Finding the basis of a free module over a commutative ring

Question

We have that $R \subseteq M_2(\mathbb{C})$ with $$R={\begin{bmatrix} w&{-z}\\\bar{z}&\bar{w}\end{bmatrix}}$$ for $w,z\in \mathbb{C}$ This is a commutative ring and hence R is a field so I believe that $M_2(\mathbb{C})$ is a free module for the subring R. I want to find a basis to show this but am not sure of one?

Answer 1

[$R$] is a commutative ring

No it isn't.

For example, $\begin{bmatrix}0&-1\\1&0\end{bmatrix}\begin{bmatrix}0&i\\i&0\end{bmatrix}\neq \begin{bmatrix}0&i\\i&0\end{bmatrix}\begin{bmatrix}0&-1\\1&0\end{bmatrix}$

$R$ is isomorphic to Hamilton's quaternions $\mathbb H$ which is the O.G. of division rings. (You may as well just replace $-z$ with $z$ to get the more commonly expressed representation $\begin{bmatrix}w&z\\-\bar{z}&\bar w\end{bmatrix}$.)

But nevertheless, every module over a division ring is a free module, and so it should have a basis. Let's pick a side and say we are talking about right $\mathbb H$ modules.

Since $M_2(\mathbb C)$ is $4$ dimensional over $\mathbb C$ and $\mathbb H$ is $2$ dimensional over $\mathbb C$ you can expect $M_2(\mathbb C)$ to be $2$ dimensional over $\mathbb H$.

We are free to put the identity matrix in the basis. What's left over after that?

For arbitrary $a,b,c,d$, we have $\begin{bmatrix}a&b\\c&d\end{bmatrix}-\begin{bmatrix}a&b\\-\bar{b}&\bar{a}\end{bmatrix}=\begin{bmatrix}0&0\\c-\bar{b}&d-\bar{a}\end{bmatrix}$, so this suggests a good candidate for the second basis element would be $\begin{bmatrix}0&0\\0&1\end{bmatrix}$ so that

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}a&b\\-\bar{b}&\bar{a}\end{bmatrix}+\begin{bmatrix}0&0\\0&1\end{bmatrix}\begin{bmatrix}\bar{d}-a&b-\bar{c}\\c-\bar{b}&d-\bar{a}\end{bmatrix}$$

It's not hard to see these two elements are $\mathbb H$ independent, so I leave that to you.