The Stacks project

Lemma 37.3.4. Let $(f, f') : (X \subset X') \to (Y \to Y')$ be a morphism of thickenings. Assume $f$ and $f'$ are locally of finite type and $X = Y \times _{Y'} X'$. Then

  1. $f$ is locally quasi-finite if and only if $f'$ is locally quasi-finite,

  2. $f$ is finite if and only if $f'$ is finite,

  3. $f$ is a closed immersion if and only if $f'$ is a closed immersion,

  4. $\Omega _{X/Y} = 0$ if and only if $\Omega _{X'/Y'} = 0$,

  5. $f$ is unramified if and only if $f'$ is unramified,

  6. $f$ is a monomorphism if and only if $f'$ is a monomorphism,

  7. $f$ is an immersion if and only if $f'$ is an immersion,

  8. $f$ is proper if and only if $f'$ is proper, and

  9. add more here.

Proof. The properties $\mathcal{P}$ listed in the lemma are all stable under base change, hence if $f'$ has property $\mathcal{P}$, then so does $f$. See Schemes, Lemmas 26.18.2 and 26.23.5 and Morphisms, Lemmas 29.20.13, 29.29.2, 29.32.10, 29.35.5, 29.41.5, and 29.44.6. Hence in each case we need only to prove that if $f$ has the desired property, so does $f'$.

A morphism is locally quasi-finite if and only if it is locally of finite type and the scheme theoretic fibres are discrete spaces, see Morphisms, Lemma 29.20.8. Since the underlying topological space is unchanged by passing to a thickening, we see that $f'$ is locally quasi-finite if (and only if) $f$ is. This proves (1).

Case (2) follows from case (5) of Lemma 37.3.1 and the fact that the finite morphisms are precisely the integral morphisms that are locally of finite type (Morphisms, Lemma 29.44.4).

Case (3). This follows immediately from Morphisms, Lemma 29.45.7.

Case (4) follows from the following algebra statement: Let $A$ be a ring and let $I \subset A$ be a locally nilpotent ideal. Let $B$ be a finite type $A$-algebra. If $\Omega _{(B/IB)/(A/I)} = 0$, then $\Omega _{B/A} = 0$. Namely, the assumption means that $I\Omega _{B/A} = 0$, see Algebra, Lemma 10.131.12. On the other hand $\Omega _{B/A}$ is a finite $B$-module, see Algebra, Lemma 10.131.16. Hence the vanishing of $\Omega _{B/A}$ follows from Nakayama's lemma (Algebra, Lemma 10.20.1) and the fact that $IB$ is contained in the Jacobson radical of $B$.

Case (5) follows immediately from (4) and Morphisms, Lemma 29.35.2.

Proof of (6). As $f$ is a monomorphism we have $X = X \times _ Y X$. We may apply the results proved so far to the morphism of thickenings $(X \subset X') \to (X \times _ Y X \subset X' \times _{Y'} X')$. We conclude $\Delta _{X'/Y'} : X' \to X' \times _{Y'} X'$ is a closed immersion by (3). In fact $\Delta _{X'/Y'}$ is a bijection on underlying sets, hence $\Delta _{X'/Y'}$ is a thickening. On the other hand $\Delta _{X'/Y'}$ is locally of finite presentation by Morphisms, Lemma 29.21.12. In other words, $\Delta _{X'/Y'}(X')$ is cut out by a quasi-coherent sheaf of ideals $\mathcal{J} \subset \mathcal{O}_{X' \times _{Y'} X'}$ of finite type. Since $\Omega _{X'/Y'} = 0$ by (5) we see that the conormal sheaf of $X' \to X' \times _{Y'} X'$ is zero by Morphisms, Lemma 29.32.7. In other words, $\mathcal{J}/\mathcal{J}^2 = 0$. This implies $\Delta _{X'/Y'}$ is an isomorphism, for example by Algebra, Lemma 10.21.5.

Proof of (7). If $f : X \to Y$ is an immersion, then it factors as $X \to V \to Y$ where $V \to Y$ is an open immersion and $X \to V$ is a closed immersion. Let $V' \subset Y'$ be the open subscheme whose underlying topological space is the same as $V$. Then $X' \to V'$ factors through $V'$ and we conclude that $X' \to V'$ is a closed immersion by part (3).

Case (8) follows from Lemma 37.3.1 and the definition of proper morphisms as being the quasi-compact, universally closed, and separated morphisms that are locally of finite type. $\square$

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 0BPG. Beware of the difference between the letter 'O' and the digit '0'.