Is ${\bf Z}_m$ a projective ${\bf Z}$-module?

Question

$ {\bf Z}_6 = {\bf Z}_2\oplus {\bf Z}_3$. So ${\bf Z}_2$ is a direct summand so that ${\bf Z}_2$ is a projective ${\bf Z}_6$-module.

Question : ${\bf Z}_m$ is a projective ${\bf Z}$-module ?

${\bf Z}$ has no proper direct summand. Further more

$$ (m+r)\cdot 1 = r \cdot 1 $$ That is, $ r\in {\bf Z}_m$ has no unique presentation. Hence I conclude that ${\bf Z}_m$ is not projective. Am I right ?

Answer 1

Since there is an epimorphism $\mathbf{Z}\to\mathbf{Z}_6$, this $\mathbf{Z}$-module would be projective if and only if it is a direct summand of $\mathbf{Z}$, which it isn't.

Actually, projective modules over $\mathbf{Z}$ are free. This is a consequence of the stronger fact that any subgroup of a free abelian group is again free; in particular, direct summands of free modules are free.