Module over a simple ring. Let $A$ be a simple ring. Is any $A-$module semi-simple? Why/Why not? (For simplicity, assume that the module is of finite length over $A$.)

EDIT: When is an A-module semi-simple and how?
FFR: "Any" is an ambiguous quantifier and is generally avoided in these statements.

If you mean "Is every $A$-module semisimple?" the answer is no: it is necessary for $A$ to be additionally Artinian for this property to hold for every module. 

If you mean "Does there exist a simple $A$ module?" the answer is yes because all rings (with identity) have simple modules.




  The first line of this proof states that $M_1$ and $M_2$ are each just a sum of (different number of) copies of the unique simple $R$−module $M$. How?


Let $S$ be a simple right ideal. Then the sum of all right ideals isomorphic to $S$ is an ideal, hence all of $R$. Thus $R$ is a semisimple module, and all right ideals are sums of copies of $S$ too. In particular if you have a module homomorphism $R/M\cong S'$ where $S'$ is simple, then by decomposing $R=M\oplus T$ you get that $T$ is just a copy of $S$, and $S\cong S'$. This proves every simple $R$ module is isomorphic to $S$. 

You can further look up why the modules of simple Artinian rings decompose into simple modules, and then you will have that every module is a sum of copies of the one type of simple submodule.
FFR: "Any" is an ambiguous quantifier and is generally avoided in these statements.

If you mean "Is every $A$-module semisimple?" the answer is no: it is necessary for $A$ to be additionally Artinian for this property to hold for every module. 

If you mean "Does there exist a simple $A$ module?" the answer is yes because all rings (with identity) have simple modules.




  The first line of this proof states that $M_1$ and $M_2$ are each just a sum of (different number of) copies of the unique simple $R$−module $M$. How?


Let $S$ be a simple right ideal. Then the sum of all right ideals isomorphic to $S$ is an ideal, hence all of $R$. Thus $R$ is a semisimple module, and all right ideals are sums of copies of $S$ too. In particular if you have a module homomorphism $R/M\cong S'$ where $S'$ is simple, then by decomposing $R=M\oplus T$ you get that $T$ is just a copy of $S$, and $S\cong S'$. This proves every simple $R$ module is isomorphic to $S$. 

You can further look up why the modules of simple Artinian rings decompose into simple modules, and then you will have that every module is a sum of copies of the one type of simple submodule.