Separated scheme stable under base extension. 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?
 A: 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.
