3
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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 $i$). $\endgroup$ – Tobias Kildetoft Feb 25 '16 at 8:54
  • $\begingroup$ @TobiasKildetoft Updated above: The division algebras are module homomorphisms between simple modules. $\endgroup$ – yoyostein Feb 25 '16 at 9:41
  • 2
    $\begingroup$ 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 $N_1 = N_2$). $\endgroup$ – Tobias Kildetoft Feb 25 '16 at 9:43
  • 2
    $\begingroup$ 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. $\endgroup$ – Tobias Kildetoft Feb 25 '16 at 9:44
  • 2
    $\begingroup$ 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. $\endgroup$ – Tobias Kildetoft Feb 25 '16 at 11:09
2
$\begingroup$

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 $M_2(D_1)$.

In both cases, yes, the ring is semisimple.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.