8
$\begingroup$

I was presented with a proof of the following fact and am hung up on a detail which I think is in fact wrong. Also I think I can fill the detail myself but need a bit of help.

Let $X\subset k^{m}$ and $Y\subset k^{n}$ be affine varieties(Assumed irreducible). Show that the product $X\times Y\subset k^{m+n}$ (equipped with the Zariski topology on $k^{m+n}$) is irreducible.

This is the "proof". Assume $X\times Y$ is reducible. Then there exists $Z_{1}$ and $Z_{2}$ closed such that $X\times Y=Z_{1}\cup Z_{2}$. Define $X_{i}=\{x|x\times Y\subset Z_{i}\}$. The proof shows $X=X_{1}\cup X_{2}$ then says $X_{i}$ is closed because projection is a closed map.

I don't think projection of the product under the zariski topology is a closed map. So can someone possibly explain that to me?

My idea on showing $X_{i}$ was closed was to show that $X_{i}=\cap_{y\in Y}\{x|(x,y)\in Z_{i}\}$ is a closed set. Which amounts to showing $\{x|(x,y)\in Z_{i}\}$ is closed in $X$.

$\endgroup$
  • 1
    $\begingroup$ Is the Zariski topology on $k^{m+n}$ a product topology of the corresponding Zariski's topologies on $k^m$ and $k^n$? I just started learning computational algebraic geometry so I'm curious. $\endgroup$ – Sergio Parreiras Dec 11 '13 at 5:28
  • $\begingroup$ Why do you think that the projection morphism is not closed? $\endgroup$ – user314 Dec 11 '13 at 5:40
  • 1
    $\begingroup$ Consider the circle $x^{2}+y^{2}-1$ in $\mathbb{R}^{2}$ projection onto $\mathbb{R}$ produces the interval $[0,1]$ which is not closed in the zariski topology. Since the only closed sets would be finite points or the whole set. $\endgroup$ – TheNumber23 Dec 11 '13 at 6:32
  • $\begingroup$ @SergioParreiras: no. Think about closed sets in $\mathbf{A}^1$, and in $\mathbf{A}^2$. $\endgroup$ – user64687 Dec 11 '13 at 9:25
  • 2
    $\begingroup$ @Adeel: for an example over any field, the hyperbola $xy=1$ in $\mathbf{A}^2$ projects to $\mathbf{A}^1 \setminus \{0\}$. $\endgroup$ – user64687 Dec 11 '13 at 9:26
7
$\begingroup$

Fix $y \in Y$. The map $X \to X \times Y, x \mapsto (x,y)$ is continuous (in fact it is a morphism). Hence, the preimage $\{x : (x,y) \in Z_i\}$ is closed. Since closed sets are stable under arbitrary intersections, it follows that $\{x \in X : x \times Y \subseteq Z_i\}$ is closed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.