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. $\square$

Proof. Write $S = \mathop{\mathrm{Spec}}(\Gamma (X, \mathcal{O}_ X))$ and $f : X \to S$ the morphism given in the lemma. Let $s : S \to X$ be a retraction; so $\text{id}_ X = sf$. Then $f s f = \text{id}_ S f$. Since $f$ induces an isomorphism $\Gamma (S, \mathcal{O}_ S) \to \Gamma (X, \mathcal{O}_ X)$ this means that $fs$ and $\text{id}_ S$ induce the same map on $\Gamma (S, \mathcal{O}_ S)$. Whence $f s = \text{id}_ S$ as $S$ is affine. Hence $f$ is an isomorphism and $X$ is an affine scheme, as was to be shown. $\square$

Proof. Let $M \subset \Gamma (S, \mathcal{O}_ S)$ be the submodule corresponding to the largest quasi-coherent $\mathcal{O}_ S$-module contained in the kernel of $f^\sharp $. Then any element $a \in M$ is mapped to zero by $f^\sharp $. However, $f^\sharp (a)$ is the element of

corresponding to $a$ itself! Thus $a = 0$. Hence $M = 0$ which proves the first assertion. Note that this is equivalent to the morphism $f : X \to S$ being scheme-theoretically surjective.

If $X$ is quasi-compact, then $\mathop{\mathrm{Ker}}(f^\sharp )$ is quasi-coherent by Morphisms, Lemma 29.6.3. Hence $\mathop{\mathrm{Ker}}(f^\sharp ) = 0$ and $f^\sharp $ is injective. In this case, $f$ is a dominant morphism by part (4) of Morphisms, Lemma 29.6.3. $\square$

Proof. Assume that $f$ is a closed immersion. Let $A \to B$ be an injective ring map and consider a commutative square

Then $\mathop{\mathrm{Spec}}(A) \times _ Y X \to \mathop{\mathrm{Spec}}(A)$ is a closed immersion and hence we get an ideal $I \subset A$ and a commutative diagram

We obtain a lift by Lemma 37.81.1.

Assume that $f$ has the lifting property stated in the lemma. To prove that $f$ is a closed immersion is local on $Y$, hence we may and do assume $Y$ is affine. In particular, $Y$ is quasi-compact and therefore $X$ is quasi-compact. Hence there exists a finite affine open covering $X = U_1 \cup \ldots \cup U_ n$. The source of the morphism

is affine and the induced ring map $\Gamma (X, \mathcal{O}_ X) \to \Gamma (U, \mathcal{O}_ U)$ is injective. By assumption, there exists a lift in the diagram

where $f'$ is the morphism of affine schemes corresponding to the ring map $\Gamma (Y, \mathcal{O}_ Y) \to \Gamma (X, \mathcal{O}_ X)$. It follows from the fact that $\pi $ is an epimorphism that the morphism $h$ is a retraction of the canonical morphism $X \to \mathop{\mathrm{Spec}}(\Gamma (X, \mathcal{O}_ X))$; details omitted. Hence $X$ is affine by Lemma 37.81.2. By Lemma 37.81.1 we conclude that $f$ is a closed immersion. $\square$