# Is this a semisimple algebra?

Let $A$ be the set of 2 by 2 matrices of the form $\begin{pmatrix}d_1 &d_2\\d_3 & d_4\end{pmatrix}$, where $d_i$ are in division algebras $D_i$.

Update: $D_1=Hom(N_1,N_1)$, $D_2=Hom(N_2,N_1)$, $D_3=Hom(N_1,N_2)$, $D_4=Hom(N_2,N_2)$, where $N_i$ are simple modules.

So $A$ looks something like this $\begin{pmatrix}Hom(N_1,N_1) &Hom(N_2,N_1)\\Hom(N_1,N_2) & Hom(N_2,N_2)\end{pmatrix}$

Is $A$ a semisimple algebra? And how do we see it?

My attempt: My idea is to view $A$ as isomorphic to $$\begin{pmatrix}D_1 &0\\0 & 0\end{pmatrix}\oplus\begin{pmatrix}0 &D_2\\0 & 0\end{pmatrix}\oplus\begin{pmatrix}0 &0\\D_3 & 0\end{pmatrix}\oplus\begin{pmatrix}0 &0\\0 & D_4\end{pmatrix}$$ which is isomorphic to $D_1\oplus D_2\oplus D_3\oplus D_4$, where $D_i$ are simple algebras?

We can also use Wedderburn's Structure Theorem applied to $A\cong M_1(D_1)\oplus\dots\oplus M_1(D_4)$ to conclude?

Thanks for help.

• What is the relation between the algebras? The multiplication of such matrices require you to multiply and add elements of the $D_i$ (for different $i$). – Tobias Kildetoft Feb 25 '16 at 8:54
• @TobiasKildetoft Updated above: The division algebras are module homomorphisms between simple modules. – yoyostein Feb 25 '16 at 9:41
• Ok, so now the multiplications make sense. But in general, it will not be isomorphic (as an algebra) to that direct sum (as can be seen by for example taking $N_1 = N_2$). – Tobias Kildetoft Feb 25 '16 at 9:43
• Ohh, and if $N_1$ and $N_2$ are not isomorphic then it is trivial that it is isomorphic to that sum, as it is just the diagonal ones that show up at all. – Tobias Kildetoft Feb 25 '16 at 9:44
• No, if $N_1$ and $N_2$ are not isomorphic then the off-diagonal entries are obviously $0$. And if they are isomorphic, then you get the algebra of $2\times 2$ matrices over some fixed algebra, which is not isomorphic to that direct sum. – Tobias Kildetoft Feb 25 '16 at 11:09

If the $N_j$ are simple right $R$ modules, then $\hom(N_i, N_k)$ is zero iff $i\neq k$, and otherwise it is the division ring $D_i=D_k$.
In the former case, the ring is isomorphic to $D_1\times D_2$, and in the latter case it is $M_2(D_1)$.