I didn't get the argument of Example 7.3.11, page 123, from the representation theory book of Peter Webb, available also online at:


<http://www-users.math.umn.edu/~webb/RepBook/RepBookLatex.pdf>


In this example it is shown that the trivial module $\mathbb Z$ does not have a projective cover over $ZG$-mod where $G$ is the cyclic group with two elements.


I have understood the first part where it is shown that the augmentation map $\epsilon : \mathbb ZG \rightarrow \mathbb Z$ is not an essential epimorfism (i.e., the kernel is not superfluous.)


Then indeed, it follows from a previous proposition from the same section that if it were a projective cover for $\mathbb Z$ then that projective cover should be a summand of the regular module $\mathbb ZG$. Thus one can write $$\mathbb ZG =P\oplus Q$$ and $P \rightarrow \mathbb Z$ surjects onto $\mathbb Z$ by the augmentation map.


My question is what is exactly is meant at the end by reduction modulo $2$? Indeed one has a surjective algebra homomorphism $$\pi: \mathbb ZG \rightarrow \mathbb F\_2G$$ which gives a decomposition of left ideals 


$$\mathbb F\_2G=\pi (P) + \pi(Q)$$ I also know that $\mathbb F\_2G$ is an indecomposable module over itself. But in order to get a contradiction one has to prove that the above sum is direct. The question I have is why the above sum should be a direct sum and why both terms $\pi(P)$ and $\pi(Q)$ are not zero?



Non-existence of projective covers
The mod-2 reduction is a functor from $\mathbb{Z}G$-modules to $\mathbb{F}\_2G$-modules that preserves direct sums , because for instance it can be realized as tensoring with $\mathbb{F}\_2$. Also if $M$ has $\mathbb{Z}$-rank $r$ then its mod-2 reduction has dimension at least $r$ as an $\mathbb{F}\_2$-vector space. In particular, this means a nonzero projective $\mathbb{Z}G$-module, being necessarily $\mathbb{Z}$-free, is sent to a nonzero $\mathbb{F}\_2 G$-module.


