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Let $\pi: X \to Y$ be a locally closed embedding of schemes. Then we can realize $X$ as a closed subscheme of an open subscheme $U$ of $Y$, and $\pi$ factors as $$X \to U \to Y$$ where $X \to U$ is a closed embedding and $U \to Y$ is an open embedding.

We want to show that $\pi$ is a closed embedding. Therefore, we need to show that for any open affine subset $\operatorname{Spec} B$ of $Y$, the preimage $\pi^{-1}(\operatorname{Spec} B)=X \cap \operatorname{Spec} B$ is affine, say $\operatorname{Spec} A$, and the induced map of sections $B \to A$ is surjective.

The problem I am seeing is that the open embedding $U \to Y$ need not be affine, i.e., $U \cap \operatorname{Spec} B$ need not be affine, and therefore $X \cap U \cap \operatorname{Spec} B= X\cap \operatorname{Spec} B$ need not be affine; but we need it to be.

user5826
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1 Answers1

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It is true that any locally closed immersion with closed image is actually a closed immersion. Since the property of being a closed immersion is local on the target, it suffices to show that there is a cover of $Y$ by schemes $U_i$ so that $\pi^{-1}(U_i)\to U_i$ is a closed immersion. Now we're quickly done: choose $U$, where the map is a closed immersion by assumption, and $\pi(X)^c$, where the map is the closed immersion of the empty scheme.

KReiser
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