Let $R$ be a ring with $1$ and $M$ be a right $R$-module. A homomorphism $f:P\to M$ is said to be a projective cover of $M$ if $P$ is projective and $f$ is an epimorphism and $\ker f \ll P$. Recall that $N$ is a small submodule of $K$ (written $N \ll K$) if $K=N+L \Longrightarrow K=L$.
I need to prove that the following are equivalent:
- Every nonzero cyclic right $R$-module has a projective cover
- Every nonzero simple right $R$-module has a projective cover.
It's clear that (1) $\Longrightarrow$ (2) because every simple is cyclic. Conversely. If $M_R$ is cyclic, then $M$ contains a maximal submodule $L$ (Recall that every nonzero finitely generated module has a maximal submodule). Thus, $M/L$ is a simple right $R$-module. By the hypothesis, $(M/L)_R$ has a projective cover $f:P\to M/L$. By the projectivity of $P$, the diagram
$$\require{AMScd} \def\diaguparrow#1{\smash{ \raise.6em\rlap{\scriptstyle #1} \lower.3em{\mathord{\diagup}} \raise.52em{\!\mathord{\nearrow}} }} \begin{CD} && P\\ & @VV f V \\ M @>>\pi > M/L \end{CD}$$ can be completed by a homomorphism $\alpha:P \to M$ such that $\pi\alpha=f$. Clearly, $\ker\alpha \subseteq \ker f$. Therefore, if $P=\ker\alpha+L$ for some submodule $L\subseteq P$, then $P=\ker\alpha+L\subseteq \ker f + L$ and hence $P=\ker f + L$. Since $\ker f \ll P$, we have $L=P$. Consequently, $\ker\alpha \ll P$. But is $\alpha:P\to M$ an epimorphism?!. If it's an epimorphism, then $\alpha:P\to M$ turns to be a projective cover of $M$ and we are done.
What is another alternative way to prove the statement ?
Thanks.