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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).

Tadashi
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Bjorn Poonen
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    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
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    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
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    See also http://mathoverflow.net/questions/48936/iterated-calculation-of-determinants – Bjorn Poonen Oct 31 '14 at 16:46
  • Would be interesting if we could exploit the usual (complex field) fact that $\det(A)=\wedge^n A$, and the block-matrix lemma mentioned here: http://mathoverflow.net/questions/173088/a-possible-extension-of-a-determinant-inequality by extracting wedge products of individual blocks as principal submatrices of the wedge products of the entire matrix (these relations don't seem to depend on having a field) – Suvrit Oct 31 '14 at 17:21
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    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
  • Another text reference for the theorem besides Bourbaki is Jacobson's Basic Algebra I (2nd ed.), Section 7.4. I don't have it in front of me, so I can't check if he uses induction. The paper by John Silvester that you link to has a proof by induction, which is not of direct interest to you, but what might be of interest is that in the last paragraph he mentions that he has seen an abstract version of the identity. Whatever the abstract version is might be the conceptual statement you seek (unless it's your corollary). – KConrad Oct 31 '14 at 19:16
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    Better alternative for the first reference: http://www.southalabama.edu/mathstat/personal_pages/williams/newdet.pdf . – darij grinberg Oct 31 '14 at 20:05
  • @KConrad: Jacobson uses induction... – ACL Oct 31 '14 at 22:25
  • If the scheme of N commuting matrices is reduced for all N, then the general case of the theorem follows easily from the algebraically close field one. But is it reduced? – Anton Fetisov Nov 01 '14 at 00:51
  • I'm not sure about the simultaneous diagonalization, maybe simultaneous triangulization? . If $n\gt \gt 20$ and the blocks are $20×20$ with constant diagonal and upper right 10×10 sub-block arbitrary then (evidently) the determinant depends only on the diagonal values, but the blocks come from a dimension 101 algebra. – Aaron Meyerowitz Nov 01 '14 at 03:32
  • The first Corollary (hence the second) follows from functoriality applied to the elegant Proposition 4.2 in Ch. II of Oesterle's 1984 Inventiones 78 paper "Nombres de Tamagawa..." (which gives a categorified version of the desired identity). – user27920 Nov 01 '14 at 06:36
  • @user52824: That Oesterlé reference is very nice, though he uses the first corollary (citing the same Bourbaki page that I cited) to prove his statement, so it's not an independent proof. – Bjorn Poonen Nov 02 '14 at 16:58

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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.

Dustin Clausen
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