Let $\varphi \colon X\to S_2$ be a proper dominant morphism between varieties. (This varieties has the same dimension in my example if it helps) I have another morphism $\psi\colon S_1\to S_2$ which is flat. $S_1$ is not necessary irreducible, it may has several components, but all of these component have the same dimension. Let $\varphi_\psi \colon X\times_{S_2}S_1\to S_1$ be the base change of $\varphi$.

Is it true that for any irreducible component of $X\times_{S_2}S_1$, its image under the map $\varphi_\psi$ is dense in some irreducible component of $S_1$?

It is not difficult to show that $\varphi_\psi$ is dominant. But why $S\times_{S_2}S_1\to S_1$ cannot have some components which are "contracted" under the map $\varphi_\psi$?

  • $\begingroup$ I think you mean $X$ not $S$ in your question. $\endgroup$ Oct 20, 2023 at 12:06
  • $\begingroup$ I don't understand your remark $\endgroup$ Oct 20, 2023 at 12:08
  • $\begingroup$ You have not defined $S$. $\endgroup$ Oct 20, 2023 at 12:09
  • $\begingroup$ Yes, of course. I've changed it. $\endgroup$ Oct 20, 2023 at 12:29
  • $\begingroup$ The image of an irreducible component of $X\times_{S_2} S_1$ is irreducible in $S_1$ and so cannot be dense in $S_1$ if $S_1$ is not irreducible. $\endgroup$ Oct 20, 2023 at 12:52

1 Answer 1


Here is my own answer.

Let $Y=X\times_{S_2}S_1$ and $D$ be some irreducible component of $Y$. The image $\psi_\varphi(D)$ is dense in $X$ as $\psi_\varphi$ is flat and flat morphism is open. As $\varphi$ is dominant $\varphi(\psi_\varphi(D))$ is dense in $S_2$. This means that if we restrict everything to some open subset of $U\subset S_2$ then $D$ is restricted to non-empty open subset $Z_U$ of $Z$.

Now general flatness says that over some open $U\subset S_2$ the morphism $\varphi$ is flat, so we can assume $\varphi$ to be flat. But in this case $\varphi_\psi$ is flat and hence open. The statement follows.

  • $\begingroup$ Be careful: a flat morphism is open under some finiteness conditions. Everything is fine for morphisms of varieties, but, for example, the inclusion of a generic point is a flat and non-open morphism. $\endgroup$ Oct 22, 2023 at 10:05
  • $\begingroup$ In my case everything is of finite type over a field. Does it work in this case? $\endgroup$ Oct 22, 2023 at 10:20
  • $\begingroup$ I think so. If $S_1$ is a variety, I don't see an issue. $\endgroup$ Oct 22, 2023 at 10:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .