The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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

  1. $M$ is simple,

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

  3. $M \cong R/\mathfrak m$ 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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