The Stacks project

Lemma 36.36.5. Let $f : X \to Y$ be a surjective finite locally free morphism of schemes. If $X$ has the resolution property, so does $Y$.

Proof. The condition means that $f$ is affine and that $f_*\mathcal{O}_ X$ is a finite locally free $\mathcal{O}_ Y$-module of positive rank. Let $\mathcal{G}$ be a quasi-coherent $\mathcal{O}_ Y$-module of finite type. By assumption there exists a surjection $\mathcal{E} \to f^*\mathcal{G}$ for some finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$. Since $f_*$ is exact on quasi-coherent modules (Cohomology of Schemes, Lemma 30.2.3) we get a surjection

\[ f_*\mathcal{E} \longrightarrow f_*f^*\mathcal{G} = \mathcal{G} \otimes _{\mathcal{O}_ Y} f_*\mathcal{O}_ X \]

Taking duals we get a surjection

\[ f_*\mathcal{E} \otimes _{\mathcal{O}_ Y} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(f_*\mathcal{O}_ X, \mathcal{O}_ Y) \longrightarrow \mathcal{G} \]

Since $f_*\mathcal{E}$ is finite locally free1, we conclude. $\square$

[1] Namely, if $A \to B$ is a finite locally free ring map and $N$ is a finite locally free $B$-module, then $N$ is a finite locally free $A$-module. To see this, first note that $N$ finite locally free over $B$ implies $N$ is flat and finitely presented as a $B$-module, see Algebra, Lemma 10.78.2. Then $N$ is an $A$-module of finite presentation by Algebra, Lemma 10.36.23 and a flat $A$-module by Algebra, Lemma 10.39.4. Then conclude by using Algebra, Lemma 10.78.2 over $A$.

Comments (0)

There are also:

  • 2 comment(s) on Section 36.36: The resolution property

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Your email address will not be published.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 0GTC. Beware of the difference between the letter 'O' and the digit '0'.