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Q: With ring $R$, if $I, J \subseteq R$ are ideals such that $I+J=R$, then the map $R/(I \cap J) \to R/I \times R/J$ given by $a + (I \cap J) \mapsto (a+I, a+J)$ is an isomorphism, broadly generalizing the Chinese Remainder Theorem.

Can someone help me get started on this one?

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  • $\begingroup$ what is $ (I \cap J)+ (I \cap J)$? $\endgroup$ – David Holden May 11 '15 at 20:13
  • $\begingroup$ you're right. Of course it equals out. Let me edit that part out of the post. $\endgroup$ – clay May 11 '15 at 20:16
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    $\begingroup$ Which part are you struggling to show? Need to check (1) well definition (2) ring homomorphism (3) injective (one to one) and (4) surjective (onto) $\endgroup$ – CPM May 11 '15 at 20:24
  • $\begingroup$ This looks wrong: you can find $a\in I, b\in J$ such as $a+b$ is not in $I\cap J$ which would contradict $\phi(0)=0$ if $\phi$ is the name of your morphism. Take $R=\mathbb{Z}$, $I=(2),J=(3)$ and $a=5,b=1$. $\endgroup$ – Sergio May 11 '15 at 20:42
  • $\begingroup$ @Sergio But $5\not\in (2)$, and $1\not\in (3)$ ? $\endgroup$ – GPerez May 11 '15 at 21:00
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Consider the canonical maps $\pi_{I}:R\rightarrow R/I$ and $\pi_{I}:R\rightarrow R/J$ and form the morphism $\varphi(r)=(\pi_{I}(r),\pi_{J}(r))$.

As Censi LI explained in his answer, because $I+J=R$ you can find $i,j$ such as $i+j=1$ which then gives $\varphi(i)=(0,1)$ and $\varphi(j)=(1,0)$, so that $\varphi$ is surjective.

The kernel of $\varphi$ is precisely $I\cap J$ so $\varphi$ factors, by the first isomorphism theorem, into an bijective morphism $\tilde{\varphi}:R/(I\cap J) \rightarrow R/I \times R/J$ which is precisely the one in your question.

The crucial point is the one in Censi LI's answer but I felt that showing that your morphism can be though of as a factored morphism was important.

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  • $\begingroup$ It's in your hypothesis ! If you don't suppose it, it is not necessarily true. $\endgroup$ – Sergio May 12 '15 at 21:33
  • $\begingroup$ I'm embarrassed I asked that $\endgroup$ – clay May 12 '15 at 22:34
  • $\begingroup$ Why does $\varphi(i) = (0,1)$ and $\varphi(j) = (1,0)$ imply that $\varphi$ is surjective. I got docked points for that by my professor. $\endgroup$ – clay May 13 '15 at 22:57
  • $\begingroup$ Ah well, it isn't a very good idea to use MSE to do your assignements, especially if you don't try to understand the answers you get on your own... I'm sure if you give it a little thought you'll able to answer your question above by yourself. $\endgroup$ – Sergio May 15 '15 at 20:30
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Hint: Denote the canonical map $R/(I \cap J) \to R/I \times R/J$ by $\varphi$, since $I+J=R$, we can find $i\in I$ and $j\in J$ such that $i+j=1_R$, then $\varphi(\bar i)=(0,1),\ \varphi(\bar j)=(1,0)$.

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