Find a non-commutative ring with exactly 2014 two sided-proper ideals. Find a ring with exactly 2014 pairwise non-isomorphic irreducible modules.
If it was the commutative ring i would have thought of the example $Z/2^{2014}Z$. But what to do for non-commutative case? What is the strategy for constructing such examples?
Will the same example work for having pairwise non-isomorphic irreducible modules??
If $R$ is a ring, then the two sided ideals in the ring $M_n(R)$ of $n\times n$ matrices are in bijection with the two sided ideals of $R$.
The ring $\mathbb{Z}/2^k\mathbb{Z}$ has exactly $k$ proper ideals.
If $F$ is a field, then $F\times F$ has exactly two (isomorphism classes of) simple modules (irreducible and simple are synonyms).