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I am reading through a proof of the following by induction but stuck at a very early step.

$k$ is an infinite field and let $f \in k[x_1,...,x_n]$. Then $f=0$ in $k[x_1,...,x_n]$ if and only if $f:k^n \rightarrow k$ is the zero function.

The proof starts by considering $k[x]$ and assumes $k$ is an infinite field, and that $f \in k[x]$ is a zero function. So I understand that, since $f(a)$ is zero for any $a \in k$, and because $k$ is infinite, it follows that $f$ has infinitely many roots.

But here's the statement I don't get, which is apparently the result of what I've just written

$f$ has infinitely many roots hence $f$ must be the zero polynomial.

But, why? I mean the easiest way is sure, if all the coefficients of the polynomial are $0$ then no matter what we have, we get zero. But why can we say that this is absolutely always the case?

So my question is, why does $f$ having infinitely many roots(as a function) imply that $f$ is the zero polynomial?

It would be great if someone could explain...thank you [**I was looking for "why" if a polynomial has more than $n$ roots it must be the zero polynomial]

Kydo
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    It is a standard fact from field theory that a polynomial of degree $n$ with coefficients in $k$ has at most $n$ roots in $k$, so any polynomial with an infinite number of roots must be the zero polynomial.

    http://math.stackexchange.com/questions/25822/how-to-prove-that-a-polynomial-of-degree-n-has-at-most-n-roots

     – Charlie Nov 19 '15 at 21:53
  • Voting to close as a duplicate of the question linked above--while the statement of that question talks only about the case $k=\mathbb{C}$, some of its answers work for arbitrary fields. – Eric Wofsey Nov 19 '15 at 22:05
  • Hi, thanks for answering, but I think it's slightly different though strongly related; I am working through the answer given below that "a polynomial with more roots than its degree is the zero polynomial" I think this is exactly the statement I needed. I think the duplicate proposed is something that explains the fact that a polynomial of degree $n$ has at most $n$ roots, but does not really explain if more than $n$ roots can happen and what the consequences might be(which is my case). Thank you for letting me know though – Kydo Nov 20 '15 at 00:17

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The general fact is that, over any infinite integral domain $A$, a polynomial $f(X)$ with more roots than its degree is the zero polynomial.

This is because we can consider it as a polynomial over its field of fractions $K$, and as the polynomial ring $K[X]$ is a Euclidean domain, we can make polynomial divisions. In particular for any $\alpha\in A$ (or $K$), we have $$f(X)=(X-\alpha)q(X)+r,\enspace r\in K. $$ As the remainder $r=f(\alpha)$, there results that $\alpha$ is a root of $f(X)$ if and only if $f(X)$ is divisible by $X-\alpha$.

Furthermore, over an integral domain, $\;\deg fg=\deg f+\deg g$. So, if $f$ has $r$ roots, $\alpha_1,\dots,\alpha_r$, $f(X)$ is divisible by $(X-\alpha_1)\dotsm(X-\alpha_r)$, so that, if $f\neq 0$, $\deg f\ge r$.

Bernard
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