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 $$\varphi_\psi$$?

• I think you mean $X$ not $S$ in your question. Oct 20, 2023 at 12:06
• I don't understand your remark Oct 20, 2023 at 12:08
• You have not defined $S$. Oct 20, 2023 at 12:09
• Yes, of course. I've changed it. Oct 20, 2023 at 12:29
• 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. Oct 20, 2023 at 12:52

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.
• I think so. If $S_1$ is a variety, I don't see an issue. Oct 22, 2023 at 10:32