showing Module is simple

Question

Given the following:

let $C \subset \mathbb{H}$ be a subring of the real quarternion algebra such that it contains the center of $\mathbb{H}$ = $Z(\mathbb{H})$

Also C $\cong \mathbb{C}$

Then let R = $\mathbb{H} \otimes _\mathbb{R} C$ and considering $\mathbb{H}$ as an R-module where scalar multiplication

R$\times \mathbb{H} \longrightarrow \mathbb{H}: h \otimes c, x \rightarrow hxc$

Then how would i show that this module is simple?

thanks in advance for any help!

Answer 1

An $R$ submodule of $\Bbb H$ would have to also be an $\Bbb H$ submodule by restriction of $R$'s action to the subring $\Bbb H\otimes 1\cong \Bbb H$. Thus a nontrivial $R$ submodule would yield a nontrivial $\Bbb H$ submodule, but of course $_\Bbb H\Bbb H$ is simple, so there is no nontrivial submodule.

Alternatively, you can just show that $R$ acts transitively on $\Bbb H$: that is, for any $x,y\in \Bbb H\setminus\{0\}$, there exists an $r\in R$ such that $rx=y$. Clearly you can just use $r=yx^{-1}\otimes 1$.