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I am supposed to show that $R/J=(R/I)/(J/I)$ for two ideals $I\subseteq J\subset R$. How can I do this? I tried to construct an explicit isomorphism, but that did not work out (for me). Is there a more elegant solution?

Thank you in advance,

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I think you mean to write $(R/I)/(J/I)$. – Dylan Moreland Nov 15 '11 at 22:02
This is sometimes known as "the third isomorphism theorem." If you get stuck, just type that into a search engine (or look it up in a textbook) and see what comes up. – Gerry Myerson Nov 15 '11 at 22:05
yes, thanks, now it actually makes more sense. that typo was in my homework. thanks also for the name of the theorem. – Marie. P. Nov 15 '11 at 22:10
up vote 3 down vote accepted

If you know it for groups, you can show that this works at the group level and then just verify that the resulting map is a homomorphism of rings.

As for an elegant method, there is an obvious map $\frac{R}{I}\to\frac{R}{J}$ given by $a+I \longmapsto a+J$ (which is well defined since $I\subseteq J$). Verify that the kernel is precisely $\frac{J}{I}$, and apply the First Isomorphism Theorem.

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