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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?
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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Feb 25 '13 at 12:39
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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.

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  • $\begingroup$ Thanks for answering both of my questions so quickly =) I thought there would be a simple argument! $\endgroup$ – user63912 Feb 25 '13 at 12:50
  • $\begingroup$ @user63912, you're welcome! $\endgroup$ – Andreas Caranti Feb 25 '13 at 12:54
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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.

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If our field $k$ is algebraically closed then this follows from the Nullstellensatz:

If $p$ is identically 0 then $V(p) = k^n$, so $\sqrt{\langle p \rangle} = I(V(p)) = 0$, so $p = 0$.

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  • $\begingroup$ How do you show I(k^n) = 0? Isn't that exactly the statement that "a polynomial which vanishes identically is zero"? $\endgroup$ – Brendan Murphy Nov 23 at 23:30
  • $\begingroup$ As $k$ is algebraically closed, the Nullstellensatz gives a one-to-one correspondence between radical ideals and algebraic varieties, where $k^n$ corresponds to the zero ideal of $k[x_1, ... ,x_n]$. $\endgroup$ – James Nov 24 at 14:40

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