# Finding the basis of a free module over a commutative ring

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?

[$$R$$] 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.