Product topology of Affine Varieties

Question

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 $\phi: X \times Y \to Y$.

Answer 1

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.)

Answer 2

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.