# Why $\mathbb Z/ 2 \mathbb Z$ is not a free module?

I am reading some abstract algebra notes about free modules. It says that not all modules are free and the example to illustrate this is $\mathbb Z/ 2\mathbb Z$ (as a $\mathbb Z$-module) is not a free module. How can I show this? More concretely, how can I prove this module has no basis?

• Do you know the category theory formulation in terms of maps? – Henno Brandsma Oct 15 '14 at 4:15
• It doesn't ring a bell that name, but maybe I've read about it with other terminology, I haven't learned category theory (yet). – user16924 Oct 15 '14 at 4:20

Clearly, the only candidate for a basis is $\{1\}$. However, is $\{1\}$ a linearly independent set? That is, is it true that for all $n \in \Bbb Z$, $n\cdot 1 = 0 \iff n = 0$?
• Why this is false? Only $n$ is $0$? – Ninja Jan 3 '17 at 10:26
• @Ninja note that $2 \cdot 1 = 0$ in $\Bbb Z_2$ – Omnomnomnom Jan 3 '17 at 14:22
• Oh, it seems like you forget the $2$ – Ninja Jan 3 '17 at 15:40
I'll show $\mathbb Z_n$ is not projective for every $n>1$ (hence it can't be free for every free module is projective).
Let $\jmath:n\mathbb Z\longrightarrow \mathbb Z$ be the inclusion and let $p:\mathbb Z\longrightarrow \mathbb Z_n$ be the canonical projection . Then one gets a short exact sequence of $\mathbb Z$-modules: $$0\longrightarrow n\mathbb Z\stackrel{\jmath}{\longrightarrow} \mathbb Z\stackrel{p}{\longrightarrow}\mathbb Z_n\longrightarrow 0$$ If $\mathbb Z_n$ were projective this sequence would split hence there would be a $\mathbb Z$-linear map $q:\mathbb Z_n\longrightarrow \mathbb Z$ such that $pq=1_{\mathbb Z_n}$. But $\textrm{Hom}_{\mathbb Z}(\mathbb Z_n, \mathbb Z)=\{0\}$ (to see this it suffices to apply the correspondence theorem for modules) so that $1_{\mathbb Z_n}=0$, a contradiction.