Quotient of free module

Let $R$ be a commutative ring with $1$ and let $J$ be a proper ideal of $R$ such that $R/J \cong R^n$ as $R$-modules where $n$ is some natural number. Does this imply that $J$ is the trivial ideal?

Basically I am trying to prove/disprove that if $J$ is a proper ideal of $R$ and $R/J$ is free then $J=0$ and above is my work.

Solution 1:

Correct me if I'm wrong, but isn't it obvious that if $J \neq 0$, $R/J$ can't be a free $R$-mod because anything in $J$ acts by $0$ on $R/J$? Therefore an equation like $jr = 0$ holds for $j \neq 0$, which can't happen if $R/J$ was free.

Solution 2:

Yes. A nice way to see this is via the annihilator $\operatorname{ann}(M)$ of a module $M$: it is the set of all $x \in R$ such that $xm = 0$ for all $m \in M$. One shows immediately that $\operatorname{ann}(M)$ is an ideal of $R$ and that isomorphic modules have equal annihilators.

If you take annihilators of both sides of your isomorphism $R/J \cong R^n$, you'll get the desired conclusion. I could say more, but I'll leave it up to you for now because this is a very important and enlightening exercise.