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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.

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    Just a comment: it was sort of a shame that Zev deleted his answer: it was not correct, but it was not correct in a very instructive way (and coming off of teaching a commutative algebra class, let me say that plenty of graduate students want to answer the way he did). Let me encourage him to parlay his answer into a cautionary tale, if he so chooses...2011-06-08
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    @Pete: You're right, it should be an educational example. I've made it CW to avoid getting any points though. I also clearly need to get some sleep...2011-06-08
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    The essential problem is one of notation: we allow ourselves to talk about isomorphisms without specifying the category (usually this would be a waste of time but sometimes, as here, it is essential).2011-06-08

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