Product topology of Affine Varieties I want to prove that $X \times Y \subseteq \Bbb A^{2n}$ is an affine variety, given that $X,Y \subseteq \Bbb A^n$ are affine varieties. 
Is this proof correct?
Since both $X$ and $Y$ are affine varieties, then they are closed subsets. This means that their projection maps are continuous and so $X\times Y$ is also closed. 
Edit: their projection maps would be say $\pi$: $X \times Y \to X$, and $\phi: X \times Y \to Y$.
 A: Let me rephrase your argument:  the product $X \times Y$ is the intersection
of $X\times \mathbb A^n$ and $\mathbb A^n \times Y$; thus to show that 
$X \times Y$ is closed in $\mathbb A^{2n}$, it suffices to show that
each of these latter sets is closed.  By symmetry, it obviously suffices to show
that $X \times \mathbb A^n$ is closed, and since this is the preimage of $X$ under
the projection to $\mathbb A^n$, it suffices to show that the projection map
$\mathbb A^{2n} \to \mathbb A^n$ is continuous.  
So, assuming you know that the projection is continuous (and its seems that
you do), this approach using projections is correct.  (But you have to phrase
it correctly.)
A: Note that the Zariski topology on $\mathbb A^{2n}$ is not the same as taking two copies of $\mathbb A^n$ with the Zariski topology and taking the product of those with the product topology. From point set topology we know that a topological space $X$ is Hausdorff if and only if the diagonal $\Delta=\{\,(x,x)\,\mid\, x\in X\,\}$ is closed in $X\times X$. Take $X=\mathbb A^1$ with the Zariski topology, we know that $X$ is not Hausdorff but since $x=y$ defines an affine variety in $\mathbb A^{2}$ the diagonal $\Delta\subseteq \mathbb A^{2}$ is closed in $Y=\mathbb A^{2}$ with the Zariski topology. Thus, this can't be the product topology.
To properly approach your question you need to actually use the definition of an affine variety.
