## 26.8 Closed subspaces of affine schemes

Example 26.8.1. Let $R$ be a ring. Let $I \subset R$ be an ideal. Consider the morphism of affine schemes $i : Z = \mathop{\mathrm{Spec}}(R/I) \to \mathop{\mathrm{Spec}}(R) = X$. By Algebra, Lemma 10.17.7 this is a homeomorphism of $Z$ onto a closed subset of $X$. Moreover, if $I \subset \mathfrak p \subset R$ is a prime corresponding to a point $x = i(z)$, $x \in X$, $z \in Z$, then on stalks we get the map

$\mathcal{O}_{X, x} = R_{\mathfrak p} \longrightarrow R_{\mathfrak p}/IR_{\mathfrak p} = \mathcal{O}_{Z, z}$

Thus we see that $i$ is a closed immersion of locally ringed spaces, see Definition 26.4.1. Clearly, this is (isomorphic) to the closed subspace associated to the quasi-coherent sheaf of ideals $\widetilde I$, as in Example 26.4.3.

Lemma 26.8.2. Let $(X, \mathcal{O}_ X) = (\mathop{\mathrm{Spec}}(R), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$ be an affine scheme. Let $i : Z \to X$ be any closed immersion of locally ringed spaces. Then there exists a unique ideal $I \subset R$ such that the morphism $i : Z \to X$ can be identified with the closed immersion $\mathop{\mathrm{Spec}}(R/I) \to \mathop{\mathrm{Spec}}(R)$ constructed in Example 26.8.1 above.

Proof. This is kind of silly! Namely, by Lemma 26.4.5 we can identify $Z \to X$ with the closed subspace associated to a sheaf of ideals $\mathcal{I} \subset \mathcal{O}_ X$ as in Definition 26.4.4 and Example 26.4.3. By our conventions this sheaf of ideals is locally generated by sections as a sheaf of $\mathcal{O}_ X$-modules. Hence the quotient sheaf $\mathcal{O}_ X / \mathcal{I}$ is locally on $X$ the cokernel of a map $\bigoplus _{j \in J} \mathcal{O}_ U \to \mathcal{O}_ U$. Thus by definition, $\mathcal{O}_ X / \mathcal{I}$ is quasi-coherent. By our results in Section 26.7 it is of the form $\widetilde S$ for some $R$-module $S$. Moreover, since $\mathcal{O}_ X = \widetilde R \to \widetilde S$ is surjective we see by Lemma 26.7.8 that also $\mathcal{I}$ is quasi-coherent, say $\mathcal{I} = \widetilde I$. Of course $I \subset R$ and $S = R/I$ and everything is clear. $\square$

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