The Stacks project

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. $\square$