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, restricted there, 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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