Lemma 42.68.44. Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak m$. Let $M$ be a finite $R$-module, and let $\psi : M \to M$ be an $R$-module map. Assume that

1. $\mathop{\mathrm{Ker}}(\psi )$ and $\mathop{\mathrm{Coker}}(\psi )$ have finite length, and

2. $\dim (\text{Supp}(M)) \leq 1$.

Write $\text{Supp}(M) = \{ \mathfrak m, \mathfrak q_1, \ldots , \mathfrak q_ t\}$ and denote $f_ i \in \kappa (\mathfrak q_ i)^*$ the element such that $\det _{\kappa (\mathfrak q_ i)}(\psi _{\mathfrak q_ i}) : \det _{\kappa (\mathfrak q_ i)}(M_{\mathfrak q_ i}) \to \det _{\kappa (\mathfrak q_ i)}(M_{\mathfrak q_ i})$ is multiplication by $f_ i$. Then we have

$\text{length}_ R(\mathop{\mathrm{Coker}}(\psi )) - \text{length}_ R(\mathop{\mathrm{Ker}}(\psi )) = \sum \nolimits _{i = 1, \ldots , t} \text{ord}_{R/\mathfrak q_ i}(f_ i).$

Proof. Recall that $H^0(M, 0, \psi ) = \mathop{\mathrm{Coker}}(\psi )$ and $H^1(M, 0, \psi ) = \mathop{\mathrm{Ker}}(\psi )$, see remarks above Definition 42.2.2. The lemma follows by combining Proposition 42.68.43 with Lemma 42.68.17.

Alternative proof. Reduce to the case $\text{Supp}(M) = \{ \mathfrak m, \mathfrak q\}$ as in the proof of Proposition 42.68.43. Then directly combine Lemmas 42.68.41 and 42.68.42 to prove this specific case of Proposition 42.68.43. There is much less bookkeeping in this case, and the reader is encouraged to work this out. Details omitted. $\square$

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