Prove that $\det (AB)=\det A \det B$ when entries of the matrices are in a non-commutative ring $R$

Question

It is well-known that $\det (AB)=\det A \det B$ for real square matrices, but what if the matrices have entries in a non-commutative ring $R$?
By the Principle of Permanence, it suffices to prove the identity in the ring $\mathbb{Z}[\{x_{ij}\},\{y_{ij}\}]$, because there is a unique homomorphism from $\mathbb{Z}[\{x_{ij}\},\{y_{ij}\}]$ to $R$. But it seems that it makes the problem even harder. I have no idea how to prove using this approach. Any help would be appreciated.

Answer 1

As egreg noted in a comment, it's not clear what "determinant" should mean over non-commutative rings, but let me suppose that you arbitrarily decided that $$ \det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc, $$ even though other choices like $da-cb$ would seem equally good. What I write below can easily be adjusted to work with these alternative definitions.

Consider two elements $a,b$ in your ring that don't commute. Then for the matrices $$ A=\begin{pmatrix}1&0\\ 0&a\end{pmatrix} \text{ and } B=\begin{pmatrix}b&0\\0&1\end{pmatrix}, $$ we have $$ AB=\begin{pmatrix}b&0\\0&a\end{pmatrix}, $$ and so $\det AB=ba$ while $\det A\cdot\det B=a\cdot b$.