Is $f_B:X_B\to \operatorname{Spec}B$ separated?

Question

Let $X$ be a Noetherian scheme. Given two morphisms of schemes $f: X\to \operatorname{Spec}A$ and $\pi: \operatorname{Spec}B\to\operatorname{Spec}A$, if $\pi$ is flat, is $f_B:X_B\to \operatorname{Spec}B$ separated?

Answer 1

Here is an explanation of my comment.

(1) Every morphism of affine schemes is separated.

First, recall that the diagonal $\Delta: Spec(B) \rightarrow Spec(B \otimes_{A} B)$ is given by the multiplication map $m:B \otimes_{A} B \rightarrow B$ which is surjective. By the surjectivity of $m$, we know that $V(ker(m)) \simeq Spec(B \otimes_{A} B / ker(m)) \simeq Spec(B)$ via $\Delta$. Furthermore, the surjectivity of $m$ tells us that the morphism of structure sheaves $\mathscr{O}_{Spec(B \otimes_{A} B)} \rightarrow \Delta_{\ast}\mathscr{O}_{Spec(B)}$ is surjective as well. Thus, $\Delta$ is a closed immersion.

(2) The base change of a separated morphism is separated.

Suppose we have morphisms $X \rightarrow S$ and $Z \rightarrow S$, and that $X \rightarrow S$ is separated. We would like to show that $X' = X \times_{S} Z \rightarrow Z$ is separated as well. To do so, simply observe that

$$ \Delta : X \times_{S} Z \rightarrow (X \times_{S} Z) \times_{Z} (X \times_{S} Z) \simeq (X \times_{S} X) \times_{S} Z $$

is just the base change of the closed immersion $X \rightarrow X \times_{S} X$. Since closed immersions are stable under base change, we are done. A proof that closed immersions are stable under base change involves making categorical arguments using pullback squares; a nice proof by Zhen Lin is given here Separated scheme stable under base extension.