$Z\pi \cong Z\oplus Z$ and so $P \cong Z$

Question

If $\pi = Z$, then the augmentation ideal $P$ is projective and $0 \to P \to Z\pi \to Z \to 0$ is a projective resolution.

Here we know that $P$ has basis $\mathbb{Z}-$ ${0}$ Then how to show that it is projective?

Any help would be appreciated!

Answer 1

You can try to do the following exercise : as a $\mathbb Z\pi$-module, $P$ is freely generated by $g-1$, where $g$ is a generator for $\pi$.

What's true regardless of the group $\pi$ is that for any system of generators $g_1,...,g_n$ of $\pi$, $g_1-1,...,g_n-1$ generate $P$, you can try to prove that as well.

The point will then be that in this specific case, $g-1$ is a free generator of the ideal

(Maybe it's easier to see if you know that $\mathbb Z\pi \cong \mathbb Z[t,t^{-1}]$ )