Lemma 37.81.1. A morphism $f : X \to Y$ of affine schemes is a closed immersion if and only if for every injective ring map $A \to B$ and commutative square
there exists a lift $\mathop{\mathrm{Spec}}(A) \to X$ making the two triangles commute.
Proof. Let the morphism $f$ be given by the ring map $\phi : R \to S$. Then $f$ is a closed immersion if and only if $\phi $ is surjective.
First, we assume that $\phi $ is surjective. Let $\psi : A \to B$ be an injective ring map, and suppose we are given a commutative diagram
Then we define a lift $S \to A$ by $s \mapsto \alpha (r)$, where $r \in R$ is such that $\phi (r) = s$. This is well-defined because $\psi $ is injective and the square commutes. Since taking the ring spectrum defines an anti-equivalence between commutative rings and affine schemes, the desired lifting property for $f$ holds.
Next, we assume that $\phi $ has lifts against all injective ring maps $\psi : A \to B$. Note that $\phi (R)$ is a subring of $S$, so we obtain a commutative square
in which a lift $S \to \phi (R)$ exists. Hence, the inclusion $\phi (R) \to S$ must be an isomorphism, which shows that $\phi $ is surjective, and we win. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)