# Proving $IH^n(X,\mathcal{F}/I\mathcal{F})=0$

Let $$X\rightarrow \operatorname{Spec}A$$ be a morphism of schemes, $$I$$ an ideal of $$A$$ and $$\mathcal{F}$$ an $$O_X-$$module. The morphism gives a morphism $$A\rightarrow O_X(X)$$ and so for any open subset $$U\subset X$$, we've got a morphism $$A\rightarrow O_X(U)$$ and so $$\mathcal{F}(U)$$ can be equipped with an $$A-$$module structure and so we can consider the $$O_X-$$module $$I\mathcal{F}:U\mapsto I\mathcal{F}(U)$$ which is a submodule of $$\mathcal{F}$$.
I want to show that $$IH^n(X,\mathcal{F}/I\mathcal{F})$$ where $$H^n$$ denotes sheaf cohomology. I'd like to check whether my answer is correct, and I have a feeling I'm complicating things so if anyone has an easier solution I would appreciate it!

For any $$x\in X$$, $$(\mathcal{F}/I\mathcal{F})_x$$ is an $$O_{X,x}-$$module and as an $$A-$$module it is annihilated by $$I$$ so $$(\mathcal{F}/I\mathcal{F})_x$$ is an $$A/I-$$module.
Let $$(\mathcal{F}/I\mathcal{F})_x\rightarrow J_x$$ be a monomorphism to an injective $$A/I-$$module. Now we consider the sheaf of abelian groups $$\mathcal{J}$$ defined by $$\mathcal{J}(U)=\prod_{x\in U}J_x.$$ We can check that this is an injective sheaf of abelian groups and that the canonical morphism $$\mathcal{F}/I\mathcal{F}\rightarrow \mathcal{J}$$ is injective. Each $$\mathcal{J}(U)$$ is an $$A/I-$$module so we can construct an injective resolution $$(\mathcal{J}^k)$$ of $$\mathcal{F}/I\mathcal{F}$$ in the category of sheaves of abelian groups such that $$\mathcal{J}^k(U)$$ is an $$A/I-$$module for all $$k$$ and $$U\subset X.$$ So the cohomology $$H^n(X,\mathcal{F}/I\mathcal{F})$$ of the complex $$(\mathcal{J}^k(X))$$ is an $$A/I-$$module so $$IH^n(X,\mathcal{F}/I\mathcal{F})=0.$$

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This might be equivalent to what you did if you go through all the proofs (and indeed you have understood the point, which is that an $$A/I$$-module is just an $$A$$-module killed by $$I$$). But it might still be helpful for you. Let us write down the setup in fancier symbols. We have a morphism of schemes $$f: X \to \mathrm{Spec} A$$, and a quasicoherent sheaf $$\mathcal{F}$$ on $$X$$. We also have a closed embedding $$\iota: \mathrm{Spec}(A/I) \to \mathrm{Spec} A$$. Let $$\iota': X \times_A A/I \to X$$ be the closed embedding induced by base-change to $$X$$ via $$f$$ (this is basically the preimage in $$X$$ of the closed subscheme cut out by $$I$$). Note that $$\iota'$$ is an affine morphism. The definition of the sheaf $$\mathcal{F}/I\mathcal{F}$$ considered as a quasicoherent sheaf on $$X$$ is $$\mathcal{F}/I\mathcal{F} := (\iota')_*(\iota')^*\mathcal{F}$$ (the pullback quotients by $$I$$ and the pushforward restricts scalars from $$A/I$$ to $$A$$ in order to consider what is naturally a sheaf on the closed subscheme $$X \times_A A/I$$ as a sheaf on all of $$X$$). Now we may apply the fact that $$H^n(X, \mathcal{F}/I\mathcal{F}) = H^n(X, (\iota')_*(\iota')^*\mathcal{F}) \cong H^n(X \times_A A/I, (\iota')^*\mathcal{F})$$ as $$A$$-modules (use e.g. Vakil 18.1(v)). In particular, the guy on the right is an $$A/I$$-module (which is where the $$A$$-module structure comes from), so we conclude that the $$A$$-module on the left is killed by $$I$$, as desired.