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I read in passing that for a commutative ring $R$ and an ideal $I$, then $R/I$ is decomposable if and only if there exist proper ideals $J$ and $K$ such that $J+K=R$, and $J\cap K=I$.

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Dear Nastassja: Suggestion: We can assume $I=0$. If $J,K$ are ideals, then the equivalence $$R=J\oplus K\iff[J+K=R\text{ and }J\cap K=0]$$ holds almost by definition. – Pierre-Yves Gaillard Mar 1 '12 at 8:31
@Pierre-YvesGaillard Why exactly can we assume $I=0$? Does this give a way to get past the unsure step in the answer I posted? Thank you. – Nastassja Mar 1 '12 at 19:32
Dear Nastassja: Question $1$: The map $J\mapsto J/I$ from the set of ideals $R$ containing $I$ to the set of ideals of $R/I$ is a bijection satisfying $$\frac{J+K}{I}=\frac{J}{I}+\frac{K}{I},\quad\frac{J\cap K}{I}=\frac{J}{I}\cap\frac{K}{I}\quad.$$ Question $2$: Yes. --- The map $E\mapsto\pi^{-1}(E)$ is the inverse of $J\mapsto J/I$. So, it satisfies $$\pi^{-1}(E+F)=\pi^{-1}(E)+\pi^{-1}(F),\quad\pi^{-1}(E\cap F)=\pi^{-1}(E)\cap \pi^{-1}(F).$$ – Pierre-Yves Gaillard Mar 1 '12 at 21:04
up vote 1 down vote accepted

I wrote down the following while trying to justify it, and would appreciate if anyone could confirm if this is the correct idea.

Suppose $R/I$ is decomposable. Let $R/I=E_1\oplus E_2$, where $R/I\neq E_1$ or $E_2$. Let $\pi\colon R\to R/I$ be the canonical projection. Set $\pi^{-1}(E_1)=J$ and $\pi^{-1}(E_2)=K$, which are proper ideals of $R$, since $E_1$ and $E_2$ are proper ideals of $R/I$.

Clearly $J+K\subset R$. If $r\in R$, then $\pi(r)\in E_1\oplus E_2$, so $\pi(r)=\pi(x)+\pi(y)=\pi(x+y)$ for some $x\in J$ and $y\in K$. So $r\in\pi^{-1}(E_1+E_2)$, which implies $r\in J+K$. (I'm unsure if this step is valid, is there a more explicit way to show $R\subset J+K$?)

If $x\in I$, then $\pi(x)=0$, so $x\in\pi^{-1}(E_1)\cap\pi^{-1}(E_2)=J\cap K$. So $I\subset J\cap K$. If $x\in J\cap K$, then $\pi(x)\in E_1\cap E_2=0$, so $x\in\ker\pi=i$, so $I=J\cap K$.

Conversely, suppose there exist proper ideals $J$ and $K$ with the desired properties. Let $E_1=\pi(J)$ and $E_2=\pi(K)$, neither of which is equal to $R/I$ since $J$ and $K$ are proper. Then $R/I=E_1+E_2$ is immediate. If $x\in E_1\cap E_2$, then $x=\pi(r),\pi(s)$ for $r\in J$, and $s\in K$. Then $\pi(r-s)=0$, so $r-s\in I$, which implies $r\in K$ and $s\in J$. Thus $\pi^{-1}(x)\in I$, so $x=0$, and $R/I=E_1\oplus E_2$, and is decomposable.

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I don't think that step is valid: What is true is that the image of $r$ in the quotient that is a direct sum of $E_1$ and $E_2$ is of the form $(x,y)$ where $x \in E_1, y \in E_2$... – user38268 Mar 1 '12 at 7:41
I'm accepting this so that Community doesn't bump it unnecessarily. – Nastassja Apr 29 '12 at 0:02

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