# Determinant of a determinant

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).

• These are easier to prove if $A$ is an integral domain, but because of the commuting blocks condition it's not clear that one can reduce to this case. If $A$ is an integral domain, one can embed $A$ in an algebraically closed field and simultaneously diagonalize the blocks. And transitivity of norm for finite field extensions can be proved by using the definition of norm as a product of conjugates (raised to the inseparable degree). – Bjorn Poonen Oct 31 '14 at 16:38
• There's a beautiful construction in Deligne's SGA 4 Exp XVII $\S$6.3 where he puts this in a geometric context to define the trace morphism with general coefficients (and right before proving that the Hilbert Scheme of points in a curve is represented by the symmetric powers). I'm afraid however that the proof goes down to the local situation and uses Bourbaki's lemma. It's worth a look though. – Reimundo Heluani Oct 31 '14 at 16:44
• – Bjorn Poonen Oct 31 '14 at 16:46
• This reminds me the proof by S. Rosset of Amitsur-Levitski Theorem. At the beginning, one has an $n\times n$ matrix $A$ with entries in $\Lambda^1(k)$, a non-commutative algebra. But $A^2$ has entries in $\Lambda^2(k)$, a commutative algebra. After checking that the traces of $A^{2p}$ all vanish, one deduces $A^{2n}=0_n$, which gives the polynomial identity. – Denis Serre Oct 31 '14 at 17:37
• Better alternative for the first reference: southalabama.edu/mathstat/personal_pages/williams/newdet.pdf . – darij grinberg Oct 31 '14 at 20:05

## 1 Answer

Here's another proof of the corollary. I think it's not the kind of proof you're looking for, especially since it secretly uses induction on m, but I think it is conceptual in a way.

1. First reduction: we're trying to prove an equality between two maps of $A$-schemes $Res_{B/A} Mat_{m\times m}\rightarrow \mathbb{A}^1$. But $Res_{B/A}GL_m$ is Zariski-dense in $Res_{B/A} Mat_{m\times m}$. Thus we can assume our $\phi$ is invertible.

2. First expansion: every pair $(V,\phi)$ as in the statement, with $\phi$ invertible, defines an element of the Quillen K-group $K_1(B)$. Now recall that the determinant extends to a map $det_B:K_1(B)\rightarrow B^\ast$, and likewise $det_A:K_1(A)\rightarrow A^\ast$. Recall also that there is a natural "transfer" map $tr_{B/A}:K_1(B)\rightarrow K_1(A)$ coming from the obvious forgetful functor on module categories. Then we can try to prove a more general claim: for any $x\in K_1(B)$, we have

$$N_{B/A} det_B(x) = det_A tr_{B/A} (x).$$

1. Both sides of this equation are elements of $A$. Since $A$ injects into the product of its localizations at all maximal ideals, and all the operations above commute with base change, we can thereby reduce to the case where $A$ is local, and hence $B$ is semi-local.

2. For a general commutative ring $R$ the determinant map $K_1(R)\rightarrow R^\ast$ has an obvious section, coming from viewing elements of $R^\ast$ as endomorphisms of the unit $R$-module. When $R$ is semi-local, these maps are actually mutually inverse isomorphisms (see http://www.math.rutgers.edu/~weibel/Kbook/Kbook.III.pdf, Lemma 1.4). Since $B$ is semi-local, we can therefore reduce to the case where $\phi$ is given by an element of $B^\ast$ acting on $B$. Then the claim is simply the definition of $N_{B/A}$.

I guess the moral of the story is that, thanks to the referenced lemma, the Zariski sheafification of $K_1$ identifies with $\mathbb{G}_m$. And this identification can't help but intertwine the transfer with the norm.