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 $J_1 \cong J_2$?
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

closed as off topic by Mariano SuárezAlvarez♦, Mark Sapir, Bruce Westbury, Qiaochu Yuan, Andreas Blass Jul 14 '11 at 18:14
Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.
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. 

