Lemma 30.13.2. Let $f : Y \to X$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent sheaf on $Y$. Let $\mathcal{I}$ be a quasi-coherent sheaf of ideals on $X$. If the morphism $f$ is affine then $\mathcal{I}f_*\mathcal{F} = f_*(f^{-1}\mathcal{I}\mathcal{F})$.

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

$I(M_ A) = ((IB)M)_ A$

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

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