I need to prove that $\mathbb Z_p$ has no projective cover when seen as a module over $\mathbb Z$. I would appreciate your help.
A projective cover of $M$ is a projective module $P$ along with a surjective morphism $f:P\rightarrow M$ so that the kernel of $f$ is superfluous in $P$.
I'm not sure how to proceed. Since $\mathbb Z$ is PID the projective cover would have to be free, so of form $\bigoplus\limits_{i\in I}\mathbb Z$. And I think that the only options for the kernel are $p\mathbb Z+\bigoplus\limits_{i\in I,i\neq j}\mathbb Z$. Which is clearly not superfluous in the whole free module.
I think this "proof" can be vastly improved (even if we assume it is correct).
Let $P$ be a projective cover of $\mathbb Z/n\mathbb Z$, $n\ge2$, and $f:P\to\mathbb Z/n\mathbb Z$ a surjective homomorphism of $\mathbb Z$-modules. Then $P/\ker f\simeq \mathbb Z/n\mathbb Z$. But $\ker f$ is superfluous in $P$, so for any other submodule $H\subset P$ from $\ker f+H=P$ we get $H=P$. It follows $\ker f\subseteq J(P)=PJ(\mathbb Z)=0$. ($J(P)$ denotes the Jacobson radical of $P$.) Thus $f$ is an isomorphism, so $\mathbb Z/n\mathbb Z$ is a free $\mathbb Z$-module, a contradiction.