Lemma 36.36.5 (0GTC)—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 free ¹, 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

$A$ .

The Stacks project

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