Lemma 42.3.4. Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $(M, \varphi , \psi )$ be a $(2, 1)$-periodic complex over $R$ with $M$ finite and with cohomology groups of finite length over $R$. Let $x \in R$ be such that $\dim (\text{Supp}(M/xM)) \leq 0$. Then
and
Proof. We will only prove the first formula as the second is proved in exactly the same manner. Let $M' = M[x^\infty ]$ be the $x$-power torsion submodule of $M$. Consider the short exact sequence $0 \to M' \to M \to M'' \to 0$. Then $M''$ is $x$-power torsion free (More on Algebra, Lemma 15.88.4). Since $\varphi $, $\psi $ map $M'$ into $M'$ we obtain a short exact sequence
of $(2, 1)$-periodic complexes. Also, we get a short exact sequence $0 \to M' \cap \mathop{\mathrm{Im}}(\varphi ) \to \mathop{\mathrm{Im}}(\varphi ) \to \mathop{\mathrm{Im}}(\varphi '') \to 0$. We have $e_ R(M', \varphi , \psi ) = e_ R(M', x\varphi , \psi ) = e_ R(M' \cap \mathop{\mathrm{Im}}(\varphi ), 0, x) = 0$ by Lemma 42.2.5. By additivity (Lemma 42.2.3) we see that it suffices to prove the lemma for $(M'', \varphi '', \psi '')$. This reduces us to the case discussed in the next paragraph.
Assume $x : M \to M$ is injective. In this case $\mathop{\mathrm{Ker}}(x\varphi ) = \mathop{\mathrm{Ker}}(\varphi )$. On the other hand we have a short exact sequence
This together with (42.2.2.1) proves the formula. $\square$
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