TITLE: Quotient of free module
QUESTION [10 upvotes]: 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.

REPLY [11 votes]: 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.