Let $M$ noetherian as general hypothesis, then $M$ is semiartinian if and only if $M$ artinian.
We have that a module $M$ is semiartinian if for every non zero quotient of it contains a simple submodule. For proving $M$ is artinian assuming $M$ is noetherian I star by consider $M=M/ \lbrace 0 \rbrace $, and this one has a simple submodule lets say $S_{1}'= S_{1}$, following this line $M / S_{1}$ has also a simple submodule lets say $S_{2} / S_{1}= S_{2}'$ and so on we have that $M / S_{n}$ has a simple submodule $S_{n+1}'= S_{n+1} / S_{n}$. But I dont see why $S_{i} \subset S_{i+1}$ and why this ascending chain stops.
For proving $M$ semiartinian by assuming $M$ is artininan Im out of ideas since I dont know how to relate artinian modules with simple submodules. Any help will be aprecciated, thanks.
Any artinian module is semiartinian. The hypothesis of $M$ being noetherian is only needed for the converse.
Why is that? If $M$ is an artinian module and $L$ is a proper submodule thereof, then $M/L$ is artinian. Hence it has a simple submodule, because every set of submodules of an artinian module has minimal elements. Take the set of nonzero submodules.
What about the converse? Start with a simple submodule $S_0$ of $M$. If $S_0=M$, you're done. Otherwise there is a submodule $S_1$ such that $S_1/S_0$ is simple.
Suppose you have found $S_n$ such that $S_n/S_{n-1}$ is simple. If $S_n\ne M$, then you can find $S_{n+1}$ such that $S_{n+1}/S_n$ is simple.
OK, at each stage you get a proper ascending chain $S_0\subset S_1\subset\dots\subset S_n$. Also, at each stage $S_n$ is artinian (induction). Hence $M$ is artinian, because the chain must eventually stop.