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Let $R$ be an integral domain. Let $I$ and $J$ be non-zero ideals of $R$. Is this statement always true: $$R\oplus(I\cap J)\cong I\oplus J\ ?$$

I regarded the short exact sequence $0\to I\cap J\to I\oplus J\to I+J\to0$ with the maps $(\cdot,\;-\,\cdot)$ and $\mathrm{pr}_1+\mathrm{pr}_2$. It splits if $I+J$ is a principal ideal and $I\oplus J\cong (I+J)\oplus(I\cap J)\cong R\oplus(I\cap J)$. In general, however, this is not the case. I cannot find any counter-example, either.

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    $R\oplus (I\cap J)$ Contains $R$ as an ideal. Why would $I\oplus J$ do so? – Tobias Kildetoft May 16 '15 at 20:01
  • An example from number theory: try with $R=\Bbb{Z}[\sqrt{-5}]$, $I=(2)$, $J=(1+\sqrt{-5})$. I think this would work, since $R$ is not an UFD. – Crostul May 16 '15 at 21:47
  • @Crostul I don't get why $R\oplus (I\cap J)\not\simeq R^2$, and what has this to do with UFDs. – user26857 May 16 '15 at 21:50
  • @user26857 Actually I'm trying to prove it on my own: this is the reason why I didn't post this on an answer. – Crostul May 16 '15 at 21:51
  • @Crostul Never mind. I've proved and post it as an answer. (In fact, this is a good example showing why in a Dedekind domain we can't replace $IJ$ in the Steinitz Isomorphism Theorem, which says $R\oplus IJ\simeq I\oplus J$, by $I\cap J$.) – user26857 May 17 '15 at 07:37

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If $R$ is a commutative ring, and $L\subset R$ is an ideal such that $R\oplus L\simeq R^2$ then $L\simeq R$.

We have $\bigwedge^2(R\oplus L)\simeq\bigwedge^2R^2$, so $L\simeq R$. (Here I've used that $L$ is a projective rank one module, and this gives us $\bigwedge^2L=0$.)

This shows that the example suggested by Crostul works: let $R=\mathbb Z[\sqrt{-5}]$, $I=(2)$, and $J=(1+\sqrt{-5})$. If $R\oplus(I\cap J)\simeq I\oplus J$ then $R\oplus(I\cap J)\simeq R\oplus R$. We have to prove that $I\cap J$ is not principal, and this is left to the reader.

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In this answer you can find two principal ideals $I,J$ such that $I\cap J$ is not finitely generated. Then you can't have $R\oplus (I\cap J)\simeq I\oplus J$ since then $R\oplus (I\cap J)\simeq R\oplus R$, and thus $I\cap J$ is isomorphic to a quotient module of $R\oplus R$ and therefore it's finitely generated.

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