Let $R$ be a commutative Noetherian local ring. Let $I\subseteq R$ be an ideal containing a non zero divisor. Is $I$ an indecomposable $R$-module? Is it true at least if $\dim(R)=1$? I see this is true if $R$ is a domain: if $I\simeq J\oplus K$ for ideals $J,K$, then $JK\subseteq J\cap K=0$. Thanks.
Is every ideal containing a non zero divisor a indecomposable module
abstract-algebracommutative-algebra
Best Answer
In general the answer is no. Example: if $F$ is a field, then $\big[F[x,y]\big/(xy)\big]_{(\overline{x},\overline{y})}$ is Noetherian of dimension $1$ and local, with unique maximal ideal $M=\left(\overline{x}\big/\overline{1},\overline{y}\big/\overline{1}\right)$. Then $M$ contains the non-zero divisor $(\overline{x}+\overline{y})\big/\overline{1}$, but we can write $M=\left(\overline{x}\big/\overline{1}\right)\oplus\left(\overline{y}\big/\overline{1}\right)$, so $M$ is decomposable.