On modules with finite uniform dimension Is it true that if a module $M$ has finite uniform dimension then the same is true for its homomorphic images ?
 A: Here is a (very easy) counterexample.  
The rationals $\mathbb{Q}$ form a uniform module over $\mathbb{Z}$, and thus $\mathbb{Q}$ has uniform dimension $1$, but
$\mathbb{Q}/\mathbb{Z}$ is an infinite direct sum of its $p$-torsion parts and thus has infinite uniform dimension.
A: Here is a counterexample.
Let $\Bbbk$ be a field and let $A=\Bbbk [x_0,x_1,x_2,\ldots]$ be the polynomial ring in countably many variables with its natural grading. Consider the ideal $I$ generated by $\{x_i x_j \mid i,j\in \mathbb N, i \neq j\}$, $\{x_i^2-x_j^2 \mid i,j\in \mathbb N, i \neq j\}$ and all monomials of degree three. In short $$I=\langle x_i x_j, x_i^2-x_j^2,A_3\rangle.$$
Then $M=A/I$ is an $A$-module with an induced grading $M=M_0 \oplus M_1 \oplus M_2$, and all submodules of $M$ have a non-zero intersection with the simple socle $\operatorname{Soc} M=M_2$. Therefore $M$ has uniform dimension one.
On the other hand $M/\operatorname{Soc} M$ is isomorphic to $A/\langle A_2 \rangle$ which has an infinite dimensional socle $M_1=A_1$. Therefore $M/\operatorname{Soc} M$ has infinite uniform dimension.
A: Of Course the answer is NO. As before answers, there are a lot of counterexamples show that the factor of a module with finite uniform dimension has no finite uniform dimension, in general. This kind of modules are said to be  QFD. Means modules whose every factor of them has finite uniform dimension. There are a lot of research works in this field. 
