Lemma 10.51.10. Let $R$ be a ring. Let $M$ be an $R$-module. The following are equivalent:

$M$ is simple,

$\text{length}_ R(M) = 1$, and

$M \cong R/\mathfrak m$ for some maximal ideal $\mathfrak m \subset R$.

Lemma 10.51.10. Let $R$ be a ring. Let $M$ be an $R$-module. The following are equivalent:

$M$ is simple,

$\text{length}_ R(M) = 1$, and

$M \cong R/\mathfrak m$ for some maximal ideal $\mathfrak m \subset R$.

**Proof.**
Let $\mathfrak m$ be a maximal ideal of $R$. By Lemma 10.51.6 the module $R/\mathfrak m$ has length $1$. The equivalence of the first two assertions is tautological. Suppose that $M$ is simple. Choose $x \in M$, $x \not= 0$. As $M$ is simple we have $M = R \cdot x$. Let $I \subset R$ be the annihilator of $x$, i.e., $I = \{ f \in R \mid fx = 0\} $. The map $R/I \to M$, $f \bmod I \mapsto fx$ is an isomorphism, hence $R/I$ is a simple $R$-module. Since $R/I \not= 0$ we see $I \not= R$. Let $I \subset \mathfrak m$ be a maximal ideal containing $I$. If $I \not= \mathfrak m$, then $\mathfrak m /I \subset R/I$ is a nontrivial submodule contradicting the simplicity of $R/I$. Hence we see $I = \mathfrak m$ as desired.
$\square$

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