Is the cartesian product of connected algebraic varieties a connected algebraic variety? I'm wondering about the following question:
Let $X \subset k^n$ and $Y \subset k^m$ be two algebraic varieties that are connected (in the Zariski topology). Is the cartesian product $X \times Y \subset k^{n+m}$ connected in the Zariski topology of $k^{n+m}$? This would obviously hold in the product topology of $k^n$ and $k^m$ but the product topology in $k^{n+m}$ is not the same as the Zariski topology. 

Suppose $X\times Y=S\cup T$, with $S,T$ clopen disjoint subsets. Then every fiber $X\times \{y\}$ and $\{x\}\times Y$ must lie entirely inside one of either $S$ or $T$: we can write each fiber as the disjoint union of the clopen sets given by intersecting with $S$ and $T$, but each fiber is connected. So, WLOG, there exists some $y\in Y$ so the fiber $X\times \{y\}\subset S$, and then for every $x\in X$, we get that $\{x\}\times Y\subset S$ as $(x,y)\in S$. So we have that $S=X\times Y$ and $T=\emptyset$, so $X\times Y$ is connected.
This is one of those times where working in the "naive" land of varieties inside $k^n$ actually makes this very easy: in general, this result is false ($\operatorname{Spec} \Bbb R[x]/(x^2+1) \times_\Bbb R \operatorname{Spec} \Bbb C$, for instance), and the circumstances for when it's true (depending on what space you take your product over) can get interesting!
