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    • CommentRowNumber1.
    • CommentAuthorTodd_Trimble
    • CommentTimeSep 6th 2015
    • (edited Sep 6th 2015)

    I started an article bijective proof.

    In a section on polynomial identities, I give a proof that polynomials in several variables are uniquely determined by their values at natural number arguments, an intuitively obvious statement if there ever was one. Without bothering to look up whether there are standard nice proofs, I cooked up a proof myself. Please let me know if you know of nicer proofs.

    Edit: Having written this, it’s painfully obvious how to prove more general statements even more simply. Ah well. I still invite comments.

    • CommentRowNumber2.
    • CommentAuthorZhen Lin
    • CommentTimeSep 6th 2015

    It suffices to prove the following:

    Let AA be an infinite integral domain and let BB be an infinite subset of AA. Given p(x 1,,x n)A[x 1,,x n]p (x_1, \ldots, x_n) \in A [x_1, \ldots, x_n], if p(b 1,,b n)=0p (b_1, \ldots, b_n) = 0 for all (b 1,,b n)B n(b_1, \ldots, b_n) \in B^n, then p=0p = 0.

    We proceed by induction on nn. The case n=1n = 1 is well known (at least if AA is an algebraically closed field – but we can embed an integral domain in the algebraic closure of its fraction field). If the claim holds for n=kn = k, then given p(x 1,,x k+1)A[x 1,,x k+1]p (x_1, \ldots, x_{k+1}) \in A [x_1, \ldots, x_{k+1}] such that p(b 1,,b k+1)=0p (b_1, \ldots, b_{k+1}) = 0 for all (b 1,,b k+1)B k+1(b_1, \ldots, b_{k+1}) \in B^{k+1}, we know that p(x 1,,x k,b)p (x_1, \ldots, x_k, b) is the zero polynomial for all bBb \in B; but then p(x 1,,x k,x k+1)p (x_1, \ldots, x_k, x_{k+1}) is a one-variable polynomial over A[x 1,,x k]A [x_1, \ldots, x_k] that vanishes at infinitely many points, so p=0p = 0.

    • CommentRowNumber3.
    • CommentAuthorTodd_Trimble
    • CommentTimeSep 6th 2015

    Yep, thanks. I would have added the (obvious) remark that we first embed A[x 1,,x k]A[x_1, \ldots, x_k] into its field of fractions KK, and then use well-known facts about K[x]K[x].

    • CommentRowNumber4.
    • CommentAuthorTodd_Trimble
    • CommentTimeSep 6th 2015

    So I went ahead and supplanted the earlier clunky proof with a version of Zhen Lin’s (thanks again).

    • CommentRowNumber5.
    • CommentAuthorDavidRoberts
    • CommentTimeSep 6th 2015

    Doron Zeilberger is fond of proving identities by checking them at the smallest few nontrivial inputs, eg 1,2 and 3, where these are trivial to calculate.

    • CommentRowNumber6.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 20th 2016

    I added some details to some examples that had been there. Will probably write (and rewrite) more later.

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