Definition 26.23.1. A morphism of schemes is called a monomorphism if it is a monomorphism in the category of schemes, see Categories, Definition 4.13.1.
26.23 Monomorphisms
Lemma 26.23.2. Let $j : X \to Y$ be a morphism of schemes. Then $j$ is a monomorphism if and only if the diagonal morphism $\Delta _{X/Y} : X \to X \times _ Y X$ is an isomorphism.
Proof. This is true in any category with fibre products. $\square$
Lemma 26.23.3. A monomorphism of schemes is separated.
Proof. This is true because an isomorphism is a closed immersion, and Lemma 26.23.2 above. $\square$
Lemma 26.23.4. A composition of monomorphisms is a monomorphism.
Proof. True in any category. $\square$
Lemma 26.23.5. The base change of a monomorphism is a monomorphism.
Proof. True in any category with fibre products. $\square$
Lemma 26.23.6. Let $j : X \to Y$ be a morphism of schemes. If $j$ is injective on points, then $j$ is separated.
Proof. Let $z$ be a point of $X \times _ Y X$. Then $x = \text{pr}_1(z)$ and $\text{pr}_2(z)$ are the same because $j$ maps these points to the same point $y$ of $Y$. Then we can choose an affine open neighbourhood $V \subset Y$ of $y$ and an affine open neighbourhood $U \subset X$ of $x$ with $j(U) \subset V$. Then $z \in U \times _ V U \subset X \times _ Y X$. Hence $X \times _ Y X$ is the union of the affine opens $U \times _ V U$. Since $\Delta _{X/Y}^{-1}(U \times _ V U) = U$ and since $U \to U \times _ V U$ is a closed immersion, we conclude that $\Delta _{X/Y}$ is a closed immersion (see argument in the proof of Lemma 26.21.2). $\square$
Lemma 26.23.7. Let $j : X \to Y$ be a morphism of schemes. If
$j$ is injective on points, and
for any $x \in X$ the ring map $j^\sharp _ x : \mathcal{O}_{Y, j(x)} \to \mathcal{O}_{X, x}$ is surjective,
then $j$ is a monomorphism.
Proof. Let $a, b : Z \to X$ be two morphisms of schemes such that $j \circ a = j \circ b$. Then (1) implies $a = b$ as underlying maps of topological spaces. For any $z \in Z$ we have $a^\sharp _ z \circ j^\sharp _{a(z)} = b^\sharp _ z \circ j^\sharp _{b(z)}$ as maps $\mathcal{O}_{Y, j(a(z))} \to \mathcal{O}_{Z, z}$. The surjectivity of the maps $j^\sharp _ x$ forces $a^\sharp _ z = b^\sharp _ z$, $\forall z \in Z$. This implies that $a^\sharp = b^\sharp $. Hence we conclude $a = b$ as morphisms of schemes as desired. $\square$
Lemma 26.23.8. An immersion of schemes is a monomorphism. In particular, any immersion is separated.
Proof. We can see this by checking that the criterion of Lemma 26.23.7 applies. More elegantly perhaps, we can use that Lemmas 26.3.5 and 26.4.6 imply that open and closed immersions are monomorphisms and hence any immersion (which is a composition of such) is a monomorphism. $\square$
Lemma 26.23.9. Let $f : X \to S$ be a separated morphism. Any locally closed subscheme $Z \subset X$ is separated over $S$.
Proof. Follows from Lemma 26.23.8 and the fact that a composition of separated morphisms is separated (Lemma 26.21.12). $\square$
Example 26.23.10. The morphism $\mathop{\mathrm{Spec}}(\mathbf{Q}) \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is a monomorphism. This is true because $\mathbf{Q} \otimes _{\mathbf{Z}} \mathbf{Q} = \mathbf{Q}$. More generally, for any scheme $S$ and any point $s \in S$ the canonical morphism is a monomorphism.
\[ \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}) \longrightarrow S \]
Lemma 26.23.11. Let $k_1, \ldots , k_ n$ be fields. For any monomorphism of schemes $X \to \mathop{\mathrm{Spec}}(k_1 \times \ldots \times k_ n)$ there exists a subset $I \subset \{ 1, \ldots , n\} $ such that $X \cong \mathop{\mathrm{Spec}}(\prod _{i \in I} k_ i)$ as schemes over $\mathop{\mathrm{Spec}}(k_1 \times \ldots \times k_ n)$. More generally, if $X = \coprod _{i \in I} \mathop{\mathrm{Spec}}(k_ i)$ is a disjoint union of spectra of fields and $Y \to X$ is a monomorphism, then there exists a subset $J \subset I$ such that $Y = \coprod _{i \in J} \mathop{\mathrm{Spec}}(k_ i)$.
Proof. First reduce to the case $n = 1$ (or $\# I = 1$) by taking the inverse images of the open and closed subschemes $\mathop{\mathrm{Spec}}(k_ i)$. In this case $X$ has only one point hence is affine. The corresponding algebra problem is this: If $k \to R$ is an algebra map with $R \otimes _ k R \cong R$, then $R \cong k$ or $R = 0$. This holds for dimension reasons. See also Algebra, Lemma 10.107.8 $\square$
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