Remark 51.22.5. Given a pair $(M, n)$ consisting of an integer $n \geq 0$ and a finite $A/I^ n$-module $M$ we set $M^\vee = \mathop{\mathrm{Hom}}\nolimits _{A/I^ n}(M, A/I^ n)$. Given a pair $(\mathcal{F}, n)$ consisting of an integer $n$ and a coherent $\mathcal{O}_{Y_ n}$-module $\mathcal{F}$ we set

Given $(M, n)$ as above, there is a canonical map

Namely, if we choose a presentation $(A/I^ n)^{\oplus s} \to (A/I^ n)^{\oplus r} \to M \to 0$ then we obtain a presentation $\mathcal{O}_{Y_ n}^{\oplus s} \to \mathcal{O}_{Y_ n}^{\oplus r} \to p^*M \to 0$. Taking duals we obtain exact sequences

and

Pulling back the first sequence by $p$ we find the desired map $can$. The construction of this map is functorial in the finite $A/I^ n$-module $M$. The kernel and cokernel of $can$ are scheme theoretically supported on $Y_ c$ if $M$ is an $(A, n, c)$-module. Namely, in that case for $a \in I^ c$ the map $a : M \to M$ factors through a finite free $A/I^ n$-module for which $can$ is an isomorphism. Hence $a$ annihilates the kernel and cokernel of $can$.

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