I would like to find the best proof of the following fact: If a polynomial $p\in k[x,y]$ (where $k$ is an algebraically closed field) is such that $p(a,b)\neq 0$ on the set $\{(a,b):b\neq 0\}$, then in fact $p\in k[y]$ (that is, $p$ does not depend on $x$). I do know how to prove this using an argument with a Vandermonde matrix, but I feel that there should be a more direct way to do it.
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Write \begin{align*} p(x, y) = a_n(y) x^n + \cdots + a_1(y) x + a_0(y) \end{align*} for some $a_i\in k[y]$. For any fixed nonzero $b\in k$, the polynomial $p(x, b)\in \overline{k}[x]$ has no zeros and must be constant. Thus the $a_i(b)$ for $i\not =0$ all vanish for $b\not = 0$. But $\overline{k}$ is algebraically closed and thus infinite, forcing all $a_1, \dots, a_n$ to vanish identically.
anomaly
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1Yes, that's it, thank you. I was working with $p(x,y)=p_0(x)+p_1(x)y+\cdots + p_m(x)y^m$ instead, that makes a proof possible but more complicated. – Math101 Dec 17 '19 at 00:49
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@Math101: No thanks are necessary, but I'm glad it helped. – anomaly Dec 17 '19 at 00:49
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1Excellent answer: +1. – Georges Elencwajg Jan 27 '22 at 16:44
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@GeorgesElencwajg: Thanks! – anomaly Jan 27 '22 at 17:12
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As an amusing complement to anomaly's excellent answer, notice that the polynomial $p(x,y)=a_0(y)$ being nonzero for $y\neq0$ is necessarily of the form $f(x,y)=cy^n$ where $c\in k^*$ and $n\in \mathbb N$.
Notice also that the theorem is false for non algebraically closed fields, as witnessed by the polynomial $x^2+1\in \mathbb R[x,y]$ .
Georges Elencwajg
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