Lemma 30.23.5. Let $X$ be a Noetherian scheme and let $\mathcal{I} \subset \mathcal{O}_ X$ be a quasi-coherent sheaf of ideals. Let $\mathcal{F}$, $\mathcal{G}$ be coherent $\mathcal{O}_ X$-modules. Set $\mathcal{H} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{G}, \mathcal{F})$. Then

$\mathop{\mathrm{lim}}\nolimits H^0(X, \mathcal{H}/\mathcal{I}^ n\mathcal{H}) = \mathop{\mathrm{Mor}}\nolimits _{\textit{Coh}(X, \mathcal{I})} (\mathcal{G}^\wedge , \mathcal{F}^\wedge ).$

Proof. To prove this we may work affine locally on $X$. Hence we may assume $X = \mathop{\mathrm{Spec}}(A)$ and $\mathcal{F}$, $\mathcal{G}$ given by finite $A$-module $M$ and $N$. Then $\mathcal{H}$ corresponds to the finite $A$-module $H = \mathop{\mathrm{Hom}}\nolimits _ A(M, N)$. The statement of the lemma becomes the statement

$H^\wedge = \mathop{\mathrm{Hom}}\nolimits _{A^\wedge }(M^\wedge , N^\wedge )$

via the equivalence of Lemma 30.23.1. By Algebra, Lemma 10.97.2 (used 3 times) we have

$H^\wedge = \mathop{\mathrm{Hom}}\nolimits _ A(M, N) \otimes _ A A^\wedge = \mathop{\mathrm{Hom}}\nolimits _{A^\wedge }(M \otimes _ A A^\wedge , N \otimes _ A A^\wedge ) = \mathop{\mathrm{Hom}}\nolimits _{A^\wedge }(M^\wedge , N^\wedge )$

where the second equality uses that $A^\wedge$ is flat over $A$ (see More on Algebra, Lemma 15.64.4). The lemma follows. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).