is the one defined by G1, which is a polynomial of degree d−2. The common zeroes of F1and F2are then in one-to-one correspondence with the distinct linear factors of G1, and the Milnor numbers are the corresponding multiplicities. Thus in this case we get Ps i=1 µi=d−2 = ((d−1)2−1)/d. We assume now that the lemma is true for n−1⩾2, and prove it for n. Choose (α1, . . . , αn−1)∈ ¯ kn−1such that none of the points P1, . . . , Psis contained in the hyperplane xn−Pn−1 j=1 αjxj= 0. We construct the polynomials F0 1, . . . , F0 ngiven by F0 i(x1, . . . , xn−1, xn) = Fi(x1, . . . , xn−1, xn+Pn−1 j=1 αjxj)+ +αiFn(x1, . . . , xn−1, xn+Pn−1 j=1 αjxj) for i= 1, . . . , n −1 F0 n(x1, . . . , xn−1, xn) = Fn(x1, . . . , xn−1, xn+Pn−1 j=1 αjxj) Then the schemes Sdefined by the ideal (F1, . . . , Fn) and S1defined by (F0 1, . . . , F0 n) correspond to each other via the automorphism ϕof Pn−1 ¯ kgiven by ϕ(x1, . . . , xn−1, xn)=(x1, . . . , xn−1, xn+ Pn−1 j=1 αjxj). In particular the sums of the Milnor numbers at the points of Sand S1are the same. 7 Antonio Rojas-Le´ on Moreover, we have Pn i=1 xi·F0 i(x1, . . . , xn−1, xn) = =Pn−1 i=1 xi·(Fi(x1, . . . , xn−1, xn+Pn−1 j=1 αjxj)+ +αiFn(x1, . . . , xn−1, xn+Pn−1 j=1 αjxj))+ +xn·Fn(x1, . . . , xn−1, xn+Pn−1 j=1 αjxj) = =Pn−1 i=1 xi·Fi(x1, . . . , xn−1, xn+Pn−1 j=1 αjxj)+ +(xn+Pn−1 i=1 αixi)·Fn(x1, . . . , xn−1, xn+Pn−1 j=1 αjxj) = 0. If P= (x1, . . . , xn) is a common zero of F0 1, . . . , F0 n, then ϕ(P) = (x1, . . . , xn−1, xn+Pn−1 j=1 αjxj) is a common zero of F1, . . . , Fnso, by the choice of the αi,ϕ(P) is not contained in the hyperplane xn−Pn−1 j=1 αjxj= 0. Hence Pis not contained in the hyperplane xn= 0. Therefore we can assume, and we will, that none of the common zeroes of F1, . . . , Fnis contained in the hyperplane xn= 0. Under this assumption, we claim that F1, . . . , Fn−1form a regular sequence in ¯ k[x1, . . . , xn] (compare [AS00b], Lemma 5.1). Otherwise, the subscheme defined by them in Pn−1 ¯ kwould have an irreducible component Yof dimension at least 1. From (2) we deduce that Yis contained in the hypersurface xnFn= 0. Being irreducible, it must be contained either in xn= 0 or in Fn= 0. Furthermore, since it has dimension ⩾1, its intersections with both xn= 0 and Fn= 0 are nonempty. So in either case, the intersection of F1, . . . , Fn−1, Fnand xn= 0 would be non-empty, in contradiction with the assumption made above. Denote by S1the subscheme of Pn−1 ¯ kdefined by (F1, . . . , Fn−1). The support of S1is the disjoint union of the points P1, . . . , Ps, which are contained in Fn= 0, and the points Ps+1, . . . , Ps+r, which are contained in xn= 0. Let ν1, . . . , νs+rbe the corresponding Milnor numbers (i.e. νi= dim¯ kOS1,Pi). Since F1, . . . , Fn−1form a regular sequence of polynomials of degree d−1, S1is a zero-dimensional complete intersection of degree (d−1)n−1, therefore s+r X i=1 νi= dim¯ kΓ(S1,OS1) = (d−1)n−1. For every i= 1, . . . , s,xnis invertible in the local ring OPn−1,Pi. So from (2) we deduce that Fn is contained in the ideal generated by F1, . . . , Fn−1in this local ring. Therefore OS,Pi=OPn−1,Pi/(F1, . . . , Fn−1, Fn) = =OPn−1,Pi/(F1, . . . , Fn−1) = OS1,Pi and in particular νi=µi. On the other hand, for i= 1, . . . , r,Fnis invertible in the local ring OPn−1,Ps+i, so xnis contained in the ideal generated by F1, . . . , Fn−1in this local ring. Let Gj=Fj(x1, . . . , xn−1,0), S2the subscheme of Pn−2 ¯ k(which we identify with the hyperplane xn= 0 in Pn−1 ¯ k) defined by (G1, . . . , Gn−1). The points Q1, . . . , Qrof S2are in one-to-one correspondence with Ps+1, . . . , Ps+r via the inclusion Pn−2(¯ k),→Pn−1(¯ k), and OS2,Qi=OPn−2,Qi/(G1, . . . , Gn−1) = =OPn−1,Ps+i/(F1, . . . , Fn−1, xn) = =OPn−1,Ps+i/(F1, . . . , Fn−1) = OS1,Ps+i, so the Milnor numbers are the same. 8 Purity of exponential sums on An Now G1, . . . , Gn−1fall under the hypotheses of the lemma, so we can apply the induction hypothesis and deduce that Ps+r i=s+1 νi= ((d−1)n−1−(−1)n−1)/d. Therefore s X i=1 µi= s X i=1 νi= s+r X i=1 νi− s+r X i=s+1 νi= = (d−1)n−1−((d−1)n−1−(−1)n−1)/d = ((d−1)n−(−1)n)/d. Thus, under the hypotheses of Theorem 11, L(T, Lψ(f))(−1)n+1 is a polynomial of degree (d− 1)n−(d−d0)((d−1)n−(−1)n)/d = (d0(d−1)n+ (−1)n(d−d0))/d. 4. End of the proof of Theorem 2 Part (3) of the theorem is a direct consequence of the previous two parts via the trace formula and Deligne’s theorem. So it suffices to prove (1) and (2). Fix a positive integer d0< d prime to psuch that d0/d > p/(p+ (p−1)2). Denote by Pd,d0the affine space of all polynomials in k[x1, . . . , xn] of degree ⩽dwhose homogeneous component of degree iis zero for all d0< i < d. Let π1:Pd,d0×An k→ Pd,d0be the projection and ev :Pd,d0×An k→A1 kthe evaluation map. Let K∈ Db c(Pd,d0,¯ Q`) be the object Rπ1!ev?Lψ[n+ dim Pd,d0]. Lemma 13.The object Kis perverse and pure of weight n+ dim Pd,d0. Proof. For d0=d−1 (i.e. when Pd,d0is the affine space of all polynomials of degree ⩽d) this is ([Kat04], Part (1) of Theorem 3.1.2). We will see that the same proof works in general. There is a natural finite map τ:An k→ˆ Pd,d0. Namely, for every t∈An(¯ k), τ(t)∈ˆ Pd,d0(¯ k) is the evaluation map at t,ev(−, t) : Pd,d0(¯ k)→¯ k. Since ¯ Q`[n] is perverse and pure of weight non An k, so is τ?¯ Q`[n] on ˆ Pd,d0. Its Fourier transform Tψ(τ?¯ Q`[n]) ∈ Db c(Pd,d0,¯ Q`) with respect to ψis K(by the very definition of K). Therefore Kis perverse and pure of weight n+ dim Pd,d0(cf. [KL85], Section 2 or [KW01], Section III.8 for the definition and main properties of the Fourier transform). Notice that for every finite extension k0/k and every f∈ Pd,d0(k0), the trace of the geometric Frobenius element in Gal(¯ k/k0) acting on the stalk of Kat a geometric point over fis the sum (−1)n+dim Pd,d0X x∈k0n ψ(Tracek0/k f(x)). Let U⊂ Pd,d0be the maximal dense open set on which Khas lisse cohomology sheaves. Then Hi(K)|U= 0 for i6=−dim Pd,d0and F:= H−dim Pd,d0(K) = Rnπ1!ev?Lψis lisse and pure of weight non U. Thus, for different finite extensions k0/k and polynomials f∈U(k0), the exponential sums Px∈k0nψ(Tracek0/k f(x)) are pure of weight nand the same rank as F. Let V⊂ Pd,d0(resp. W⊂ Pd,d0) be the dense open set of all polynomials fsuch that fddefines a non-singular hypersurface on Pn−1 ¯ k(resp. the dense open set of all fsuch that ∂fd ∂x1,..., ∂fd ∂xnhave a finite number of common zeroes in Pn−1 ¯ kand the hypersurface fd0= 0 does not contain any of them). We know that i) For every f∈V(k), we have Hi c(An ¯ k,Lψ(f)) = 0 for i6=n, n+1. For i > n+1, this is Proposition 5. For i < n it is just Poincar´e duality, since An ¯ kis smooth and Lψ(f)is lisse. ii) For every f∈W(k), the L-function L(T, Lψ(f))(−1)n+1 = 2n Y i=0 det(1 −T·F|Hi c(An ¯ k,Lψ(f)))(−1)n+i 9