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I know that a nonzero single-variable polynomial over a finite field can vanish identically e.g. take the product $\prod_a(x-a)$ for every $a$ in the field. But I know that for an infinite field this cannot happen since a degree $d$ polynomial has at most $d$ roots. My questions are:

  1. Why does a nonzero two-variable or higher polynomial over $\mathbb{R}$ not vanish identically? (In this case I know they can't but I don't know why)
  2. What about nonzero multivariate polynomials over other infinite fields?
user63912
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    Have you tried induction on the number of variables? If your polynomial has $n$ variables, view it as polynomial with coefficients from the polynomial ring in $n-1$ variables. By induction hypothesis at least one of the coefficients does not vanish for some choice of values. Then use the base case, $n=1$, that you already know to justify the induction step. – Jyrki Lahtonen Feb 25 '13 at 12:39

2 Answers2

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Let $F$ be an infinite field, and $f(x, y) \in F[x, y]$ a nonzero polynomial.

Regard $f$ as a polynomial $g(y) = f(x, y) \in (F(x))[y]$. This is a polynomial in $y$, with coefficients in the infinite field $F(x)$. Since it has a finite number of distinct roots, there is a $b \in F$ such that $0 \ne g(b) = f(x, b) \in F[x]$. Now apply the result for the univariate case.

Andreas Caranti
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    Thanks for answering both of my questions so quickly =) I thought there would be a simple argument! – user63912 Feb 25 '13 at 12:50
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    @user63912, you're welcome! – Andreas Caranti Feb 25 '13 at 12:54
  • I feel like this answer is brushing over a couple of subtleties; please set me straight if I am misunderstanding. First, I think we should consider the possibility that $g$ is the zero polynomial in $(F[x])[y]$. Only upon showing that this would make $f$ the zero polynomial of $F[x,y]$ can we discount it. – ummg Jan 03 '25 at 01:07
  • More importantly, if $g$ is not the zero polynomial of $(F[x])[y]$ then the usual theorem says that $g$ has finitely many roots in $F[x]$, not in $F$. Of course, identifying $F$ as a subfield of $F[x]$ we find that $g$ also only has finitely many roots in $F$. – ummg Jan 03 '25 at 01:07
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For the first part, view a two-variable polynomial in x and y as a single variable polynomial in x, with y as a parameter. So if we treat y as fixed, and let x vary, then it is identically zero if and only if all the coefficients (which are polynomials in y) are zero. But seeing as the coefficients are themselves single variable polynomials in y, they are identically zero if and only if all their coefficients are zero, i.e. we have the zero polynomial.

user61496
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