From: achava@hotmail.com Subject: Re: chevalley's theorem Date: Tue, 26 Oct 1999 18:00:20 GMT Newsgroups: sci.math Keywords: Artin, Chevalley: homogeneous forms in enough variables vanish In article <7v499e$v9f$1@nnrp1.deja.com>, holmes4@my-deja.com wrote: > I'm looking for a proof of the following theorem, attributed to Artin & > Chevalley: > > Given a finite field and a polynomial f(x1,x2,...,xr) of degree n in r > variables, with n < r. Then if f(0,0,...,0) = 0 then there is some > (a1,a2,...,ar) != (0,0,...,0) such that f(a1,a2,...,ar) = 0. > > This appears to be well-known, but I'm having trouble finding a proof, > and can't think of one myself. > > Thanks, > James The Chevalley theorem is that a form (= homogeneous polynomial) over a finite field has at least one non-trivial zero is the number of variables is greater than the degree. Warning's theorem states that the number of solutions to F(x1, x2, ..., xr) = 0 over a finite field is divisible by the characteristic of the field if the number of variables is greater than the degree. Chevalley's theorem is an immediate corollary of this more general result, but I am pretty sure it came earlier. I believe that Artin's role here was that he conjectured Chevalley's result and then Chevalley proved it. If Artin didn't see a proof immediately, I would not feel too bad about not seeing one immediately myself. This leaves open the possibility that some polynomial over a finite field whose degree is smaller than the number of variables still has no solution. I don't have an example to give off the top of my head, but I suspect that there is one. Proofs of these statements can be found in Number Theory by Borevich and Shafarevich right near the beginning. The exercises to this section have more interesting facts connected to Warning's theorem. Hope this helps, Achava Sent via Deja.com http://www.deja.com/ Before you buy. ============================================================================== From: jeanfi61 Subject: Re: chevalley's theorem Date: Thu, 28 Oct 1999 04:15:39 +0200 Newsgroups: sci.math holmes4@my-deja.com a écrit : > I'm looking for a proof of the following theorem, attributed to Artin & > Chevalley: > > Given a finite field and a polynomial f(x1,x2,...,xr) of degree n in r > variables, with n < r. Then if f(0,0,...,0) = 0 then there is some > (a1,a2,...,ar) != (0,0,...,0) such that f(a1,a2,...,ar) = 0. You could find a proof of the Chevalley-Warning theorem in the book "Cours d'arithmétique" by Jean-Pierre Serre, possibly translated in english under the title "A course in arithmetic" (not quite sure, though). jeanfi61 ============================================================================== From: wcw@math.psu.edu (William C Waterhouse) Subject: Re: chevalley's theorem Date: 1 Nov 1999 22:10:53 GMT Newsgroups: sci.math In article <7v499e$v9f$1@nnrp1.deja.com>, holmes4@my-deja.com writes: > I'm looking for a proof of the following theorem, attributed to Artin & > Chevalley: > > Given a finite field and a polynomial f(x1,x2,...,xr) of degree n in r > variables, with n < r. Then if f(0,0,...,0) = 0 then there is some > (a1,a2,...,ar) != (0,0,...,0) such that f(a1,a2,...,ar) = 0. > > This appears to be well-known, but I'm having trouble finding a proof, > and can't think of one myself. There have been a couple of references, but it really isn't too hard just to write out a proof. Let q = p^b be the size of the field, F. Lemma 1.The sum of x^c over all x in F is 0 for c = 0 (where we interpret x^0 as 1 always), -1 for c > 0 divisible by q-1, 0 for other positive c. Proof. The first is obvious. The second is true because all the nonzero x give value 1, so we get sum q-1 = -1. In the third case, we must have some r nonzero in F with r^c different from 1. (Specifically, we can take an r where r^{q-1} is the first power equal to one; we know there are such numbers.) Now the sum of all x^c is the same as the sum of all (rx)^c; but this is r^c times the original sum. Hence the sum is 0. Lemma 2. Let g(x_1,...,x_r) be a polynomial over F of degree < r(q-1). Then the sum of values of g over all r-tuples from F is 0. Proof. It is enough to prove this for each monomial x_1^{e_1}...x_r^{e_r} where the sum of the e_i is < r(q-1). The sum of values can be rewritten as (product over i) (sum over x_i in F of x_i^{e_i}). It is thus 0 unless each e_i is positive and divisible by q-1. But that would make the degree at least r(q-1). Lemma 3. Let f be a polynomial over F in r variables. Suppose there are exactly M r-tuples for which it is zero. Then the sum of the values of f^{q-1} over all r-tuples is -M in F. Proof. We get M terms equal to 0 and q^r - M others all equal to 1. Theorem. Let f be a polynomial over F of degree < r in r variables. Then the number M of r-tuples from F making it zero is divisible by p. Proof. Lemmas 2 and 3 show that the sum of values of f^{q-1} is both 0 and -M. William C. Waterhouse Penn State