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.
For affine schemes, closed immersions correspond to ideals.
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.
$\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (1)
Comment #1105 by Evan Warner on