## 26.4 Closed immersions of locally ringed spaces

We follow our conventions introduced in Modules, Definition 17.13.1.

Definition 26.4.1. Let $i : Z \to X$ be a morphism of locally ringed spaces. We say that $i$ is a *closed immersion* if:

The map $i$ is a homeomorphism of $Z$ onto a closed subset of $X$.

The map $\mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is surjective; let $\mathcal{I}$ denote the kernel.

The $\mathcal{O}_ X$-module $\mathcal{I}$ is locally generated by sections.

Lemma 26.4.2. Let $f : Z \to X$ be a morphism of locally ringed spaces. In order for $f$ to be a closed immersion it suffices that there exists an open covering $X = \bigcup U_ i$ such that each $f : f^{-1}U_ i \to U_ i$ is a closed immersion.

**Proof.**
Omitted.
$\square$

Example 26.4.3. Let $X$ be a locally ringed space. Let $\mathcal{I} \subset \mathcal{O}_ X$ be a sheaf of ideals which is locally generated by sections as a sheaf of $\mathcal{O}_ X$-modules. Let $Z$ be the support of the sheaf of rings $\mathcal{O}_ X/\mathcal{I}$. This is a closed subset of $X$, by Modules, Lemma 17.5.3. Denote $i : Z \to X$ the inclusion map. By Modules, Lemma 17.6.1 there is a unique sheaf of rings $\mathcal{O}_ Z$ on $Z$ with $i_*\mathcal{O}_ Z = \mathcal{O}_ X/\mathcal{I}$. For any $z \in Z$ the stalk $\mathcal{O}_{Z, z}$ is equal to a quotient $\mathcal{O}_{X, i(z)}/\mathcal{I}_{i(z)}$ of a local ring and nonzero, hence a local ring. Thus $i : (Z, \mathcal{O}_ Z) \to (X, \mathcal{O}_ X)$ is a closed immersion of locally ringed spaces.

Definition 26.4.4. Let $X$ be a locally ringed space. Let $\mathcal{I}$ be a sheaf of ideals on $X$ which is locally generated by sections. The locally ringed space $(Z, \mathcal{O}_ Z)$ of Example 26.4.3 above is the *closed subspace of $X$ associated to the sheaf of ideals $\mathcal{I}$*.

Lemma 26.4.5. Let $f : X \to Y$ be a closed immersion of locally ringed spaces. Let $\mathcal{I}$ be the kernel of the map $\mathcal{O}_ Y \to f_*\mathcal{O}_ X$. Let $i : Z \to Y$ be the closed subspace of $Y$ associated to $\mathcal{I}$. There is a unique isomorphism $f' : X \cong Z$ of locally ringed spaces such that $f = i \circ f'$.

**Proof.**
Omitted.
$\square$

Lemma 26.4.6. Let $X$, $Y$ be locally ringed spaces. Let $\mathcal{I} \subset \mathcal{O}_ X$ be a sheaf of ideals locally generated by sections. Let $i : Z \to X$ be the associated closed subspace. A morphism $f : Y \to X$ factors through $Z$ if and only if the map $f^*\mathcal{I} \to f^*\mathcal{O}_ X = \mathcal{O}_ Y$ is zero. If this is the case the morphism $g : Y \to Z$ such that $f = i \circ g$ is unique.

**Proof.**
Clearly if $f$ factors as $Y \to Z \to X$ then the map $f^*\mathcal{I} \to \mathcal{O}_ Y$ is zero. Conversely suppose that $f^*\mathcal{I} \to \mathcal{O}_ Y$ is zero. Pick any $y \in Y$, and consider the ring map $f^\sharp _ y : \mathcal{O}_{X, f(y)} \to \mathcal{O}_{Y, y}$. Since the composition $\mathcal{I}_{f(y)} \to \mathcal{O}_{X, f(y)} \to \mathcal{O}_{Y, y}$ is zero by assumption and since $f^\sharp _ y(1) = 1$ we see that $1 \not\in \mathcal{I}_{f(y)}$, i.e., $\mathcal{I}_{f(y)} \not= \mathcal{O}_{X, f(y)}$. We conclude that $f(Y) \subset Z = \text{Supp}(\mathcal{O}_ X/\mathcal{I})$. Hence $f = i \circ g$ where $g : Y \to Z$ is continuous. Consider the map $f^\sharp : \mathcal{O}_ X \to f_*\mathcal{O}_ Y$. The assumption $f^*\mathcal{I} \to \mathcal{O}_ Y$ is zero implies that the composition $\mathcal{I} \to \mathcal{O}_ X \to f_*\mathcal{O}_ Y$ is zero by adjointness of $f_*$ and $f^*$. In other words, we obtain a morphism of sheaves of rings $\overline{f^\sharp } : \mathcal{O}_ X/\mathcal{I} \to f_*\mathcal{O}_ Y$. Note that $f_*\mathcal{O}_ Y = i_*g_*\mathcal{O}_ Y$ and that $\mathcal{O}_ X/\mathcal{I} = i_*\mathcal{O}_ Z$. By Sheaves, Lemma 6.32.4 we obtain a unique morphism of sheaves of rings $g^\sharp : \mathcal{O}_ Z \to g_*\mathcal{O}_ Y$ whose pushforward under $i$ is $\overline{f^\sharp }$. We omit the verification that $(g, g^\sharp )$ defines a morphism of locally ringed spaces and that $f = i \circ g$ as a morphism of locally ringed spaces. The uniqueness of $(g, g^\sharp )$ was pointed out above.
$\square$

Lemma 26.4.7. Let $f : X \to Y$ be a morphism of locally ringed spaces. Let $\mathcal{I} \subset \mathcal{O}_ Y$ be a sheaf of ideals which is locally generated by sections. Let $i : Z \to Y$ be the closed subspace associated to the sheaf of ideals $\mathcal{I}$. Let $\mathcal{J}$ be the image of the map $f^*\mathcal{I} \to f^*\mathcal{O}_ Y = \mathcal{O}_ X$. Then this ideal is locally generated by sections. Moreover, let $i' : Z' \to X$ be the associated closed subspace of $X$. There exists a unique morphism of locally ringed spaces $f' : Z' \to Z$ such that the following diagram is a commutative square of locally ringed spaces

\[ \xymatrix{ Z' \ar[d]_{f'} \ar[r]_{i'} & X \ar[d]^ f \\ Z \ar[r]^{i} & Y } \]

Moreover, this diagram is a fibre square in the category of locally ringed spaces.

**Proof.**
The ideal $\mathcal{J}$ is locally generated by sections by Modules, Lemma 17.8.2. The rest of the lemma follows from the characterization, in Lemma 26.4.6 above, of what it means for a morphism to factor through a closed subspace.
$\square$

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