Projective cover of modules

Question

Give me a hand please, how can i prove this sentence.

"Show $\mathbb{Z}_{2}$ doesn't have projective cover as $\mathbb{Z}$-module."

Answer 1

I think it's helpful to use the fundamental lemma for projective covers.

Suppose $p\colon P\to\mathbb{Z}_2$ is a projective cover. There is also the canonical surjection $\pi:\mathbb{Z}\to\mathbb{Z}_2$ given by reduction mod $2$. Since $\mathbb{Z}$ is projective, the Fundamental Lemma of Projective Covers gives that there exists a decomposition $\mathbb{Z}=P'\oplus P''$, where $P\simeq P'$, $\pi(P'')=0$, and the restriction $\pi|_{P'}\colon P'\to\mathbb{Z}_2$ is a projective cover.

But $\mathbb{Z}$ is indecomposable, so in any case $\mathbb{Z}=P'$. Hence one would get $\pi\colon\mathbb{Z}\to\mathbb{Z}_2$ is a projective cover, but this is a contradiction since $\ker\pi=2\mathbb{Z}$ is not superfluous in $\mathbb{Z}$, since $2\mathbb{Z}+3\mathbb{Z}=\mathbb{Z}$, for instance.