Let $R$ be a commutative ring with a multiplicative identity. If $a\in R, a \not = 0$, prove that $R/(a)$ is not a free $R$-module.
My initial attempt was to say that R/(a) is not torsion free, thus it cannot be free, but then I realized that the theorem "M is a free module if and only if it is torsion free" only works for finitely generated modules over PIDs. So I'm pretty lost as to how to prove this. Any help would be appreciated!
On
I'm sorry I'm late for this question, but since it is of my interest I would like to answer it anyway. If I understood correctly $(a)$ denote de $R$-module generated by $a\in R$, so $(a)=aR$. Suppose $[x]\in R/(a)$ is any element, so we have $[x]a=[xa]=[0]$, since $xa\in aR=(a)$. If I am not mistaken this proves that any subset of $R/(a)$ can not be linearly independent over $R$. Thus we conclude that $R/(a)$ can not be free over $R$.
Can anyone confirm if this is correct? Thanks.
You have the right idea, but you want to look at annihilators of the whole module, not of individual elements.