# Isomorphism between quotient modules [closed]

Is it true for a commutative ring $R$ and its ideals $I$ and $J$ that if the quotient $R$-modules $R/I$ and $R/J$ are isomorphic then $I=J$?

-

## closed as off-topic by user26857, choco_addicted, qwr, USER91500, gebruikerJun 11 at 8:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, choco_addicted, qwr, USER91500, gebruiker
If this question can be reworded to fit the rules in the help center, please edit the question.

No, just take a polynomial ring $k[X]$ for some field $k$ (which is a commutative domain), and consider the ideal generated by $X$ and the ideal generated by $X-1$. The quotient rings are both (isomorphic to) $k$.

This answer asserts that $R/I \simeq R/J$ (as rings) does not imply that $I = J$. However, the OP asks about $R/I \simeq R/J$ (as $R$-modules), which is covered in the other answer.

-
I don't understand why the OP has accepted this answer, and also why this answer treats the quotients $R/I$ and $R/J$ as being isomorphic as rings when the question says clearly: "the quotient $R$-modules $R/I$ and $R/J$ are isomorphic". – user26857 Apr 2 at 21:36
Yup. For example the polynomial $X$ acts as identity on one quotient module and as zero on the other. – Jyrki Lahtonen Apr 11 at 15:04

Problem 22 of Chapter 4 of Steven Roman's Advanced Linear Algebra asks to prove this question in the affirmative when $R/I\simeq R/J$ are isomorphic as $R$-modules, and then asks about the case when $R/I\simeq R/J$ as rings.

The nice existing answers show it is not necessarily true that $I=J$ when the quotients are isomorphic as rings. However, suppose $R/I\simeq R/J$ as $R$-modules with the standard $R$-module structure.

Let $\tau\colon R/I\to R/J$ be an $R$-module isomorphism. Then for any $j\in J$, $$\tau(j+I)=\tau(j\cdot 1+I)=j\cdot\tau(1+I)=0+J$$ since $\tau(1+I)\in R/J$, and any element of $R/J$ is annihilated by multiplication by elements of $J$. Thus $j+I\in\ker\tau=\{I\}$, the equality following since $\tau$ is injective. So $j+I=I$, thus $j\in I$, and $J\subseteq I$. The reverse containment follows similarly, by looking at $\tau^{-1}$ say. So $I=J$.

-
In fact isomorphic modules have equals annihilators. (+1) – user26857 Apr 2 at 21:39