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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... –  Pete L. Clark Jun 8 '11 at 14:56
    
@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... –  Zev Chonoles Jun 8 '11 at 15:18
    
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). –  Qiaochu Yuan Jun 8 '11 at 18:37
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4 Answers

Here is my original answer:

No; let $R=k[x_1,x_2,\ldots]$ for any field $k$ (a polynomial ring in infinitely many variables). Do you see a non-trivial ideal $J\subset R$ such that $R/J\cong R$? (There are a lot).

What I was aiming at was that, for any infinite set $S\subseteq\mathbb{N}$, choosing $J=(\{x_i\mid i\notin S\})$ gives a ring isomorphism $R/J\cong R$. The problem with my answer was that this is not the same as an isomorphism of $R$-modules. The definition of $R$-module is an abelian group that is acted on by $R$ (by scalar multiplication); the fact that $R/J\cong R$ as rings includes the fact that they are isomorphic as abelian groups (under addition), which is part of what is necessary for an isomorphism of $R$-modules, but as all the other (correct) answers point out, the essential problem lies in the scalar multiplication aspect: $J$ annihilates $R/J$ (i.e., scaling $R/J$ by any element of $J$ gives the zero map), while $R^n$ has trivial annihilator, so $J$ must be trivial.

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You have an epi $f:R\to R^n$. Tensoring it with $k=R/\mathfrak m$ for some maximal ideal $\mathfrak m\subset R$, we get an epi $k\to k^n$, so $n$ must be equal to $1$. Now the short exact sequence $$0\to J\to R\xrightarrow{\;f\;} R\to 0$$ must split, because the rightmost $R$ is projective.

Can you see how to finish this?

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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.

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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 $j.r = 0$ holds for $j \neq 0$, which can't happen if $R/J$ was free.

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That is precisely what Pete wrote, no? –  Mariano Suárez-Alvarez Jun 8 '11 at 15:01
    
Well, the words are different. :) Also what I wrote was (intentionally) slightly oblique, and this explains things more plainly. –  Pete L. Clark Jun 8 '11 at 15:05
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Sorry, I think I was writing up my answer before yours was posted... –  qwert Jun 8 '11 at 15:07
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You have my sympathy, qwert! It has happened to me quite a few times, because I am rather slow and do a lot of checking before posting. I know too well the feeling of slight embarrassment of discovering one second after having posted that the key idea is already there in another answer! –  Georges Elencwajg Jun 8 '11 at 21:00
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