Lemma 29.2.1. Let $i : Z \to X$ be a morphism of schemes. The following are equivalent:

1. The morphism $i$ is a closed immersion.

2. For every affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$, there exists an ideal $I \subset R$ such that $i^{-1}(U) = \mathop{\mathrm{Spec}}(R/I)$ as schemes over $U = \mathop{\mathrm{Spec}}(R)$.

3. There exists an affine open covering $X = \bigcup _{j \in J} U_ j$, $U_ j = \mathop{\mathrm{Spec}}(R_ j)$ and for every $j \in J$ there exists an ideal $I_ j \subset R_ j$ such that $i^{-1}(U_ j) = \mathop{\mathrm{Spec}}(R_ j/I_ j)$ as schemes over $U_ j = \mathop{\mathrm{Spec}}(R_ j)$.

4. The morphism $i$ induces a homeomorphism of $Z$ with a closed subset of $X$ and $i^\sharp : \mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is surjective.

5. The morphism $i$ induces a homeomorphism of $Z$ with a closed subset of $X$, the map $i^\sharp : \mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is surjective, and the kernel $\mathop{\mathrm{Ker}}(i^\sharp )\subset \mathcal{O}_ X$ is a quasi-coherent sheaf of ideals.

6. The morphism $i$ induces a homeomorphism of $Z$ with a closed subset of $X$, the map $i^\sharp : \mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is surjective, and the kernel $\mathop{\mathrm{Ker}}(i^\sharp )\subset \mathcal{O}_ X$ is a sheaf of ideals which is locally generated by sections.

Proof. Condition (6) is our definition of a closed immersion, see Schemes, Definitions 26.4.1 and 26.10.2. So (6) $\Leftrightarrow$ (1). We have (1) $\Rightarrow$ (2) by Schemes, Lemma 26.10.1. Trivially (2) $\Rightarrow$ (3).

Assume (3). Each of the morphisms $\mathop{\mathrm{Spec}}(R_ j/I_ j) \to \mathop{\mathrm{Spec}}(R_ j)$ is a closed immersion, see Schemes, Example 26.8.1. Hence $i^{-1}(U_ j) \to U_ j$ is a homeomorphism onto its image and $i^\sharp |_{U_ j}$ is surjective. Hence $i$ is a homeomorphism onto its image and $i^\sharp$ is surjective since this may be checked locally. We conclude that (3) $\Rightarrow$ (4).

The implication (4) $\Rightarrow$ (1) is Schemes, Lemma 26.24.2. The implication (5) $\Rightarrow$ (6) is trivial. And the implication (6) $\Rightarrow$ (5) follows from Schemes, Lemma 26.10.1. $\square$

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