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Let I, J two right ideals of a ring R such that I+J =R. Show thath the direct sum of I and J is isomorphic to the direct sum of R and the intersection of I and J.

Can anyone please give me at least a ring morphism to prove this via first isomorphism theorem or there is another way to prove or attack this problem. Thanks

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  • $\begingroup$ There is a short exact sequence $I\cap J\to I\oplus J\to I+J$. Not sure what to do with it. If $R$ has no idempotents other than $0$ and $1$, then $R\oplus(I\cap J)$ will have a nonzero idempotent whereas $I\oplus J$ will not. Are you sure you aren't looking for an isomorphism of $R$-modules? $\endgroup$ – blue May 29 '14 at 4:23
  • $\begingroup$ mmm ok, the morphism from $$ I\oplus J\to I+J $$ is f(x,y)= x-y but what about the other morphism ?? :p $\endgroup$ – user1410 May 29 '14 at 4:29
  • $\begingroup$ They're $a\mapsto (a,-a)$ and $(x,y)\mapsto x+y$ respectively. $\endgroup$ – blue May 29 '14 at 4:29
  • $\begingroup$ Hey boyz! I understood it to be an isomorphism of $R$-modules since the ideals are one-sided. Dig? A one-sided ideal $I$ is a module over $R$, $R/I$ is a right module, not a ring unless $I$ is two sided. So, like I said, the context seems to me to tacitly imply module-ness. $\endgroup$ – Robert Lewis May 29 '14 at 4:37
  • $\begingroup$ Ah, yes. That makes sense. Thanks babe. $\endgroup$ – blue May 29 '14 at 4:40
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There is an s.e.s. $I\cap J\to I\oplus J\to I+J$ given by $a\mapsto(a,-a)$ and $(x,y)\mapsto x+y$.

We want to show this splits, i.e. find a map $I+J\to I\oplus J$ such that $I+J\to I\oplus J\to I+J$ is the identity. Write $1=i+j$ and identify $I+J$ with $R$. The map $R\to I\oplus J$ is $r\mapsto (ir,jr)$.

Thus, the right $R$-module isomorphism $(I\cap J)\oplus R\mapsto I\oplus J$ is $(a,r)\mapsto (a+ir,-a+jr)$.

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  • $\begingroup$ Yeep, thanks !! $\endgroup$ – user1410 May 29 '14 at 4:48
  • $\begingroup$ My sentiments exactly, +1! $\endgroup$ – Robert Lewis May 29 '14 at 5:07

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