Lemma 10.102.3. In Situation 10.102.1. Suppose $R$ is a local Noetherian ring with maximal ideal $\mathfrak m$. Assume $\mathfrak m \in \text{Ass}(R)$, in other words $R$ has depth $0$. Suppose that $0 \to R^{n_ e} \to R^{n_{e-1}} \to \ldots \to R^{n_0}$ is exact at $R^{n_ e}, \ldots , R^{n_1}$. Then the complex is isomorphic to a direct sum of trivial complexes.
Proof. Pick $x \in R$, $x \not= 0$, with $\mathfrak m x = 0$. Let $i$ be the biggest index such that $n_ i > 0$. If $i = 0$, then the statement is true. If $i > 0$ denote $f_1$ the first basis vector of $R^{n_ i}$. Since $xf_1$ is not mapped to zero by exactness of the complex we deduce that some matrix coefficient of the map $R^{n_ i} \to R^{n_{i - 1}}$ is not in $\mathfrak m$. Lemma 10.102.2 then allows us to decrease $n_ e + \ldots + n_1$. Induction finishes the proof. $\square$
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