TITLE: Every module over a field is free. Is every commutative ring whose modules are all free a field?
QUESTION [10 upvotes]: Let $A$ denote a commutative ring. Then if $A$ is a field, we may deduce that every $A$-module is free. Does the converse hold? i.e. If every $A$-module is free, can we deduce that $A$ is a field?

REPLY [3 votes]: Assume that $R$ is a commutative ring with $1$ whose every module is free. Let $I\subseteq R$ be an ideal. Note that $R/I$ is an $R$-module and therefore free. If $I\neq\{0\}$, then every element of $R/I$ is a torsion element (since we can act by any non-zero element of I to obtain $0$ in $R/I$). Since $R/I$ is also free, $R/I=\{0\}$. Therefore, $I=R$. So, the only ideals are $\{0\}$ and $R$, meaning $R$ is a field.