I expect this to be true and proven, but I can't find any proofs of this. So anyone can confirm or deny this?


Let $R$ be a commutative ring, and let $M$ be a $kn\times kn$ matrix, which can be split into $n^{2}$ block of dimension $k\times k$. Assuming the block matrices form a solvable Lie algebra. Then the determinant of $M$ can be found by treating each block as a single element of the matrix ring and use the usual formula for the determinant on that to get a matrix, then calculate the determinant of that matrix.


Thank you.



Determinant of block matrix
It is true when $R$ is reduced, without $\mathbb{Z}$-torsion. If your blocks are $(M\_{i,j})\_{1 \leq i,j \leq n}$ and if 
$$N = \sum\_{\sigma \in \mathfrak{S}\_n} \epsilon(\sigma) M\_{1,\sigma(1)} \dots M\_{n,\sigma(n)},$$
then $\mathrm{det}(M) = \mathrm{det}(N)$.


Indeed, if $R$ is reduced without $\mathbb{Z}$-torsion then it can be embedded into a product of algebraically closed fields of characteristic $0$ (take algebraic closures of the residue fields at the minimal primes), so that one can assume that $R$ is an algebraically closed field of characteristic $0$. Then by Lie's theorem on can assume that the blocks $(M\_{i,j})$ are all upper-triangular. By replacing $R$ with an algebraic closure of $R(T)$, and $M\_{1,1}$ by $M\_{1,1} + T I\_k$ if necessary, one can assume that $M\_{1,1}$ is invertible. The proof is by induction on $n$, but I will detail only the case $n=2$. In this case one has 
$$
\begin{pmatrix} M\_{1,1} & M\_{1,2} \\ M\_{2,1} & M\_{2,2} \end{pmatrix} \begin{pmatrix} I\_k & - M\_{1,1}^{-1} M\_{1,2} \\ 0 & I\_k \end{pmatrix} = \begin{pmatrix} M\_{1,1} & 0 \\ M\_{2,1} & M\_{2,2} - M\_{2,1} M\_{1,1}^{-1} M\_{1,2} \end{pmatrix},
$$
so that $\mathrm{det}(M) = \mathrm{det}(M\_{1,1}) \mathrm{det}(M\_{2,2} - M\_{2,1} M\_{1,1}^{-1} M\_{1,2}) $. Since the blocks are upper triangular, the determinants are computed by taking the product of diagonal entries, so that this is equal to $\mathrm{det}(M\_{1,1}M\_{2,2} - M\_{2,1} M\_{1,2})$.


The hypothesis "without $\mathbb{Z}$-torsion" is necessary. Indeed, consider $R = F(T)$ where $F$ is a finite field and consider the $R$-vector space $V = R^F = \mathrm{Maps}(F,R)$. Consider the endomorphisms given by
$$
a : f \in V \mapsto (x \mapsto f(x+1)) \\
b : f \in V \mapsto (x \mapsto xf(x)) .
$$
Then $[a,b] = a, [a^{-1},b] = - a^{-1}$, so that the Lie algebra generated by $a,a^{-1},b$ is solvable. Consider the block matrix
$$
\begin{pmatrix} a & Tb \\ b & a^{-1} \end{pmatrix}.
$$
By the computation above, its determinant is 
$$\mathrm{det}(\mathrm{id} - aba^{-1}(Tb)) = \mathrm{det}(\mathrm{id} - (b+1)bT)) = \prod\_{x \in F} (1 - x(x+1) T).$$
This is a polynomial of degree $|F| - 2$. However,
$$
\mathrm{det}(a a^{-1} - b (Tb)) = \prod\_{x \in F} (1 - x^2 T)
$$
has degree $|F| - 1$, hence is not equal to the determinant of our block matrix.


