The Stacks project

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25.10 Immersions of schemes

In Lemma 25.9.2 we saw that any open subspace of a scheme is a scheme. Below we will prove that the same holds for a closed subspace of a scheme.

Note that the notion of a quasi-coherent sheaf of $\mathcal{O}_ X$-modules is defined for any ringed space $X$ in particular when $X$ is a scheme. By our efforts in Section 25.7 we know that such a sheaf is on any affine open $U \subset X$ of the form $\widetilde M$ for some $\mathcal{O}_ X(U)$-module $M$.

Lemma 25.10.1. Let $X$ be a scheme. Let $i : Z \to X$ be a closed immersion of locally ringed spaces.

  1. The locally ringed space $Z$ is a scheme,

  2. the kernel $\mathcal{I}$ of the map $\mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is a quasi-coherent sheaf of ideals,

  3. for any affine open $U = \mathop{\mathrm{Spec}}(R)$ of $X$ the morphism $i^{-1}(U) \to U$ can be identified with $\mathop{\mathrm{Spec}}(R/I) \to \mathop{\mathrm{Spec}}(R)$ for some ideal $I \subset R$, and

  4. we have $\mathcal{I}|_ U = \widetilde I$.

In particular, any sheaf of ideals locally generated by sections is a quasi-coherent sheaf of ideals (and vice versa), and any closed subspace of $X$ is a scheme.

Proof. Let $i : Z \to X$ be a closed immersion. Let $z \in Z$ be a point. Choose any affine open neighbourhood $i(z) \in U \subset X$. Say $U = \mathop{\mathrm{Spec}}(R)$. By Lemma 25.8.2 we know that $i^{-1}(U) \to U$ can be identified with the morphism of affine schemes $\mathop{\mathrm{Spec}}(R/I) \to \mathop{\mathrm{Spec}}(R)$. First of all this implies that $z \in i^{-1}(U) \subset Z$ is an affine neighbourhood of $z$. Thus $Z$ is a scheme. Second this implies that $\mathcal{I}|_ U$ is $\widetilde I$. In other words for every point $x \in i(Z)$ there exists an open neighbourhood such that $\mathcal{I}$ is quasi-coherent in that neighbourhood. Note that $\mathcal{I}|_{X \setminus i(Z)} \cong \mathcal{O}_{X \setminus i(Z)}$. Thus the restriction of the sheaf of ideals is quasi-coherent on $X \setminus i(Z)$ also. We conclude that $\mathcal{I}$ is quasi-coherent. $\square$

Definition 25.10.2. Let $X$ be a scheme.

  1. A morphism of schemes is called an open immersion if it is an open immersion of locally ringed spaces (see Definition 25.3.1).

  2. An open subscheme of $X$ is an open subspace of $X$ which is a scheme by Lemma 25.9.2 above.

  3. A morphism of schemes is called a closed immersion if it is a closed immersion of locally ringed spaces (see Definition 25.4.1).

  4. A closed subscheme of $X$ is a closed subspace of $X$ which is a scheme by Lemma 25.10.1 above.

  5. A morphism of schemes $f : X \to Y$ is called an immersion, or a locally closed immersion if it can be factored as $j \circ i$ where $i$ is a closed immersion and $j$ is an open immersion.

It follows from the lemmas in Sections 25.3 and 25.4 that any open (resp. closed) immersion of schemes is isomorphic to the inclusion of an open (resp. closed) subscheme of the target. We will define locally closed subschemes below.

Remark 25.10.3. If $f : X \to Y$ is an immersion of schemes, then it is in general not possible to factor $f$ as an open immersion followed by a closed immersion. See Morphisms, Example 28.3.4.

Lemma 25.10.4. Let $f : Y \to X$ be an immersion of schemes. Then $f$ is a closed immersion if and only if $f(Y) \subset X$ is a closed subset.

Proof. If $f$ is a closed immersion then $f(Y)$ is closed by definition. Conversely, suppose that $f(Y)$ is closed. By definition there exists an open subscheme $U \subset X$ such that $f$ is the composition of a closed immersion $i : Y \to U$ and the open immersion $j : U \to X$. Let $\mathcal{I} \subset \mathcal{O}_ U$ be the quasi-coherent sheaf of ideals associated to the closed immersion $i$. Note that $\mathcal{I}|_{U \setminus i(Y)} = \mathcal{O}_{U \setminus i(Y)} = \mathcal{O}_{X \setminus i(Y)}|_{U \setminus i(Y)}$. Thus we may glue (see Sheaves, Section 6.33) $\mathcal{I}$ and $\mathcal{O}_{X \setminus i(Y)}$ to a sheaf of ideals $\mathcal{J} \subset \mathcal{O}_ X$. Since every point of $X$ has a neighbourhood where $\mathcal{J}$ is quasi-coherent, we see that $\mathcal{J}$ is quasi-coherent (in particular locally generated by sections). By construction $\mathcal{O}_ X/\mathcal{J}$ is supported on $U$ and equal to $\mathcal{O}_ U/\mathcal{I}$. Thus we see that the closed subspaces associated to $\mathcal{I}$ and $\mathcal{J}$ are canonically isomorphic, see Example 25.4.3. In particular the closed subspace of $U$ associated to $\mathcal{I}$ is isomorphic to a closed subspace of $X$. Since $Y \to U$ is identified with the closed subspace associated to $\mathcal{I}$, see Lemma 25.4.5, we conclude that $Y \to U \to X$ is a closed immersion. $\square$

Let $f : Y \to X$ be an immersion. Let $Z = \overline{f(Y)} \setminus f(Y)$ which is a closed subset of $X$. Let $U = X \setminus Z$. The lemma implies that $U$ is the biggest open subspace of $X$ such that $f : Y \to X$ factors through a closed immersion into $U$. We define a locally closed subscheme of $X$ as a pair $(Z, U)$ consisting of a closed subscheme $Z$ of an open subscheme $U$ of $X$ such that in addition $\overline{Z} \cup U = X$. We usually just say “let $Z$ be a locally closed subscheme of $X$” since we may recover $U$ from the morphism $Z \to X$. The above then shows that any immersion $f : Y \to X$ factors uniquely as $Y \to Z \to X$ where $Z$ is a locally closed subspace of $X$ and $Y \to Z$ is an isomorphism.

The interest of this is that the collection of locally closed subschemes of $X$ forms a set. We may define a partial ordering on this set, which we call inclusion for obvious reasons. To be explicit, if $Z \to X$ and $Z' \to X$ are two locally closed subschemes of $X$, then we say that $Z$ is contained in $Z'$ simply if the morphism $Z \to X$ factors through $Z'$. If it does, then of course $Z$ is identified with a unique locally closed subscheme of $Z'$, and so on.

Comments (3)

Comment #1783 by Keenan Kidwell on

In the discussion following the proof of 01IQ, the first word of the sentence "If we define a..." should be removed (or something should be added to the sentence).

Comment #2519 by Ammar Y. Kılıç on

Why the data of a closed subscheme consists of only a topological object?

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