Definition: An $R$-module $M$ is called locally Noetherian if, any finitely generated submodule of $M$ is Noetherian.
Question: Let $M$ be a locally Noetherian module. Then, either $M$ is uniform or it has a uniform direct summand?
I don't have any idea about the existence or non-existence of above statement. Please guide me how to initiate.
Best Answer
Any module over a Noetherian ring is locally Noetherian. Let $K$ be a field and $R=K[x,y]/(x^2,xy)$. We will abuse notation and denote the cosets of $x$ and $y$ by $x$ and $y$ respectively. This is a famous example of a Noetherian ring with an embedded prime: the set of zero divisors in $R$ is precisely the ideal $(x,y)$. We'll show $R$ is indecomposable as a module over itself but not uniform, thus showing the OP's question has a negative answer.
First suppose $R=I\oplus J$ for nonzero ideals $I,J$. Then $IJ\subseteq I\cap J=0$, so $I,J$ both consist of zero divisors. By the statement above, this implies $I+J\subseteq (x,y)$, and so we cannot have $R=I\oplus J$. Next note that the ideal $(x)$ equals $Kx$ and the ideal $(y)$ equals $yK[y]$, whence $(x)\cap(y)=0$. Thus ${}_RR$ is not uniform.