Are Noetherian Rings Goldie Rings? We know that a ring $R$ is said to be Right-Goldie if $R$, as a right $R$ module, satisfies:
(i) $R$ has finite uniform dimension;
(ii) every ascending chain of right annihilators of $R$ terminate. 

Suppose that $R$ is a Noetherian ring. I want to prove that $R$ is Right-Goldie. 

It is pretty obvious that it satisfies the second property. But how can I prove that it has finite uniform dimension? 
Let's recall what it is meant for a right $R$-module $M$ to have finite uniform dimension: 


*

*We say that a submodule $N$ of $M$ is essential when its intersections with non-zero submodules of $M$ are also nonzero;

*We say that a $R$ module $M$ is uniform when all its nonzero submodules are essential;

*We say that a $R$ module $M$ have finite uniform dimension if it contains a essential submodule $N$ which is (or is isomorphic to) a finite direct sum of uniform modules.
So, can anyone please help me? I know it is a possibly stupid question, but i am not managing to solve it.

It's actually really easy. I've thought a solution through:
Let $R$ be a Noetherian ring. If it is uniform, there's nothing to prove. If it is not uniform, then there is a (non-trivial) direct sum of ideals contained in $R$, say $R_{1} \oplus R_{1}'$. So, now, either $R_{1}$ is uniform or there is a (nontrivial) direct sum of ideals in $R_{1}$, say $R_{2} \oplus R_{2}'$. Keeping our track in this direction, two are our possibilities: either we find a uniform submodule of $R$ and our result is achieved, or we find an infinite direct sum of ideals of $R$, namely $R_{1}' \oplus R_{2}' \oplus R_{3}' \ldots$. However, this gives us an infinite ascending chain of ideals which never terminates, for the ideals $R_{i}´$ are never trivial. This is impossible since our ring is Noetherian. The result is thus proved.
