Let $R$ be a commutative ring with identity such that every submodule of every free $R$-module is free. As part of an exercise, I'm trying to show that $R$ is an integral domain.
Aiming for a contradiction, I tried looking at the submodule $Ra$, where $a$ is a zero divisor, but couldn't get anywhere. Is that a good approach, or should I be looking elsewhere?
Best Answer
One definition of a free module is "any module which has a basis". But the zero-divisor module you constructed cannot have a basis: if $\{e\}$ were a basis then $e$ would be a multiple of $a$ and therefore a zero divisor, so that $ed=0$ for some nonzero $d\in R$. This means precisely that the set $\{e\}$ is not $R$-linearly independent, so that it cannot be a basis.