Why is $\mathbb{Z}_{6}$ a free $\mathbb{Z}_{6}$ module?

Question

$\mathbb{Z}_{6}=\{\bar{0},\bar{1},\bar{2},\bar{3},\bar{4},\bar{5}\}$. I know that in order for a module to be free, it has to have a basis. Further, I know that if $b_{1},...,b_{n}$ is a basis for $\mathbb{Z}_{6}$, then $r_{1}b_{1}+...+r_{n}b_{n}=0$, $r_{1}=...=r_{n}=0$. I do not know how I can use this to show that $\mathbb{Z}_{6}$ is a free $\mathbb{Z}_{6}$-module.

Answer 1

Observe that $\;\{\overline1\}\;$ , or $\;\{\overline 5\}\;$, are free basis of $\;\Bbb Z_6\;$ over itself, because $\;m\cdot\overline 1=0\iff m=0\pmod6\;$ (and the same is true with $\;\overline5\;$)

This is the reason $\;\Bbb Z_6\;$ is a free module of rank $\;1\;$ over itself.

Answer 2

Indeed, a commutative unital ring $R$ is always a free $R$-module of rank one because (as an $R$-module) we have that $R = R \langle 1_R \rangle,$ i.e., every element of $R$ can be written as a linear combination of $1_R$ — namely, for every $r$ in $R,$ we have that $r = r \cdot 1_R.$