About free modules

Question

I study the properties of the free modules, but I can't understand why $\mathbb Z/n\mathbb Z$ is not a free $\mathbb Z$-module?

Thanks for any help.

Answer 1

That is because it has torsion: no congruence class $\overline a$ can be free since $\; n\,\overline{a}=\overline{0}$.

Answer 2

A free module $F$ has a subset $B$ such that

for each module $M$ and each function $f\colon B\to M$, there exists a unique homomorphism $\hat{f}\colon F\to M$ such that $\hat{f}(x)=f(x)$, for every $x\in B$.

Take a prime $p$ such that $p$ does not divide $n$. Then the only homomorphism $\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}$ is the zero constant map.

Indeed, if $g$ is such a homomorphism and $x\in \mathbb{Z}/n\mathbb{Z}$, then $nx=0$ and so $ng(x)=0$; however, also $pg(x)=0$ and, as $\gcd(n,p)=1$, we conclude that $g(x)=0$.

This implies that no nonempty subset $B$ of $\mathbb{Z}/n\mathbb{Z}$ can satisfy the property above for $M=\mathbb{Z}/p\mathbb{Z}$. Thus the only possible $B$ can be the empty set, but in this case the free module is the trivial module $\{0\}$.

Why is that? Suppose the subset $B$ is empty; if $F\ne\{0\}$, then there are at least two distinct homomorphisms $F\to F$, namely the trivial homomorphism mapping everything to $0$ and the identity map. Both extend the unique map $\emptyset\to F$.