Let $F$ is a field, and $R = M_n(F)$. Prove that
(i) Every minimal left ideal of $R$ are isomorphic to each other
(ii) Prove every simple $R$-module are isomorphic to each other
I have if $M$ is a simple $R$-module, then there is a maximal left ideal $I$ of $R$ such that $M \cong R/I$. Now the big problem of mine is that I can't determine how the maximal ideal of $R$ look like? Why $R$ is semi-simple?
The ring $R$ is simple (why?) and artinian (why?).
A simple artinian ring is semisimple: indeed, the intersection of the maximal left ideals is the zero ideal (it can't be anything else) and so $0$ is the intersection of a finite family of maximal left ideals. Thus $0=\mathfrak{m}_1\cap\mathfrak{m}_2\cap\dots\mathfrak{m}_n$ and $$ R\hookrightarrow\bigoplus_{1\le i\le n}R/\mathfrak{m}_i $$ proves $R$ is a semisimple left module.
Let $S$ be a simple left $R$-module; take an epimorphism $f\colon R\to S$; this splits, so $R\cong S\oplus S'$ and therefore $S$ is isomorphic to a minimal left ideal of $R$. Conversely, any minimal left ideal is a simple module.
If $S$ and $T$ are nonisomorphic minimal left ideals of $R$, then $ST=0$; indeed, if $ST\ne0$, there is $t\in T$ with $St\ne0$; but $St\subseteq T$, so the two left ideals would be isomorphic. Thus $S$ is contained in the annihilator of the simple module $T$, forcing $S=0$, a contradiction.