Lemma 10.52.10. 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 10.52.10. 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 10.52.6 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.
$\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #4887 by Peng DU on

Comment #5164 by Johan on

There are also: