Consider an $mn \times mn$ 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 $A$.

**Corollary:** If $B$ is an $A$-algebra that is finite and free as an $A$-module, $V$ is a finite free $B$-module, and $\phi \in \operatorname{End}_B V$, one can view $\phi$ also as an $A$-linear endomorphism, and then $\det_A \phi = N_{B/A}(\det_B \phi)$, where $N_{B/A}$ denotes norm.

**Corollary:** For finite free extensions $A \subset B \subset C$, we have $N_{B/A} \circ N_{C/B} = N_{C/A}$.

Question:Does the theorem (or either corollary) follow from some more conceptual statement, say some exterior power identities? Is there at least a proof that does not use induction on $m$?

Other references containing a proof of the theorem: http://dx.doi.org/10.2307/2589750 and http://www.ee.iisc.ernet.in/new/people/faculty/prasantg/downloads/blocks.pdf (thanks to Andrew Sutherland for pointing out the former).

7more comments