I am trying to prove this lemma about the determinant of a block matrix, taken from this MO question.
Consider an mn×mn matrix over a commutative ring $A$, divided into $n\times n$ blocks that commute pairwise. One can pretend that each of the $m^2$ blocks is a number and apply the $m\times m$ determinant formula to get a single block, and then take the $n\times n$ determinant to get an element of $A$. Or one can take the big $mn\times mn$ determinant all at once.
Theorem (cf. Bourbaki, Algebra III.9.4, Lemma 1): These two procedures give the same element of $A$.
This is just a huge polynomial identity over an arbitrary commutative ring, so my thought was to try using the method of "universal identities" from here.
So if this polynomial identity holds over $\mathbb C$ with arbitrary variables adjoined, then it holds over $\mathbb Z$ with variables adjoined, and by the evaluative map into any commutative ring, it holds in general.
Over $\mathbb C$ it can be proved with simultaneous triangularisability and row operations, but there is the problem that the pairwise commuting condition means this is actually a polynomial identity on the set of points that satisfy the pairwise commuting condition, which is a variety(?).
Is there still a way to show the polynomial over $\mathbb C$ is uniquely determined even though it isn't defined on an open set in the enveloping space? This feels like a algebraic geometry issue, which I have very little knowledge about.
Any help or nudges in the right direction would be very appreciated.