Lemma 29.3.1 (07RK)—The Stacks project

Lemma 29.3.1. Let $Z \to Y \to X$ be morphisms of schemes.

If $Z \to X$ is an immersion, then $Z \to Y$ is an immersion.
If $Z \to X$ is a quasi-compact immersion and $Y \to X$ is quasi-separated, then $Z \to Y$ is a quasi-compact immersion.
If $Z \to X$ is a closed immersion and $Y \to X$ is separated, then $Z \to Y$ is a closed immersion.

Proof. In each case the proof is to contemplate the commutative diagram

\[ \xymatrix{ Z \ar[r] \ar[rd] & Y \times _ X Z \ar[r] \ar[d] & Z \ar[d] \\ & Y \ar[r] & X } \]

where the composition of the top horizontal arrows is the identity. Let us prove (1). The first horizontal arrow is a section of $Y \times _ X Z \to Z$, whence an immersion by Schemes, Lemma 26.21.11. The arrow $Y \times _ X Z \to Y$ is a base change of $Z \to X$ hence an immersion (Schemes, Lemma 26.18.2). Finally, a composition of immersions is an immersion (Schemes, Lemma 26.24.3). This proves (1). The other two results are proved in exactly the same manner. $\square$

Comments (5)

Comment #733 by Kestutis Cesnavicius on June 24, 2014 at 02:15

Is the analogue of 3) for algebraic spaces stated anywhere in the SP? (The proof is the same and based on http://stacks.math.columbia.edu/tag/03KP ). Also, it seems to me that section 28.2. would be a more natural place for part 3).

Comment #739 by Johan on June 26, 2014 at 01:02

OK, I added a similar lemma to stacks-morphisms.tex You can find it in this commit. It will appear on the web soonish. Thanks!

Comment #4342 by Kazuki Masugi(馬杉和貴) on August 11, 2019 at 12:54

in (2), " $Z\to Y$ is quasi-separated" should be " $Y\to X$ is quasi-separated".

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