Lemma 26.10.4. 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 26.4.3. 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 26.4.5, we conclude that $Y \to U \to X$ is a closed immersion. $\square$

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