Proof of Wedderburn's Theorem I've been going through a proof of Wedderburn's theorem:

and I'm stuck on the very last part, where the author refers to example 2.1.4. (linked below). I don't understand what $D^n$ means, or why it would be isomorphic to $L$, or why this follows from the example 2.1.4. The chain of isomorphisms he refers to therefore makes absolutely no sense to me. If anyone could clear things up it would be much appreciated!

 A: In my opinion, this explanation underevaluates the difficulty of the task and glosses over too much. That's why it winds up hard to understand.
You can find a much clearer explanation online in Martin Isaacs' book starting on page 189. Start reading section 13E there that ends with this:

"It is also true that $D$ is uniquely determined by $R$ (up to isomorphism), but this is more difficult to prove (and therefore more interesting.)

That is followed by two key lemmas and the proof of uniqueness.

I think the wording of "the induced map from $\bigoplus I_r\to M$" is just not helpful. 
The point is that all simple $M_n(D)$ modules are isomorphic to each other. So if $S$ is a simple $M_n(D)$ module, then it is isomorphic to each and every $I_r$, and every other simple submodule of $\bigoplus I_r$.
We can only guess that the isomorphism they are interested in is "induced from" the projection homomorphism $\phi:\bigoplus I_r\twoheadrightarrow S$ by the first homomorphism theorem, but it's unclear why they want to jump from this to an isomorphism with a particular $I_r$. It's true that $\bigoplus I_r /\ker(\phi)$ is simple, but even if you write $\bigoplus I_r =\ker(\phi)\oplus N$ and mod out by $\ker(\phi)$ on both sides, there's no guarantee $N$ is one of the $I_r$. 
At any rate, it's not important. Just be sure to prove for yourself that the simple right (or left) modules of a simple artinian ring are all mutually isomorphic.
A: Example $2.1.4$ says in the last three lines that $M\simeq I_1^n=I_1\oplus \cdots \oplus I_1$. This is the isomorphism which you asked about. Hence we have $L\simeq D^n$, and analogously $L\simeq (D')^m$. If you want to have more details, see for example here: http://www.math.hawaii.edu/~lee/algebra/wedder.pdf‎.
