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Given two (algebraic) varieties $X,Y$ (not necessarily irreducible) and $D,E$ dense subsets of $X$ and $Y$ respectively, I've read something saying that the product $D \times E$ is dense in $X \times Y$ (given the Zariski topology of course).

Is it true ? And how do you prove it ?

I've tried to look for it in different books but I couldn't find anything.

Thank you !

1 Answers 1

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Suppose $X,Y$ are varieties over an algebraically closed field $k$.
Claim
If $D\subset X$ and $ E\subset Y$ are dense subsets, then $D\times E\subset X\times Y$ is dense too.
Proof
Let $Z$ be the closure of $D\times E$ in $X\times Y$.
Fix a poind $d_0\in D$ and consider the closed subvariety $\lbrace d_0\rbrace \times Y \subset X\times Y$.
The subset $\lbrace d_0\rbrace \times E \subset \lbrace d_0\rbrace \times Y$ has closure $\lbrace d_0\rbrace \times Y \;$ [ because $\lbrace d_0\rbrace \times Y$ is isomorphic to $Y$ and $E$ is dense in $Y$]
so that $\lbrace d_0\rbrace \times Y \subset Z $ . We have proved the

Partial Result:
For all $d_0\in D $ we have $\lbrace d_0\rbrace \times Y \subset Z$

End of proof:
Consider an arbitrary $y\in Y$ .
For any $d\in D$ we know, thanks to the Partial Result, that $(d,y)\in Z$. Hence $D\times \lbrace y \rbrace \subset Z$.
Since the closure of $D\times \lbrace y \rbrace $ is $X\times \lbrace y \rbrace $ [ because $X \times \lbrace y \rbrace $ is isomorphic to $X$ and $D$ is dense in $X$], we have proved that $X\times \lbrace y \rbrace \subset Z$.
Since $y\in Y$ was arbitrary, this implies that $Z=X\times Y $ i.e. that $D\times E$ is dense in $X\times Y $

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    Tha$n$k you a lot Georges (and also Qi). I almost had all what was needed to $p$ro$v$e the result, but I couldn't put it into good order. I'm glad I learnt something :)2011-12-13