Suppose that $R$ is a ring, $I$ and $J$ are ideals in $R,$ and $R/I\cong R/J$ as rings. When does $I\cong J$ as $R$-modules hold?
If ideal quotients of a ring are isomorphic, are these ideals isomorphic?
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abstract-algebra
ideals
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1In the case when $R$ is noetherian, and one of the ideals is contained in the other, we do have an equality. – 2015-12-31
2 Answers
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We can also consider $\left(\prod_{n=1}^\infty\mathbb{Z}\right)/\mathbb{Z}\cong\left(\prod_{n=1}^\infty\mathbb{Z}\right)/(\mathbb{Z}\times\mathbb{Z}).$ Clearly $\mathbb{Z}\not\cong\mathbb{Z}\times\mathbb{Z}$ since $\mathbb{Z}$ only has one generator while $\mathbb{Z}\times\mathbb{Z}$ has two generators.
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$\Bbb R[X,Y]/(X)\simeq\Bbb R[X,Y]/(Y)$ but $(X)\neq(Y)$. The question should be if $I$ and $J$ are isomorphic as $R$-modules.
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0Now the question is as you wished. Unfortunately, the answer has became obsolete. – 2016-05-05