A ring $R$ is left-perfect if every left $R$-module $M$ admits a projective cover, i.e., an epimorphism $\varphi:P\to M$ with $P$ projective, such that $\ker \varphi\subset P$ is superfluous; i.e. for every submodule $N\subset P$, we have $\ker \varphi + N=P \Rightarrow N=P$.
There are some equivalences with this condition which are very interesting. However, I would like to stick with the formulation above.
Why is $\mathbb{Z}$ not a perfect ring?
Here's my attempt. Let's consider $\mathbb{Z}_2=\mathbb{Z}/2\mathbb{Z}$. Over $\mathbb{Z}$, projectives are free; let's then consider an epimorphism $\varphi:F\to \mathbb{Z}_2$ with $F$ a free abelian group.
Surjectivity guarantees the existence of $x\in F$ such that $\varphi(x)=1$. Then, $\varphi(3x)=1$ too, and in fact $F=\ker \varphi + \langle 3x\rangle$. Indeed, if $u\in F$, then:
- If $\varphi(u)=0$, then of course $u\in \ker \varphi$ and it's done.
- If $\varphi(u)=1$, then $u=(u-3x)+3x$ with $3x\in \langle 3x\rangle$ and $u-3x\in \ker \varphi$ wince $\varphi(u-3x)=\varphi(u)-3\varphi(x)=1-1=0$.
If $\ker \varphi$ were superfluous in $F$, it would mean $F=\langle 3x\rangle$.
So...?