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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?

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  • $\begingroup$ 7.3.11 is on page 116. The answer is perfect. $\endgroup$ – მამუკა ჯიბლაძე Jan 28 '17 at 18:50
  • $\begingroup$ Sorry, it was the page of my pdf file. $\endgroup$ – muser17 Jan 28 '17 at 20:51
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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, it preserves the rank over the coefficient ring, that is, if $M$ has $\mathbb{Z}$-rank 3 then its mod-2 reduction has dimension 3 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.

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  • $\begingroup$ Thank you very much for your answer! Indeed, I didn't realize that is just tensoring with $\mathbb F_2$ and therefore preserves direct sums. $\endgroup$ – muser17 Jan 28 '17 at 20:43

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