Lemma 10.62.3. Suppose that $R$ is a Noetherian local ring with maximal ideal $\mathfrak m$. Let $M$ be a nonzero finite $R$-module. Then $\text{Supp}(M) = \{ \mathfrak m\} $ if and only if $M$ has finite length over $R$.

**Proof.**
Assume that $\text{Supp}(M) = \{ \mathfrak m\} $. It suffices to show that all the primes $\mathfrak p_ i$ in the filtration of Lemma 10.62.1 are the maximal ideal. This is clear by Lemma 10.62.2.

Suppose that $M$ has finite length over $R$. Then $\mathfrak m^ n M = 0$ by Lemma 10.52.4. Since some element of $\mathfrak m$ maps to a unit in $R_{\mathfrak p}$ for any prime $\mathfrak p \not= \mathfrak m$ in $R$ we see $M_{\mathfrak p} = 0$. $\square$

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