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> Let $A$ be an Artinian algebra. Let $S$ be a simple module over $A$. Let $\pi: S \rightarrow I$ be the injective hull and $\tau: P \rightarrow S$ be the projective cover of $S$. Then $I$ and $P$ must be indecomposable.
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This statement was merely implied in a paper I was reading. After thinking about it I don't find it clear at all.
I know that this statement must be true if $A$ was a finite dimension algebra but this is not the case.
Does anyone know why this holds true? Any help is highly appreciated!



injective hull and projective cover of simple modules are indecomposable
One [definition of "projective cover"](https://en.wikipedia.org/wiki/Projective_cover) of $S$ is that it is a projective module $P$, together with an epimorphism $\phi\colon P\to S$ such that the kernel $K$ is a [superfluous submodule](https://en.wikipedia.org/wiki/Essential_extension) of $P$, meaning that for any submodule $H$ of $P$, $K+H=P$ implies $H=P$. Suppose that $P$ is a direct sum $P=M+N$. Consider the restriction of $\phi$ to $M$. Since $S$ is simple, this restriction is either $0$ or an epimorphism. If it is $0$, then $K$ contains $M$, and therefore is not superfluous, since then $K+N=P$, while $N\neq P$. If, on the other hand, the restriction of $\phi$ is an epimorphism, then it is easy to see that $K+M=P$. Indeed, for every $p\in P$, there then exists an $m\in M$ such that $\phi(p)=\phi(m)$, so that $p-m\in K$. Thus, in this case again $K$ is not superfluous, since $K+M=P$, but $M\neq P$.


As you see, finite-dimensionality plays no role in this argument. The argument for injective hull is "dual" to this.


