The first problem I have is finding a counter example to the following problem: Suppose for all $i\in I$, $N_i$ are submodules of M. I know that if $I$ is finite then $M/\cap_{i\in I} N_i$ is Noetherian(resp. Artinian) iff $M/N_i$ is Noetherian(resp. Artinian). But what if I is infinite? I don't think the statement still holds since an infinite direct sum of Noetherian modules need not be Noetherian. Does anyone have any counter examples?
The second counterexample is for the following problem: I know that $J=\cap\{\text{maximal left ideals of R}\}=\cap_{S\text{ simple}}Ann(S)$. From this we can conclude that if M is semisimple then for all $j\in J$ and for all $m\in M, jm=0 $. However, is the converse true? i.e. if for all $j\in J$ and for all $m\in M, jm=0 $, then M is semisimple. Again I can't seem to come up with a concrete counter example.
Any hints and answers appreciated.
Take $M$ be an infinite dimensional vector space over a field $k$ with basis $\{e_i\mid i\in\mathbb N\}$, and take $N_j=\mathrm{span}(e_i\mid i\neq j)$. Then each $M/N_j$ is one dimensional, so Noetherian, but $\bigcap_jN_j=0$, so $M/\bigcap _jN_j=M$ is not Noetherian.
Take $R=\mathbb Z$ and $M=\mathbb Z$. Then $J=0$, so $JM=0$, but $M$ is not semisimple.