Let $M$ be a locally Noetherian module. Then, either $M$ is uniform or it has a uniform direct summand? 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.
 A: 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.
A: Let $k$ be a field with an endomorphism $\phi$ such that $[k:\phi(k)]= n >1$.  Consider the twisted polynomial ring $k[x;\phi]$ where $xa:=\phi(a)x$ for all $a\in k$.  Let $R$ be the quotient ring by the ideal $(x^2)$.
Then $R$ is Artinian, hence locally Noetherian, considered as a right module over itself. It is  also local, so it has no proper summands.  But it is also not uniform, because it has distinct simple submodules.
