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}}(\varphi ) \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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