Closed immersion

An example is the inclusion map ${\displaystyle \mathrm{Spec}(R/I)\to \mathrm{Spec}(R)}$ induced by the canonical map ${\displaystyle R\to R/I}$.

Other characterizations

The following are equivalent:

1. ${\displaystyle f:Z\to X}$ is a closed immersion.
2. For every open affine ${\displaystyle U=\mathrm{Spec}(R)\subset X}$, there exists an ideal ${\displaystyle I\subset R}$ such that ${\displaystyle f^{-1}(U)=\mathrm{Spec}(R/I)}$ as schemes over U.
3. There exists an open affine covering ${\displaystyle X=\bigcup U_j,U_j=\mathrm{Spec}{R}_j}$ and for each j there exists an ideal ${\displaystyle I_j\subset R_j}$ such that ${\displaystyle f^{-1}(U_j)=\mathrm{Spec}(R_j/I_j)}$ as schemes over ${\displaystyle U_j}$.
4. There is a quasi-coherent sheaf of ideals ${\displaystyle \mathcal{I}}$ on X such that ${\displaystyle f_{\ast }{\mathcal{O}}_Z\cong {\mathcal{O}}_X/\mathcal{I}}$ and f is an isomorphism of Z onto the global Spec of ${\displaystyle {\mathcal{O}}_X/\mathcal{I}}$ over X.

Definition for locally ringed spaces

In the case of locally ringed spaces a morphism ${\displaystyle i:Z\to X}$ is a closed immersion if a similar list of criteria is satisfied

1. The map ${\displaystyle i}$ is a homeomorphism of ${\displaystyle Z}$ onto its image
2. The associated sheaf map ${\displaystyle {\mathcal{O}}_X\to i_{\ast }{\mathcal{O}}_Z}$ is surjective with kernel ${\displaystyle \mathcal{I}}$
3. The kernel ${\displaystyle \mathcal{I}}$ is locally generated by sections as an ${\displaystyle {\mathcal{O}}_X}$-module

The only varying condition is the third. It is instructive to look at a counter-example to get a feel for what the third condition yields by looking at a map which is not a closed immersion, ${\displaystyle i:{\mathbb{G}}_m\hookrightarrow {\mathbb{A}}^1}$ where

${\displaystyle {\mathbb{G}}_m=\text{Spec}(\mathbb{Z}[x,x^{-1}])}$

If we look at the stalk of ${\displaystyle i_{\ast }{\mathcal{O}}_{{\mathbb{G}}_m}{|}_0}$ at ${\displaystyle 0\in {\mathbb{A}}^1}$ then there are no sections. This implies for any open subscheme ${\displaystyle U\subset {\mathbb{A}}^1}$ containing ${\displaystyle 0}$ the sheaf has no sections. This violates the third condition since at least one open subscheme ${\displaystyle U}$ covering ${\displaystyle {\mathbb{A}}^1}$ contains ${\displaystyle 0}$.