On different proofs of the Chevalley-Warning theorem

Discussion in 'Math Research' started by PLClark, Jul 17, 2008.

  1. PLClark

    PLClark Guest

    My question is mostly a "research" question in the librarians' sense
    of the term. It is about the origins of various proofs of the
    Chevalley-Warning theorem. Just about the only thing I know is that
    Chevalley showed that if
    P_1(t),...,P_r(t) are polynomials in F_q[t_1,...,t_n] with zero
    constant term and such that sum_i deg(P_i) < n,
    then there exists 0 =/= x in (F_q)^n such that P_1(x) = ... = P_r(x) =
    0; whereas Warning improved this by removing the assumption that they
    have zero constant term and concluding that the number of simultaneous
    solutions is divisible by p (the characteristic of F_q).

    Does anyone know what the original proofs of Chevalley and Warning
    were? In Ireland and Rosen it says that Warning's proof actually
    showed that the number of simultaneous solutions is at least q^{n-d}
    (assuming, I suppose, that there is at least one solution). Nowadays
    the proof that you almost invariably see is Ax's incredibly short
    proof, which begins with the easy observation that the number of
    simultaneous solutions is equal, as an element of F_q to
    (equivalently, is congruent mod p to)

    sum_{x in F_q^d} prod_{i=1}^n (1-P_i(x)^{q-1})

    and then just noticing that all monomial terms appearing in this sum
    have the property that the sum over all x is zero.

    But Ireland and Rosen also give a different proof of Chevalley's part
    of the theorem, which I learned in my undergraduate days and has
    always seemed more interesting. Just now I realized that this proof
    can be extended to give a proof of the Warning theorem which I
    likewise think is "more interesting" than Ax's proof. I am inclined
    to think that "my new proof" is in fact not new and am asking for
    references.

    Here it goes: (I will write F for F_q to simplify notation)

    First a bit on reduced polynomials over F: Define a monomial c*
    t_1^{a_1}*...* t_n^{a_n} to be _reduced_ if
    a_i < q for all i. Define a polynomial to be reduced if all of its
    monomial terms are reduced; equivalently, it has
    degree strictly less than q in each of its variables.

    Lemma: The map which takes a (formal!) polynomial P in F[t_1,...,t_n]
    and returns a function
    Phi(P): F^n -> F_q by Phi(P)(x) = P(x) is a homomorphism of rings.
    Moreover:
    (i) The homomorphism is surjective: every function f: F^n -> F is
    given by some polynomial.
    (ii) Every polynomial is congruent modulo Ker(Phi) to a unique reduced
    polynomial.
    (iii) The kernel of Phi is the ideal <t_1^q-t_1,...,t_n^q-t_n>.

    Proof: For any function f, define

    P_f(t) := sum_{y in F^n} f(y) prod_{i=1}^n (1-(t_i-y_i)^{q-1}).

    Then in the sum defining P_f(x), it is immediate to see that we get
    f(y)*1 when y = x and f(y)*0 otherwise. (Note
    that P_f is a reduced polynomial.) For (ii), evidently t_i^q - t_i is
    in the kernel for all i, and a finite number of
    applications of replacement of t_i^q in a polynomial by t_i will
    convert any polynomial to a reduced polynomial
    which induces the same function. There are q^n reduced monomials
    overall and these give an F-basis for the
    set of all reduced polynomials, so there are q^(q^n) reduced
    polynomials. Since this is the same as the total
    number of functions from F^n -> F and we know that every function is
    represented by a reduced monomial, the
    representation must therefore be unique. The deduction of (iii) from
    this is straightforward (and actually not used
    in the rest of the proof).

    [Remark: This lemma is fairly standard, but it seems amusing to deduce
    the uniqueness of the representation by
    reduced polynomials from the surjectivity of the evaluation map rather
    than the other way around.]

    Therefore for any function f: F^n -> F, one can speak of the reduced
    degree in each variable t_i and also the reduced total degree.
    Evidently the reduced total degree of a polynomial function P(t) is
    less than or equal to the
    degree of the polynomial (because reduction is effected by a series of
    operations each of which does not increase any monomial degree).

    Now let Z be the set of common zeros of P_1,...,P_r in F^n. Consider
    the "characteristic function"
    1_Z of Z, i.e., the (F_q-valued!) function which is 1 at x if x is in
    Z and 0 otherwise. We have two different polynomial representations
    for this function. One is:

    P(t) = prod_{i=1}^r (1-P_i(t)^{q-1)).

    The other is the representation we derived above for _any_ function:
    in this case it is

    1_Z(t) = sum_{x in F^n} prod_{i=1}^n (1-(t_i-x_i)^{q-1}).

    Now the total degree of P(t) is (q-1) sum_i deg(P_i) < (q-1)n, by
    assumption. Hence the degree of the reduced
    polynomial representing P is also strictly less than (q-1)n. But 1_Z
    is the reduced polynomial representing P,
    and it is a sum of polynomials each with a monomial term
    t_1^{q-1}*...*t_n^{q-1}. Therefore the coefficient of
    this monomial term in 1_Z is #Z, so unless this is zero in F_q the
    total degree of 1_Z is at least (q-1)*n. Done.

    Please let me know if you've seen such an argument before, and if so,
    where.
     
    PLClark, Jul 17, 2008
    #1
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  2. PLClark

    PLClark Guest

    There were a couple of minor typos in the preceding post: the
    coefficient of the final monomial is not #Z but (-1)^n #Z, and one of
    the last sums should be only over x in Z.

    I also did a bit more digging around and found some notes of Jarden on
    finite fields which seem to suggest that this argument might be very
    close to Warning's original proof! (If so, the presentation in
    Ireland and Rosen is then rather puzzling...)

    I will try to investigate this. In case anyone else has the original
    paper closer at hand and/or a better command of the German language,
    please do let me know.

    Thanks,

    Pete L. Clark
     
    PLClark, Jul 17, 2008
    #2
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