Lemma 37.81.4. Let $f: X \to Y$ be a quasi-compact morphism of schemes. Then $f$ 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 diagram commute.
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. $\square$
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