# Zariski dense implies classically dense?

I was surprised that I wasn't able to find this question already posted; if it has been posted and I just didn't find the right search terms, let me know.

Let $X$ be any complex variety. A priori, any set which is dense in the classical topology on $X$ is automatically dense in the Zariski topology on $X$, just because the Zariski topology has fewer open/closed sets.

For the converse, this question, though distinct, does shed some light: in $\mathbb A_{\mathbb C}^1$, any infinite subset is Zariski dense, but certainly not necessarily classically dense.

But in my experience, it seems that any Zariski open subset of $X$ that is Zariski dense is also classically dense. Is this true? How do you prove it?

• There are a lot of incorrect proofs of this floating around. Here is one argument, but I have run into others—for example, one can reduce to the case of curves, then show that any two points on an irreducible variety can be connected by a curve. For non-archimedean fields, the only proof I've seen uses model completeness! – Slade Mar 13 '15 at 17:40
• @Slade thanks for the link. I'm not familiar enough with analytic Nullstellensatz; do you think you could provide a reference, or expand on it in an answer? – Dustan Levenstein Mar 13 '15 at 23:28
• I'll write an answer if I come across a better reference. I suspect that the easiest proof involves inducting on the dimension, or using various facts about curves. There may also be an approach using the Implicit Function Theorem. Note that the proof for $X=\mathbb{A}^n$ is very straightforward, as we can use power series. – Slade Mar 14 '15 at 17:51

Let $T$ be a locally constructible subset of a finite type $\mathbb C$-scheme $X$. (You can take $T$ to be a (Zariski) open of a complex algebraic variety, for instance.)

Then $T$ is dense in $X$ if and only if $T(\mathbb C)$ is dense in $X(\mathbb C)$. A reference for this is Expose XII, Cor. 2.3 p. 243 of SGA 1.

http://arxiv.org/pdf/math/0206203v2.pdf

• If you use your reference to transcribe a proof to here, I'll be more likely to upvote it, accept it, and/or assign the bounty to you. Otherwise if I need to take the time to go through and translate the reference myself, when I do have the extra time, I'll just transcribe the proof into my own answer, and let the bounty go unassigned. – Dustan Levenstein Mar 17 '15 at 13:04

Claim. If $A\subsetneq\mathbb A^n_{\mathbb C}$ is algebraic, then its complement is dense.

Proof. Let $A=\{\,x\in\mathbb A^n\mid \forall f\in S\colon f(x)=0\,\}$ where the $S\subseteq \mathbb C[X_1,\ldots, X_n]$. For $a\in A$ we have to exhibit points close to $a$ that are $\notin A$. Pick $b\in\mathbb A^n\setminus A$ and $f\in S$ with $f(b)\ne 0$. Then the polynomial $g(T)=f(a+(b-a)T)\in\mathbb C[T]$ is not the zero polynomial. Hence its root at $0$ is isolated and so $g(h)\ne 0$ for all sufficiently small nonzero $h$. Then $a+h(b-a)\notin A$ for such $h$, showing the claim. $_\square$

The extension to the projective case and then to the locally quasiprojective case should be clear

• How to extend from affine space to affine varieties is not at all clear to me. For example, if $V,W$ are distinct hypersurfaces, why should $V\setminus W$ be dense in $V$? – Slade Mar 24 '15 at 1:45
• It's sufficient to prove the result for affine varieties, but not clear to me at all how the proof for $\mathbb A^n$ can be extended to that case. – Dustan Levenstein Mar 24 '15 at 15:48