The Stacks project

Proof. Note that $Y$ is locally Noetherian by Morphisms, Lemma 29.15.6. Because $f$ is surjective the fibre $Y_\xi$ is not empty. Pick $\xi ' \in Y$ mapping to $\xi$. Let $Z' = \overline{\{ \xi '\} }$. We may think of $Z' \subset Y$ as a reduced closed subscheme, see Schemes, Lemma 26.12.4. Hence the sheaf $\mathcal{F} = (Z' \to Y)_*\mathcal{O}_{Z'}$ is a coherent sheaf on $Y$ (see Lemma 30.9.9). Look at the commutative diagram

We see that $f_*\mathcal{F} = i_*f'_*\mathcal{O}_{Z'}$. Hence the stalk of $f_*\mathcal{F}$ at $\xi$ is the stalk of $f'_*\mathcal{O}_{Z'}$ at $\xi$. Note that since $Z'$ is integral with generic point $\xi '$ we have that $\xi '$ is the only point of $Z'$ lying over $\xi$, see Algebra, Lemmas 10.36.3 and 10.36.20. Hence the stalk of $f'_*\mathcal{O}_{Z'}$ at $\xi$ equal $\mathcal{O}_{Z', \xi '} = \kappa (\xi ')$. In particular the stalk of $f_*\mathcal{F}$ at $\xi$ is not zero. This combined with the fact that $f_*\mathcal{F}$ is of the form $i_*f'_*(\text{something})$ implies the lemma. $\square$

Proof. The notation means the following. Since $f^{-1}$ is an exact functor we see that $f^{-1}\mathcal{I}$ is a sheaf of ideals of $f^{-1}\mathcal{O}_ X$. Via the map $f^\sharp : f^{-1}\mathcal{O}_ X \to \mathcal{O}_ Y$ this acts on $\mathcal{F}$. Then $f^{-1}\mathcal{I}\mathcal{F}$ is the subsheaf generated by sums of local sections of the form $as$ where $a$ is a local section of $f^{-1}\mathcal{I}$ and $s$ is a local section of $\mathcal{F}$. It is a quasi-coherent $\mathcal{O}_ Y$-submodule of $\mathcal{F}$ because it is also the image of a natural map $f^*\mathcal{I} \otimes _{\mathcal{O}_ Y} \mathcal{F} \to \mathcal{F}$.

Having said this the proof is straightforward. Namely, the question is local and hence we may assume $X$ is affine. Since $f$ is affine we see that $Y$ is affine too. Thus we may write $Y = \mathop{\mathrm{Spec}}(B)$, $X = \mathop{\mathrm{Spec}}(A)$, $\mathcal{F} = \widetilde{M}$, and $\mathcal{I} = \widetilde{I}$. The assertion of the lemma in this case boils down to the statement that

where $M_ A$ indicates the $A$-module associated to the $B$-module $M$. $\square$

Proof. We will prove that under the assumptions of the lemma for any coherent $\mathcal{O}_ X$-module $\mathcal{F}$ we have $H^1(X, \mathcal{F}) = 0$. This will in particular imply that $H^1(X, \mathcal{I}) = 0$ for every quasi-coherent sheaf of ideals of $\mathcal{O}_ X$. Then it follows that $X$ is affine from either Lemma 30.3.1 or Lemma 30.3.2.

Let $\mathcal{P}$ be the property of coherent sheaves $\mathcal{F}$ on $X$ defined by the rule

We are going to apply Lemma 30.12.8. Thus we have to verify (1), (2) and (3) of that lemma for $\mathcal{P}$. Property (1) follows from the long exact cohomology sequence associated to a short exact sequence of sheaves. Property (2) follows since $H^1(X, -)$ is an additive functor. To see (3) let $Z \subset X$ be an integral closed subscheme with generic point $\xi$. Let $\mathcal{F}$ be a coherent sheaf on $Y$ such that the support of $f_*\mathcal{F}$ is equal to $Z$ and $(f_*\mathcal{F})_\xi$ is annihilated by $\mathfrak m_\xi$, see Lemma 30.13.1. We claim that taking $\mathcal{G} = f_*\mathcal{F}$ works. We only have to verify part (3)(c) of Lemma 30.12.8. Hence assume that $\mathcal{J} \subset \mathcal{O}_ X$ is a quasi-coherent sheaf of ideals such that $\mathcal{J}_\xi = \mathcal{O}_{X, \xi }$. A finite morphism is affine hence by Lemma 30.13.2 we see that $\mathcal{J}\mathcal{G} = f_*(f^{-1}\mathcal{J}\mathcal{F})$. Also, as pointed out in the proof of Lemma 30.13.2 the sheaf $f^{-1}\mathcal{J}\mathcal{F}$ is a quasi-coherent $\mathcal{O}_ Y$-module. Since $Y$ is affine we see that $H^1(Y, f^{-1}\mathcal{J}\mathcal{F}) = 0$, see Lemma 30.2.2. Since $f$ is finite, hence affine, we see that

by Lemma 30.2.4. Hence the quasi-coherent subsheaf $\mathcal{G}' = \mathcal{J}\mathcal{G}$ satisfies $\mathcal{P}$. This verifies property (3)(c) of Lemma 30.12.8 as desired. $\square$