Module $R/I$ indecomposable iff I irreducible. I want to prove that a cyclic module $R/I$ is indecomposable if and only if $I$ is irreducible. 

  
*
  
*A module is indecomposable if $M=M_{0}\oplus M_{1}$ iff $M=M_{0}$ or $M=M_{1}$. 
  
*An ideal is irreducible iff $I=I_{0}\cap I_{1}$ implies $I=I_{0}$ or $I=I_{1}$.
  

Here is my idea for the forward direction:
Assume that $R/I$ is cyclic and indecomposable. Let $I=I_{0}\cap I_{1}$. Since $R/I$ is indecomposable, if $R/I\cong R/I_{0} \oplus R/I_{1}$, then $R/I=R/I_{0}$ or $R/I=R/I_{1}$. Then $I=I_{0}$ or $I=I_{1}$. Then $I$ is irreducible. 
How is this solution? Does it make any sense or can it be done better? 
As for the backward direction, I am not sure how to approach this.

Submodules of $R/I$ are of the form $J/I$ where $J$ is an ideal containing $I$.
Say $$R/I = J_1 /I \oplus J_2/I$$ 
Then $J_1/I \cap J_2/I=0$ and hence $$J_1 \cap J_2=I$$
This proves the result $R/I$ is decomposable iff $\exists$ proper ideals $J_1, J_2$ such that $J_1+J_2=R$ and $J_1 \cap J_2=I$ 
This proves one direction. The other direction is false as given by the following counterexample.
Let $R=k[t^2,t^3]$ where $k$ is a field and consider $I=(t^2)\cap(t^3)$ 
Then clearly $I$ is not irreducible. However if $R/I$ is indecomposable  we will get two comaximal ideals $J_1,J_2$ containing $I$ such that $J_1 \cap J_2 =I$ . All polynomials in $I$ start from $t^5$ onwards and hence if a polynomial with non zero constant belongs to either of $J_1$ or $J_2$ then it becomes the whole ring. Thus no polynomial in $J_1 $ and no polynomial in $J_2$ has non zero constant and hence so does no polynomial in $J_1 + J_2$. 
The upshot is you cannot write $I$ as intersection of two comaximal ideals and hence $R/I$ is not decomposable although $I$ is irreducible.
