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Suppose $J_1$ and $J_2$ are two ideals in a ring both containing another ideal $I$. If $J_1/I \cong J_2/I$ then is $J_1 \cong J_2$?

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 Reading the FAQ will explain why this question is not really well-suited to this site (in short, it is not a question about "research math"). The FAQ also suggests other places, like math.stackexchange.com, where your question will be much more at home. Good luck! – Mariano Suárez-Alvarez Jul 14 2011 at 16:36
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 What notion of isomorphism of ideals is everyone implicitly working with here? – Qiaochu Yuan Jul 14 2011 at 18:03
@Qiaochu: I had a similar question, as I think of ideals as things that can be equal to each other, not isomorphic. One possibility is to use "isomorphic as $R$-modules"; this is Neil's interpretation below. But an ideal is really an $R$-module with a monomorphism to the rank-$1$ free module, and the category thereof is a poset. – Theo Johnson-Freyd Jul 14 2011 at 18:28
@Theo: the category thereof is that if you want it to be. In some conexts, you don't, though. For example, the ideal class group of a Dedekind domain is usefully seen as the set of isomorphism classes of ideals (as modules). Fractional ideals are, from that point of view, a kludge one uses to contruct the operation in the group. – Mariano Suárez-Alvarez Jul 14 2011 at 23:47

closed as off topic by Mariano Suárez-Alvarez, Mark Sapir, Bruce Westbury, Qiaochu Yuan, Andreas Blass Jul 14 2011 at 18:14

1 Answer

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Take $R=\mathbb{Z}[x,y]/(x^2,xy,y^2,4x,4y)$ and $J_1=(x)$ and $J_2=(2x,2y)$ and $I=(2x)$. Then $J_1/I\simeq J_2/I\simeq R/(2,x,y)$ as $R$-modules, but $2J_2=0$ and $2J_1\neq 0$, so $J_1\not\simeq J_2$.

On the other hand, if $R$ is a principal ideal domain, then the answer to your question is positive.

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