Lemma 47.5.6. Let $R$ be a ring. Let $E$ be an indecomposable injective $R$-module. Then

1. $E$ is the injective hull of any nonzero submodule of $E$,

2. the intersection of any two nonzero submodules of $E$ is nonzero,

3. $\text{End}_ R(E, E)$ is a noncommutative local ring with maximal ideal those $\varphi : E \to E$ whose kernel is nonzero, and

4. the set of zerodivisors on $E$ is a prime ideal $\mathfrak p$ of $R$ and $E$ is an injective $R_\mathfrak p$-module.

Proof. Part (1) follows from Lemma 47.5.3. Part (2) follows from part (1) and the definition of injective hulls.

Proof of (3). Set $A = \text{End}_ R(E, E)$ and $I = \{ \varphi \in A \mid \mathop{\mathrm{Ker}}(f) \not= 0\}$. The statement means that $I$ is a two sided ideal and that any $\varphi \in A$, $\varphi \not\in I$ is invertible. Suppose $\varphi$ and $\psi$ are not injective. Then $\mathop{\mathrm{Ker}}(\varphi ) \cap \mathop{\mathrm{Ker}}(\psi )$ is nonzero by (2). Hence $\varphi + \psi \in I$. It follows that $I$ is a two sided ideal. If $\varphi \in A$, $\varphi \not\in I$, then $E \cong \varphi (E) \subset E$ is an injective submodule, hence $E = \varphi (E)$ because $E$ is indecomposable.

Proof of (4). Consider the ring map $R \to A$ and let $\mathfrak p \subset R$ be the inverse image of the maximal ideal $I$. Then it is clear that $\mathfrak p$ is a prime ideal and that $R \to A$ extends to $R_\mathfrak p \to A$. Thus $E$ is an $R_\mathfrak p$-module. It follows from Lemma 47.3.3 that $E$ is injective as an $R_\mathfrak p$-module. $\square$

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