TITLE: Zariski Open Sets are Dense?
QUESTION [21 upvotes]: Is it true that any nonempty open set is dense in the Zariski topology on $\mathbb{A}^n$? I'm pretty sure it is, but I can't think of a proof! Could someone possibly point me in the right direction? Many thanks!
Note: I am not asking about the Euclidean topology at all!

REPLY [23 votes]: As an exercise, let's reduce everything to statements about polynomials. Every open set contains a basic open set $U$, which is the complement of the zero set of some nonzero polynomial $f$, so it suffices to show that these are Zariski dense. The Zariski closure of a set is the intersection of the zero sets of all polynomials vanishing on it. This is equal to $\mathbb{A}^n$ if and only if any polynomial vanishing on $U$ vanishes on $\mathbb{A}^n$. Thus the claim is equivalent to the following statement about polynomials:

Suppose a polynomial $g$ has the property that if $f(x) \neq 0$, then $g(x) = 0$. Then $g(x) = 0$ for all $x$. 

But the condition is equivalent to the claim that $f(x) g(x) = 0$ for all $x$. Can you finish the problem from here? (Note that you need to assume $k$ infinite.)