Jacobson radical annihilates simple modules

Let $R$ be a ring, let $J(R)$ be its Jacobson radical, and let $S$ be a simple right $R$ -module. Then

S J(R)=0.

Equivalent characterizations

Equivalently, every element of $J(R)$ acts as zero on $S$ , so

J(R)\subseteq \operatorname{Ann}_R(S)

for every simple right module $S$ , where the annihilator consists of the ring elements acting as zero on $S$ .

Remarks

If $R$ is commutative, every simple $R$ -module is isomorphic to $R/\mathfrak m$ for some maximal ideal $\mathfrak m$ . The theorem then gives

J(R)\subseteq \mathfrak m \quad\text{for all maximal ideals }\mathfrak m,

Examples

Local rings. If $(R,\mathfrak m)$ is a commutative local ring, then $J(R)=\mathfrak m$ , and every simple $R$ -module is isomorphic to the residue field $R/\mathfrak m$ . The ideal $\mathfrak m$ acts as zero on it.

The integers. In $R=\mathbb Z$ , $J(\mathbb Z)=0$ . The simple $\mathbb Z$ -modules are $\mathbb Z/p\mathbb Z$ , so the assertion is immediate.

Dual numbers. Let $R=k[\varepsilon]/(\varepsilon^2)$ . Then $J(R)=(\bar\varepsilon)$ , and every simple module is isomorphic to $R/(\bar\varepsilon)\cong k$ . Thus $kJ(R)=0$ .