The Stacks project

41.7 Universally injective, unramified morphisms

Recall that a morphism of schemes $f : X \to Y$ is universally injective if any base change of $f$ is injective (on underlying topological spaces), see Morphisms, Definition 29.10.1. Universally injective and unramified morphisms can be characterized as follows.

Lemma 41.7.1. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

  1. $f$ is unramified and a monomorphism,

  2. $f$ is unramified and universally injective,

  3. $f$ is locally of finite type and a monomorphism,

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

  5. $f$ is locally of finite type and $X_ s$ is either empty or $X_ s \to s$ is an isomorphism for all $s \in S$.

Proof. We have seen in More on Morphisms, Lemma 37.6.8 that being formally unramified and locally of finite type is the same thing as being unramified. Hence (4) is equivalent to (2). A monomorphism is certainly universally injective and formally unramified hence (3) implies (4). It is clear that (1) implies (3). Finally, if (2) holds, then $\Delta : X \to X \times _ S X$ is both an open immersion (Morphisms, Lemma 29.35.13) and surjective (Morphisms, Lemma 29.10.2) hence an isomorphism, i.e., $f$ is a monomorphism. In this way we see that (2) implies (1).

Condition (3) implies (5) because monomorphisms are preserved under base change (Schemes, Lemma 26.23.5) and because of the description of monomorphisms towards the spectra of fields in Schemes, Lemma 26.23.11. Condition (5) implies (4) by Morphisms, Lemmas 29.10.2 and 29.35.12. $\square$

This leads to the following useful characterization of closed immersions.

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 37.44.1 shows $f$ is finite. Then Morphisms, Lemma 29.44.15 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 29.41.6 and Schemes, Lemma 26.23.8). By Lemma 41.7.1 we see that (2) implies (3). It is clear that (3) implies (5). $\square$

Here is another result of a similar flavor.

Lemma 41.7.3. Let $\pi : X \to S$ be a morphism of schemes. Let $s \in S$. Assume that

  1. $\pi $ is finite,

  2. $\pi $ is unramified,

  3. $\pi ^{-1}(\{ s\} ) = \{ x\} $, and

  4. $\kappa (s) \subset \kappa (x)$ is purely inseparable1.

Then there exists an open neighbourhood $U$ of $s$ such that $\pi |_{\pi ^{-1}(U)} : \pi ^{-1}(U) \to U$ is a closed immersion.

Proof. The question is local on $S$. Hence we may assume that $S = \mathop{\mathrm{Spec}}(A)$. By definition of a finite morphism this implies $X = \mathop{\mathrm{Spec}}(B)$. Note that the ring map $\varphi : A \to B$ defining $\pi $ is a finite unramified ring map. Let $\mathfrak p \subset A$ be the prime corresponding to $s$. Let $\mathfrak q \subset B$ be the prime corresponding to $x$. Conditions (2), (3) and (4) imply that $B_{\mathfrak q}/\mathfrak pB_{\mathfrak q} = \kappa (\mathfrak p)$. By Algebra, Lemma 10.41.11 we have $B_{\mathfrak q} = B_{\mathfrak p}$ (note that a finite ring map satisfies going up, see Algebra, Section 10.41.) Hence we see that $B_{\mathfrak p}/\mathfrak pB_{\mathfrak p} = \kappa (\mathfrak p)$. As $B$ is a finite $A$-module we see from Nakayama's lemma (see Algebra, Lemma 10.20.1) that $B_{\mathfrak p} = \varphi (A_{\mathfrak p})$. Hence (using the finiteness of $B$ as an $A$-module again) there exists a $f \in A$, $f \not\in \mathfrak p$ such that $B_ f = \varphi (A_ f)$ as desired. $\square$

The topological results presented above will be used to give a functorial characterization of étale morphisms similar to Theorem 41.5.1.

[1] In view of condition (2) this is equivalent to $\kappa (s) = \kappa (x)$.

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