Two simple modules over a semisimple ring are isomorphic when their annihilators are equal We know that if $M_R$ and $N_R$ are isomorphic $R$-modules, then their annihilators $ann(M) = ann(N)$. 
The converse is not true in general. But if $R$ is a semisimple ring and $M_R$ and $N_R$ are simple modules, then the converse is also true. In other words, if $ann(M)=ann(N)$, then $M_R \cong N_R$. 
Since $R$ is semisimple, $R = \bigoplus_{i=1}^{n} S_i$. Then, $M \cong S_k$ and $N \cong S_l$ for some $k,l \in \{1,2,...,n\}$. Now we want to show that if $ann(S_k) = ann(S_l)$, then $S_k \cong S_l$. But I could not continue. Is there any idea?
 A: There are a few pieces of disparate background knowledge to summarize.


*

*Let $I\lhd R$ be any ideal. It's not hard to show that the modules of $R/I$ are exactly the $R$ modules annihilated by $I$. Furthermore, any two such $R/I$ modules are isomorphic iff they are isomorphic as $R$ modules. This is all a result of the natural module structure $m\cdot (r+I):=m\cdot r$.

*A simple right $R$ module $S$ is isomorphic to $R/M$ where $M$ is a maximal right ideal in $R$, and $ann(S)$ is the largest ideal of $R$ contained in $M$.

*A semisimple ring can be expressed in its Wedderburn decomposition as $\prod_{i=1}^n R_i$ where the $R_i$ are simple Artinian rings, each of which has exactly one isoclass of simple right $R$ module.

*The maximal right ideals of $\prod_{i=1}^n R_i$ are of the form $\prod_{i=1}^n T_i$ where $T_j$ is a maximal right ideal of $R_j$ for some $j$, and $T_i=R_i$ for all $i\neq j$. Something analogous can be said for maximal two-sided ideals.
Now let $S, S'$ be simple right $R$ modules which share the same annihilator. Considering $2,3$ and $4$ together, the annihilator is of the form $ann(S)=ann(S')=\prod_{i=1}^n I_i$ where $I_j=\{0\}$ for some $j$, and $I_i=R_i$ for all $i\neq j$.
By $1$ now, $S,S'$ both are simple $R/I\cong R_j$ modules, where $R_j$ is a simple Artinian ring. But as mentioned in $3$, simple Artinian rings only have one isoclass of simple right $R$ module, so $S\cong S'$ as $R_j$ modules, and hence as $R$ modules as well.
