$M=\mathbb{Z}/2018\mathbb{Z}$ is not a projective $\mathbb{Z}$-module

Question

$M=\mathbb{Z}/2018\mathbb{Z}$ is not a projective $\mathbb{Z}$-module.

I know that $\mathbb{Z}$ is a PID and therefore any projective $\mathbb{Z}$-module is also free. $M$ is not a free $\mathbb{Z}$-module, which proves the statement.

Now, I want to show this differently. Consider the natural projection $\pi: \mathbb{Z}\to M$. Claim: there is no $\sigma: M\to\mathbb{Z}$ such that $\pi\circ\sigma=\operatorname{id}_M $.

Do I have to specify $\sigma$? I feel like I should somehow find a nonzero element that is mapped to zero (or the other way around). How can I approach this?

Thanks.

Answer 1

For any homomorphism $\sigma: M\to \mathbb Z$, you would have to have $2018\sigma(x)=\sigma(2018x)=0$ for every $x\in M$. But if $0\neq \sigma(x)\in \mathbb Z$, $2018\sigma(x)\neq 0$ either.

So in fact, $\sigma=0$ in every case.

Answer 2

Projective $\mathbb Z$-modules are free, and these are never finite sets except for the trivial free module. Your module is finite and non-trivial, so it cannot be free.

Answer 3

Hint: if $M/N$ is projective, then $N$ is a direct summand of $M$.

Since $\mathbb{Z}$ has no nontrivial direct summands…