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. $\square$

 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 $A$.

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