# $R$ is isomorphic to a direct product of matrix rings over division rings

Suppose as rings, $R$ is isomorphic to a direct product of matrix rings over division rings, that is $R=R_1 \times ... \times R_n$ where $R_i$ is a two-sided ideal of $R$ and $R_i$ is isomorphic to the ring of all $n_i\times n_i$ matrices with entries in a division ring.

I've also seen the notation for the $R_i$ be presented as $M_n{_i}(D)$ where $D$ is a division ring.

We want to show the ring $R$ considered as a left $R$-module is a direct sum $R=L_1\oplus L_2\oplus ...\oplus L_n$ such that $L_i$ are simple modules (no nonzero proper submodules) and such that $L_i=Re_i$ for $e_i\in R$ with

(a) $e_ie_j=0$ if $i\neq j$

(b) $e_i^2 = e_i$ for all $i$ (idempotent)

(c) $\Sigma _{i=1}^n e_i = 1$

I have a theorem from my notes that says if $e_1,...e_n$ are orthogonal idempotents with $\Sigma _{i=1}^n e_i = 1$, then we automatically get our conclusion $R=Re_1\oplus ...\oplus Re_n$.

But I'm not sure how we obtain the $e_i$ from our assumptions that we're given.

I also have that $M_n(D)$ or each of the $R_i$ is a semisimple ring, $D^n$ is a simple left $M_n(D)$ module and $M_n(D)$ is isomorphic to $D^n\oplus \ldots \oplus D^n$ as a left $M_n(D)$ module.

So I have all these tools, but not sure how to make them work for the proof.

To prove that $M_n(D)\cong D^n\oplus\dots\oplus D^n$ as left $M_n(D)$-modules you would apply the theorem from your notes with the orthogonal idempotents $e_i:=E_{ii}$ of $M_n(D)$, where $(E_{ii})_{kl}=\delta_{ik}\delta_{il}$ where $\delta$ denotes the Kronecker delta.
If you have two rings $R$ and $S$ and you consider the ring $R\times S$, then every idempotent $e$ of $R$ gives you an idempotent $(e,0)$ of $R\times S$. Similarly every idempotent $f$ of $S$ gives you an idempotent $(0,f)$ of $R\times S$. It is easy to see that two such idempotents $(e,0)$ and $(0,f)$ are orthogonal.