The Stacks project

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

\[ \xymatrix{ \mathop{\mathrm{Spec}}(B) \ar[d] \ar[r] & X \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{..>}[ur] & Y } \]

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

\[ \xymatrix{ \mathop{\mathrm{Spec}}(B) \ar[d] \ar[r] & X \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{..>}[ur] & Y } \]

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

\[ \xymatrix{ \mathop{\mathrm{Spec}}(B) \ar[d] \ar[r] & \mathop{\mathrm{Spec}}(A/I) \ar[r] \ar[d] & X \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{..>}[ur] & \mathop{\mathrm{Spec}}(A) \ar[r] & Y } \]

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

\[ \pi : U = \coprod U_ i \longrightarrow X \]

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

\[ \xymatrix{ U \ar[r]^\pi \ar[d] & X \ar[d]^ f \\ \mathop{\mathrm{Spec}}(\Gamma (X, \mathcal{O}_ X)) \ar[r]^-{f'} \ar@{..>}[ur]^ h & Y } \]

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. $\square$

Comments (0)

Post a comment

Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0H2S. Beware of the difference between the letter 'O' and the digit '0'.