Lemma 57.12.2. Let $k$ be a field. Let $X$ be a scheme of finite type over $k$ which is regular. Let $x \in X$ be a closed point. For a coherent $\mathcal{O}_ X$-module $\mathcal{F}$ supported at $x$ choose a coherent $\mathcal{O}_ X$-module $\mathcal{F}'$ supported at $x$ such that $\mathcal{F}_ x$ and $\mathcal{F}'_ x$ are Matlis dual. Then there is an isomorphism

$\mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, M) = H^0(X, M \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{F}'[-d_ x])$

where $d_ x = \dim (\mathcal{O}_{X, x})$ functorial in $M$ in $D_{perf}(\mathcal{O}_ X)$.

Proof. Since $\mathcal{F}$ is supported at $x$ we have

$\mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, M) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X, x}}(\mathcal{F}_ x, M_ x)$

and similarly we have

$H^0(X, M \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{F}'[-d_ x]) = \text{Tor}^{\mathcal{O}_{X, x}}_{d_ x}(M_ x, \mathcal{F}'_ x)$

Thus it suffices to show that given a Noetherian regular local ring $A$ of dimension $d$ and a finite length $A$-module $N$, if $N'$ is the Matlis dual to $N$, then there exists a functorial isomorphism

$\mathop{\mathrm{Hom}}\nolimits _ A(N, K) = \text{Tor}^ A_ d(K, N')$

for $K$ in $D_{perf}(A)$. We can write the left hand side as $H^0(R\mathop{\mathrm{Hom}}\nolimits _ A(N, A) \otimes _ A^\mathbf {L} K)$ by More on Algebra, Lemma 15.74.15 and the fact that $N$ determines a perfect object of $D(A)$. Hence the formula holds because

$R\mathop{\mathrm{Hom}}\nolimits _ A(N, A) = R\mathop{\mathrm{Hom}}\nolimits _ A(N, A[d])[-d] = N'[-d]$

by Dualizing Complexes, Lemma 47.16.4 and the fact that $A[d]$ is a normalized dualizing complex over $A$ ($A$ is Gorenstein by Dualizing Complexes, Lemma 47.21.3). $\square$

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).