The Stacks project

Lemma 41.7.2. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

  1. $f$ is a closed immersion,

  2. $f$ is a proper monomorphism,

  3. $f$ is proper, unramified, and universally injective,

  4. $f$ is universally closed, unramified, and a monomorphism,

  5. $f$ is universally closed, unramified, and universally injective,

  6. $f$ is universally closed, locally of finite type, and a monomorphism,

  7. $f$ is universally closed, universally injective, locally of finite type, and formally unramified.

Proof. The equivalence of (4) – (7) follows immediately from Lemma 41.7.1.

Let $f : X \to S$ satisfy (6). Then $f$ is separated, see Schemes, Lemma 26.23.3 and has finite fibres. Hence More on Morphisms, Lemma 37.39.1 shows $f$ is finite. Then Morphisms, Lemma 29.42.15 implies $f$ is a closed immersion, i.e., (1) holds.

Note that (1) $\Rightarrow $ (2) because a closed immersion is proper and a monomorphism (Morphisms, Lemma 29.39.6 and Schemes, Lemma 26.23.8). By Lemma 41.7.1 we see that (2) implies (3). It is clear that (3) implies (5). $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 04XV. Beware of the difference between the letter 'O' and the digit '0'.