Given a separated scheme morphism $X\to Y$, and a morphism $Z \to Y$, Hartshorne proves that the extension $X\times_YZ \to Z$ is also separated, as long as the schemes involved are Noetherian. The valuative criterion he uses to prove it depends on this assumption.

Is it true for general $X, Y, Z$?

$X\to Y$ separated means that the diagonal map $X\to X\times_YX$ is a closed immersion. I have to use this somehow to prove that $$ \Delta:X\times_YZ \to (X\times_YZ)\times_Z(X\times_YZ) $$ is a closed immersion. Can that product be simplified in any way? Is there an obvious way to see this that I'm missing?

  • 2
    $\begingroup$ You'd like to leverage the fact that $\Delta_{X/Y}$ is a closed immersion and the fact that closed immersions are stable under base change. So the naive thing to do would be to put the two morphisms in question into a square and hope it's cartesian and in fact it is -- the proof has nothing to do with schemes, just diagram chasing. $\endgroup$ – Hoot May 11 '15 at 19:51
  • $\begingroup$ Can you help me to understand the answer given below.. The user who answered is Last seen Nov 27 '16 at 12:19. I understand that pull back pasting lemma says the outer triangle is a pullback diagram. He then says rectangle below is also a pull back diagram.. I do not understand why is that true $\endgroup$ – user312648 Apr 4 '17 at 10:27

As suggested in the comments, the proof is purely formal and only uses simple facts about closed immersions and pullbacks.

Assume $X \to Y$ is separated, i.e. $X \to X \times_Y X$ is a closed immersion. Consider a pullback square: $$\require{AMScd} \begin{CD} X' @>>> X \\ @VVV @VVV \\ Y' @>>> Y \end{CD}$$ The pullback pasting lemma says the outer rectangle of the following diagram is a pullback diagram, $$\begin{CD} X' \times_{Y'} X' @>>> X' @>>> X \\ @VVV @VVV @VVV \\ X' @>>> Y' @>>> Y \end{CD}$$ so the outer rectangle below is also a pullback diagram, $$\begin{CD} X' \times_{Y'} X' @>>> X \times_Y X @>>> X \\ @VVV @VVV @VVV \\ X' @>>> X @>>> Y \end{CD}$$ but the right half is a pullback square, so the left half is itself a pullback square. Thus, in the following diagram, $$\begin{CD} X' @>>> X \\ @VVV @VVV \\ X' \times_{Y'} X' @>>> X \times_Y X \\ @VVV @VVV \\ X' @>>> X \end{CD}$$ the outer rectangle and the bottom half are pullback diagrams, so the top half is a pullback square. Hence $X' \to X' \times_{Y'} X'$ is indeed a closed immersion.

A similar (but more tricky) proof shows that the class of separated morphisms is closed under composition.

  • $\begingroup$ You can also come up directly with the last diagram and proof that is cartesian passing to the category of set thanks to Yoneda lemma. $\endgroup$ – yamete kudasai Dec 16 '17 at 23:55

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