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Do we have $(R/I)/(J/I) \cong (R/J)/(I/J)$ where $R$ is a ring and $I,J$ are ideals?

If $I \subset J$ then it follows from the third isomorphism theorem.

In $R= \mathbb Z$ with $I = 3 \mathbb Z$ and $J = 5 \mathbb Z$ we have $I/J = \{ \bar{0}, \bar{3}, \bar{6}, \bar{9}, \bar{12} \} \cong R/J$ and $J/I = \{ \bar{0}, \bar{5}, \bar{10} \} \cong R/I$. Then $(R/I) / (J/I) = 0 = (R/J) / (I/J)$ so it seems to hold also.

I assume it doesn't hold in general. Or does it? Does it hold for principal ideal domains? Thanks.

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You can only speak of $I/J$ if $J\subseteq I$, so your question only makes sense if both $I\subseteq J$ and $J\subseteq I$, or equivalently $I=J$. –  Zev Chonoles Jun 28 '12 at 20:58
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In your example, neither $I/J$ nor $J/I$ make sense, because neither contains the other. –  Arturo Magidin Jun 28 '12 at 20:59
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@ZevChonoles: I guess he means $I+J / J$ and $J+I/I$. –  Michalis Jun 28 '12 at 20:59
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Right, but it's probably good to make that point clear before tackling this question. –  Dylan Moreland Jun 28 '12 at 21:02
    
Note. Even assuming that it all makes sense, $J/I\cong R/I$ does not imply $(R/I)/(J/I) = 0$! Take $R=\oplus_{i=1}^{\infty}\mathbb{Z}$, $J$ the sum of the even-indexed terms, and $I$ the sum of the multiple-of-4 index terms. Then $R/I\cong R$, $J/I\cong R$, but $(R/I)/(J/I)$ is isomorphic to $R$ again, not to $0$. –  Arturo Magidin Jun 28 '12 at 21:25
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2 Answers

up vote 6 down vote accepted

You can only speak of $I/J$ if $J\subseteq I$, so your question of whether $$(R/I)/(J/I) \cong (R/J)/(I/J)$$ only makes sense if both $I\subseteq J$ and $J\subseteq I$, or equivalently $I=J$. Of course, when that is the case, then the above statement is trivially true.

Let's use your specific example, where $R=\mathbb{Z}$, $I=3\mathbb{Z}$, and $J=5\mathbb{Z}$. There is no such thing as $I/J$, but you can talk about $$I/(I\cap J)=3\mathbb{Z}/15\mathbb{Z}=\{\bar{0},\bar{3},\bar{6},\bar{9},\bar{12}\}\cong \mathbb{Z}/5\mathbb{Z},$$ and $$J/(I\cap J)=5\mathbb{Z}/15\mathbb{Z}=\{\bar{0},\bar{5},\bar{10}\}\cong \mathbb{Z}/3\mathbb{Z}.$$ However, note that (by the third isomorphism theorem) $$(R/(I\cap J))/(I/(I\cap J))\cong R/I$$ $$(R/(I\cap J))/(J/(I\cap J))\cong R/J$$ and there is no reason for those two rings to be isomorphic in general.

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As written, what you have does not make sense. $J/I$ requires $I\subseteq J$, which is not met in your example. If both $J/I$ and $I/J$ make sense as written, then this requires $I=J$, and so in both cases we have $$(R/I)/(J/I) = (R/I)/(I/I) \cong R/I =R/J \cong (R/J)/(J/J) = (R/J)/(I/J),$$ and we are fine, but I doubt this is what you meant.

Instead, we can consider $(R/I)/((J+I)/I)$ and $(R/J)/((I+J)/J)$, since $I+J$ is the smallest ideal that contains both $I$ and $J$.

If so, then the answer is "yes", and it follows from the Isomorphism Theorems. In both cases we have that the quotient is isomorphic to $R/(I+J)$.

(However, as noted by Zev and as I failed to notice, this would not agree with your claim that "$5\mathbb{Z}/3\mathbb{Z} = \{\overline{0}, \overline{5},\overline{10}\}$", since my interpretation would be to set it equal to $\mathbb{Z}/3\mathbb{Z}$; while it also has $3$ elements, since $(I+J)/I \cong J/(I\cap J)$, it is not the "same" quotient.)

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@Clark And note that in your example $3$ and $5$ are coprime and hence $3\mathbf Z + 5\mathbf Z = \mathbf Z$. –  Dylan Moreland Jun 28 '12 at 21:09
    
@DylanMoreland Thank you. –  Matt N. Jun 28 '12 at 21:18
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@ClarkKent: The definition of quotient only applies then you have an ideal. An ideal must, among other things, be a subset. So you cannot talk about $J/I$ unless $I$ is an ideal of $J$, which requires $I$ to be a subset of $J$. What you are doing, instead, is looking at $J/(I\cap J)$, which does make sense, since $I\cap J$ is an ideal of $J$. (Note that we are talking about the quotient structure, not the equivalence relation; the equivalence relation you give is induced by $I\cap J$) –  Arturo Magidin Jun 28 '12 at 21:21
    
@ArturoMagidin Thank you. –  Matt N. Jun 28 '12 at 21:23
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