Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to prove the following: Let $f: X \rightarrow S$ be a separated morphism of schemes. Show that any section $g: S \rightarrow X$ of $f$ i.e. a morphism such that $f \circ g=\textrm{id}_{S}$ is a closed immersion.

It seems that this can be done by considering an appropriate graph morphism. Here is what I have thus far:

Taking $X$ as an $S$-scheme under $f$, we have that a section of $f$ is simply $g: S \rightarrow X$. Let us consider $\Gamma_{g}:S \rightarrow S \times_{S} X$, which is a locally closed immersion, and a closed immersion since we have that $X$ is separated. Then $\Gamma_{g}$ is canonically identified with $g$ in this situation. Is that all that is required here?

share|improve this question

1 Answer 1

up vote 4 down vote accepted

Yes this works. A bit more detailed: If $T \to X$ is a morphism of $S$-schemes and $X$ is separated over $S$, then the graph morphism $T \to T \times_S X$ is a closed immersion since the following diagram is cartesian:

$$\begin{array}{c} T & \rightarrow & T \times_S X \\ \downarrow && \downarrow \\X & \rightarrow & X \times_S X \end{array}$$

Applying this to $T=S$, we get the desired result that every section of $X \to S$ is a closed immersion.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.