**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.
$\square$

## Comments (0)

There are also: