0

Let $f:X\to Y$ be a morphism of algebraic varieties. Assume $Y$ is separated (meaning that the diagonal is closed in $Y\times Y$) and $X$ is complete (meaning that for all varieties $Z$ the projection $X\times Z\to Z$ is closed). Show that $f$ is a closed map.

I showed that the graph of $f$ is closed under the assumption that $Y$ is separated. I am thinking of writing $f$ as the composition of $g:X\to X\times Y,x\mapsto (x,f(x))$ and the projection onto $Y$, since the latter is closed by assumption. But then why should $g$ be closed?

We defined some weird topology on the product of two varieties, which I don't really understand, so to start and juggle with concrete closed sets doesn't seem like a good idea.

Any help is appreciated, either for this question or clearing up what exactly the topology is on $X\times Y$.

Dr. Heinz Doofenshmirtz
  • 2,360
  • The first half of your question about has been discussed here before, see here and here - do these resolve that portion of your question? The second bit should probably be a separate question, but the main idea is that you want $\Bbb A^n\times\Bbb A^m\cong \Bbb A^{n+m}$, and without the "weird topology" this isn't true. See for instance here for some discussion. – KReiser Nov 01 '20 at 19:33
  • @KReiser I managed to solve it, but I am still struggling with the following question. If $Z$ is a closed subvariety of a complete variety $X$, then $Z$ is complete. I have to show that $Z\times Y$ is closed in $X\times Y$ for every variety $Y$. – Dr. Heinz Doofenshmirtz Nov 02 '20 at 14:09
  • You may assume that $Y$ and $X$ are both affine (as checking that something is closed is checking that is is closed when restricted to every open set of an open cover), and thus $Z$ is also affine and it follows immediately. – Ahr Nov 02 '20 at 15:02
  • The key thing to prove your final claim is that closed immersions are stable under base change, so they are universally closed. Then apply the fact that the composition of two universally closed maps is universally closed. If this resolves your issues, please consider writing an answer to your own question to help others who have this question in the future; if not, please edit your question to make it clear what your current difficult is and I'd be happy to write up an answer. – KReiser Nov 02 '20 at 19:04
  • @Ahr I don't think this is what you want to do - an affine variety which is complete is finite, but certainly one could consider any closed subvariety of $\Bbb P^n$, for instance. – KReiser Nov 02 '20 at 19:05
  • I was only answering about the fact that $Z\times Y$ is closed in $X\times Y$ as soon as $Z$ is closed in $X$ which may be proven as I suggested, if I'm not mistaken. – Ahr Nov 02 '20 at 19:36

0 Answers0