Every submodule of free $R$-module is free; is $R$ an integral domain?

Question

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?

Answer 1

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.

Answer 2

Alternatively, if you know that $b\neq 0$ and $ab=0$, then clearly $b\in ann(a)=ann(Ra)\neq \{0\}$.

Since every free module (of a nonzero ring with identity, of course) has annihilator $\{0\}$, $Ra$ cannot be free.