The Stacks project

Lemma 38.11.5. Let $f : X \to S$ be a morphism of schemes and $U \subset S$ an open. If

  1. $f$ is separated, locally of finite type, and flat,

  2. $f^{-1}(U) \to U$ is an isomorphism, and

  3. $U \subset S$ is retrocompact and scheme theoretically dense,

then $f$ is an open immersion.

Proof. By Lemma 38.11.2 the morphism $f$ is locally of finite presentation. The image $f(X) \subset S$ is open (Morphisms, Lemma 29.25.10) hence we may replace $S$ by $f(X)$. Thus we have to prove that $f$ is an isomorphism. We may assume $S$ is affine. We can reduce to the case that $X$ is quasi-compact because it suffices to show that any quasi-compact open $X' \subset X$ whose image is $S$ maps isomorphically to $S$. Thus we may assume $f$ is quasi-compact. All the fibers of $f$ have dimension $0$, see Lemma 38.11.3. Hence $f$ is quasi-finite, see Morphisms, Lemma 29.29.5. Let $s \in S$. Choose an elementary ├ętale neighbourhood $g : (T, t) \to (S, s)$ such that $X \times _ S T = V \amalg W$ with $V \to T$ finite and $W_ t = \emptyset $, see More on Morphisms, Lemma 37.41.6. Denote $\pi : V \amalg W \to T$ the given morphism. Since $\pi $ is flat and locally of finite presentation, we see that $\pi (V)$ is open in $T$ (Morphisms, Lemma 29.25.10). After shrinking $T$ we may assume that $T = \pi (V)$. Since $f$ is an isomorphism over $U$ we see that $\pi $ is an isomorphism over $g^{-1}U$. Since $\pi (V) = T$ this implies that $\pi ^{-1}g^{-1}U$ is contained in $V$. By Morphisms, Lemma 29.25.15 we see that $\pi ^{-1}g^{-1}U \subset V \amalg W$ is scheme theoretically dense. Hence we deduce that $W = \emptyset $. Thus $X \times _ S T = V$ is finite over $T$. This implies that $f$ is finite (after replacing $S$ by an open neighbourhood of $s$), for example by Descent, Lemma 35.23.23. Then $f$ is finite locally free (Morphisms, Lemma 29.48.2) and after shrinking $S$ to a smaller open neighbourhood of $s$ we see that $f$ is finite locally free of some degree $d$ (Morphisms, Lemma 29.48.5). But $d = 1$ as is clear from the fact that the degree is $1$ over the dense open $U$. Hence $f$ is an isomorphism. $\square$

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Your email address will not be published.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 081M. Beware of the difference between the letter 'O' and the digit '0'.