Showing that the set of isomorphism class of simple left module is finite Let $R$ be an Artinian Ring, then prove that the isomorphism classes of simple left $R$-modules is finite.
Morever show that the isomorphims class of simple left modules and the isomorphism classes of simple right modules has the same cardinality.
What am I thinking is to use descending chain condition, but I am unable to use it. Suppose that there are infinitely many isomorphism classes, then I am trying to use Artinian property of $R$, but it doesn't help
 A: If $S$ is a simple left $R$-module, it is cyclic (pick any non zero element $x \in S$, by simplicity $Rx = S$), hence $S \simeq R/\mathbf{Ann}_R(x)$. Since $S$ has no non-trivial submodules, we conclude that $\mathbf{Ann}_R(x)$ is maximal. Moreover, each quotient $R/ \mathfrak m$ is simple when $\mathfrak m$ is maximal.
This tells us that the classes of isomorphism of left (resp. right) simple modules are the same as the classes of isomorphism of quotients $R/\mathfrak m$ with $\mathfrak m$ a maximal left (resp. right) ideal. However, when $R$ is commutative the distinction between side ideals is irrelevant (and so the classes of isomorphism of right and left simple modules are in bijection) and we know there are finitely many maximal ideals, in particular, finitely many simple modules up to isomorphism.
Maybe this can be adapted to the non-commutative case, which I am not familiar with.
A: The simple right modules of $R$ correspond exactly with those of $R/J(R)$ where $J(R)$ is the Jacobson radical (exercise.)
Suppose $R$ is right Artinian.  Then $R/J(R)$ is a semisimple ring, i.e. a finite product of matrix rings over division rings.
Since all right ideals are summands in such a ring, each simple right module is isomorphic to a minimal right ideal. (A simple module $S$ is isomorphic to $R/M$ for some maximal ideal, but since $M$ is a summand it’s complement is a minimal right ideal isomorphic to $S$.)
Finally it is also easy to show the minimal right ideals of a finite product of rings is as one would expect: the right ideals which are nonzero only on one fixed coordinate, and in that coordinate its values are all from a minimal right ideal in that ring.
Since matrix rings over division rings each have exactly one simple right ideal each, you only have finitely many minimal right ideals in the product, and hence only finitely many isoclasses of simple right modules.
