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For $k$ an infinite field, for integer $n\geq2$, for $p \in k[x_1,\ldots,x_n]$, if $p(\vec{a}) = 0$ for all $\vec{a} \in k^n$, then is $p$ the zero polynomial? (Here by "infinite" field we mean a field having infinitely many elements.)

My understanding so far is as follows:

  • If $k$ is algebraically closed, then $p$ is contained in all max ideals of $k[x_1,\ldots,x_n]$, but the intersection of all max ideals in this situation is 0.

  • For $k = \mathbb{R}$, from analysis we know a polynomial vanishing everywhere is the zero polynomial. Similarly we can use analysis for the $k=\mathbb{Q}$ case.

(If I'm not mistaken, an example of a nonzero polynomial vanishing everywhere with $n\geq2$ but $k$ finite can be achieved by taking $\prod_{a \in k}(x_1-a)$.)

I.A.S. Tambe
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    See my answer to https://math.stackexchange.com/questions/2789699/zeros-of-polynomial-over-an-infinite-field/2789760#2789760 – Jens Schwaiger Mar 15 '23 at 05:22
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    It's confusing that the question in the title (to which the answer is "no") is the opposite of the question inside (to which the answer is "yes"). – 2'5 9'2 Mar 15 '23 at 05:35
  • This question recurs often enough. The oldest applicable thread may be this. Unfortunately the answer there is in the comments. At the time users were reluctant to reproduce passages found in several textbooks as an answer. Standards have slipped since. I don't mind relatively new users to show their understanding by posting. Not at all. But the veterans should know :-) – Jyrki Lahtonen Mar 15 '23 at 06:30
  • @2'59'2 Thanks for noticing that, I'll edit the title to match the question inside – I.A.S. Tambe Mar 15 '23 at 14:47

1 Answers1

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The answer is yes. We proceed by induction on $n$. For $n=1$, the number of roots of a non-zero polynomial is at most its degree and in particular finite. For general $p\in k[x_1,\dots,x_n],$ we write $$p=\sum_{k=1}^m q_k(x_1,\dots,x_{n-1})x_n^k.$$ If we fix $x_1,\dots,x_{n-1}$, $p$ reduces to a polynomial in one variable and the coefficients must vanish. Since this holds for any $x_1,\dots,x_{n-1}$, the $q_k$ are zero polynomials by induction hypothesis, i.e., $p=0$.

Acrobatic
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