# Find three finite dimensional semisimple $\mathbb{R}$ algebras

I'm asked to find three finite dimensional semisimple $$\mathbb{R}$$-algebras, $$R$$, each with only one simple module $$S$$ up to isomorphism so $$R\simeq S\oplus S$$ as $$R$$-modules. I'm guessing I need to use Artin-Wedderburn. If we let $$R$$ be a finite dimensional $$\mathbb{R}$$ algebra then by Artin-Wedderburn, we know that $$R\simeq M_{n_1}(D_1)\oplus\cdots\oplus M_{n_s}(D_s)$$, where $$D_i$$ are finite dimensional $$\mathbb{R}$$ division algebras. Moreover, in this decomposition there are precisely $$s$$ simple modules up to isomorphism. Using this, it's easy to see that if $$R$$ is a candidate having the required properties then, that $$R\simeq M_n(D)$$ where $$D$$ is a finite dimensional $$\mathbb{R}$$ division algebra. By Frobenius theorem, there are only $$3$$ finite dimension $$\mathbb{R}$$ division algebras, $$\mathbb{R}$$, $$\mathbb{C}$$, $$\mathbb{H}$$. My problem is in determining which of these satisfy the condition that $$R\simeq S\oplus S$$, where $$S$$ is the sole simple $$R$$-module up to isomorphism.

It certainly can't be that $$n=1$$ and $$D=\mathbb{R}$$. But beyond this, I am stuck. Any help is appreciated.

Note that $$\dim(R) = n^2 \cdot \dim(D)$$ while $$\dim(S \oplus S) = 2 \dim(S) = 2n \cdot \dim(D)$$, so necessarily $$n = 2$$.
For a division ring $$D$$, $$M_n(D)$$ has composition length 2 iff $$n=2$$. One can see this by observing that the subset of matrices which is nonzero only on row $$i$$ is a simple right ideal.
So you're looking for three finite dimensional $$\mathbb R$$-division algebras, each of which you use for $$D$$ in $$M_2(D)$$. You already mentioned three such algebras...