injective hull and projective cover of simple modules are indecomposable

Question

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.

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!

Answer 1

One definition of "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 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.