Here is the theorem I working on:

Let $f(X), g(X) \in k[X] = k[x_1,...,x_n]$, where $k$ is an infinite field.

(i) If $f(X)$ is nonzero, then there are $a_1,...,a_n \in k$ with $f(a_1,..,a_n) \neq 0$.

(ii) If $f(a_1,...,a_n) = g(a_1,...,a_n)$ for all $(a_1,...,a_n) \in k^n$, then $f(X)=g(X)$.

And here is the author's proof of (ii): enter image description here

I already proved part (i), and I thought that proving part (ii) would involve a simple application of (i) like so:

Let $h(X) := f(X) - g(X)$. Then clearly $h(a_1,...,a_n) = 0$ for every $(a_1,...a_n) \in k^n$, and therefore $h(X)= 0(X) = 0$ or $f(X) = g(X)$.

But, as the picture shows, the author does something a bit more complicated. Is there anything wrong with the proof I gave?

  • 1
    $\begingroup$ No, and I'd say it would be the standard argument to deduce (ii) directly from (i). Unless one proves (ii) first to then deduce (i) from it. $\endgroup$ – Daniel Fischer Sep 28 '17 at 15:00
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    $\begingroup$ It looks like $f$ and $g$ are polynomials in $X$, and you've proven that for all $n$-tuples, they are equal. However, we're looking at these as polynomials, rather than as functions, so proving them equal doesn't mean that they're equal for all $n$-tuples, but that their coefficients are all equal. $\endgroup$ – Kevin Long Sep 28 '17 at 15:00
  • $\begingroup$ @KevinLong I am not sure I follow you. We are assuming they are equal for all $n$-tuples; we aren't trying to show this. $\endgroup$ – user193319 Sep 28 '17 at 15:04
  • $\begingroup$ As an example of what Kevin Long is talking about (but not quite on the nose for the theorem, for the obvious reason), Fermat's little theorem gives pairs of distinct polynomials that agree as functions. $\endgroup$ – Eric Towers Sep 28 '17 at 15:05
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    $\begingroup$ @user193319 You should have mentioned (i) in the argument itself, not only in the preface to the argument. But since you use (i) to deduce all coefficients are zero, the argument is correct (provided (i) has been proved). $\endgroup$ – Daniel Fischer Sep 28 '17 at 15:12

Your argument is correct, and is indeed the most natural argument to use in this context. The author seems to have just overlooked it for some reason.


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