# Splitting the determinant polynomial into linear factors - a Dedekind problem

Here's the question in a nutshell. For some $n\in\mathbb N$, we consider the polynomial

$\det\left(\left(X_{i,j}\right) _ {1\leq i\leq n,\ 1\leq j\leq n}\right)\in\mathbb Z\left[X_{i,j}\mid 1\leq i\leq n,\ 1\leq j\leq n\right]$

in $n^2$ indeterminates $X_{i,j}$. This is known to be irreducible over $\mathbb Z$, but is there a "nice" ring in which $\mathbb Z$ embeds and where this polynomial splits into linear factors? The ring needs not be commutative, but the variables $X_{i,j}$ are still supposed to commute with everything from this ring. For instance, if $n=1$, then the ring can be taken to be $\mathbb Z$, and if $n=2$, then it can be taken to be $M_2\left(\mathbb Z\right)$, since

$\det\left(\begin{array}{cc}X&Y \newline Z&W\end{array}\right) = \left(\begin{array}{cc}X&Y \newline Z&W\end{array}\right)\mathrm{adj}\left(\begin{array}{cc}X&Y \newline Z&W\end{array}\right)$,

and both factors on the right hand side are linear in $X$, $Y$, $Z$, $W$. For $n>2$, however, the adjoint is not linear anymore. Can we extend $M_n\left(\mathbb Z\right)$ further to make it split? (This is what I mean by "nice" - it should be a kind of natural generalization of $M_n\left(\mathbb Z\right)$. Although I have troubles constructing even a non-nice splitting ring...)

Here is the actual source of the question:

Keith Conrad, in his expository paper The Origin of Representation Theory, discusses an apparently forgotten problem that goes back to Dedekind: Given a finite group $G$, the polynomial $\det\left(\left(X_{gh^{-1}}\right)_{g\in G,\ h\in G}\right)\in\mathbb Z\left[X_g\mid g\in G\right]$ (this is a generalization of the circulant, which is obtained if $G$ is a cyclic group) is known to split into a product of irreducible factors as follows:

$\det\left(\left(X_{gh^{-1}}\right) _ {g\in G,\ h\in G}\right) = \prod\limits_{\rho\text{ is an irrep of }G\text{ over }\mathbb C} \det\left(\sum\limits_{g\in G}X_g\rho\left(g\right)\right)^{\dim\rho}$.

(Okay, apparently Keith writes $\deg \rho$ instead of $\dim \rho$, but otherwise this is in his Section 5.)

Now, some of the factors on the right hand side - those corresponding to representations of dimension $> 1$ - are nonlinear, and Dedekind tried to split them into linear factors by extending the base field. Two examples are given, and both times the extension of the field is more or less the endomorphism ring of the representation $\rho$ - but this is not surprising, because both times $\dim \rho=2$, and we have the adjoint decomposition I gave above for the case $n=2$. The actually interesting problem seems to be the $n > 2$ case. Since any irrep $\rho$ over an algebraically closed field like $\mathbb C$ has the property that the $\rho\left(g\right)$ for all $g\in G$ span the whole endomorphism ring of the irrep (this is called the density theorem, I believe), we can actually forget about the irrep and try to split the determinant of the general matrix. That's the problem above.

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Nice question!!! A small remark on "this is called the density theorem, I believe" - I think in the case of finite groups it is called Burnside theorem (one of many theorems of Burnside!), while you are right that it immediately follows from Jacobson's density theorem... –  Vladimir Dotsenko Jun 10 '10 at 11:00
Darji, do you wish to Vandermondize all matrices?! –  Wadim Zudilin Jun 10 '10 at 11:27
I'm not quite sure what you mean with Vandermondize. Certainly, in the case of cyclic groups the Vandermonde matrix (of the roots of unity) diagonalizes the group matrix (= circulant), but in the general case I'm just talking about a formula for the determinant, not necessarily an explicit diagonalization. I also don't know whether the Vandermonde determinant over noncommutative rings is still the Vandermonde determinant that we all know and love - but that's an interesting question you brought up. –  darij grinberg Jun 10 '10 at 11:58
However, my hope for my original question is that it can be solved without the help of noncommutative determinants, since this is a part of algebra I have no grasp on... –  darij grinberg Jun 10 '10 at 12:00
Darij, have you tried $M_3(\mathbb Z[i])$? (n=3 case) –  Gjergji Zaimi Jun 10 '10 at 12:50

This question can be approached as a universal problem.

Find a ring extension $f:R\to S$, where $R=\mathbb{Z}[x_{ij}]$ and $f(R)\subseteq Z(S)$, and elements $a_{ij}^k\in S$ such that

$$(*) \qquad \det[x_{ij}]=\prod_{k=1}^n (\sum_{ij}x_{ij}a_{ij}^k).$$

Clearly, there exists a universal solution, namely $S=S_n$ is the quotient of the ring of noncommutative polynomials $R\langle a_{ij}^k \rangle$ ($R$ is central) modulo the relations obtained by expanding (*) in the variables $x_{ij}$ and equating the coefficients. Now, it "only" remains to show that $f$ is an injection. I don't see any obvious way to do it — abstract nonsense can only take you so far — but this seems to be a worthwhile perspective (similar techniques have been applied to the study of PI algebras using the ring of generic matrices). I am skeptical that for $n\geq 3$, an explicit solution (i.e. a homomorphic image of $S_n$) can be found among the familiar rings.

Remark. I should also mention that something very similar is possible:

$$(**)\qquad \det[x_{ij}]\prod_{i=1}^n\xi_i= \prod_{i=1}^n \left(\sum_{j=1}^n x_{ij}\xi_j\right)= \prod_{j=1}^n \left(\sum_{i=1}^n x_{ij}\xi_i\right),$$

where $\xi_1,\ldots,\xi_n$ are the generators of the rank $n$ exterior algebra $\Lambda$ over the ring $R=\mathbb{Z}[x_{ij}]$. This is just a restatement of the row and column expansions of the determinant in terms of the exterior algebra. Both the middle and the right hand side expressions in (**) are products of linear forms in $x_{ij}.$ Of course, the product $\prod_i \xi_i$ in the left hand side is not invertible — quite to the contrary, it generates the socle of $\Lambda$ as a $\Lambda_R$-module — so you can't simply divide every linear factor in the right hand side by the correspoding $\xi_i$ and isolate the determinant.

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Thanks anyway, this clears the situation up a bit. –  darij grinberg Jun 10 '10 at 21:47
I think that the usual techniques should work, but it's a bit far from my area, I'll post an update if I can think of something. By the way, my first useful thought had been "Dirac operator", from which I eventually came to the categorical formulation. –  Victor Protsak Jun 10 '10 at 22:32

Your question extends to other homogeneous polynomials, and is interesting in Partial Differential equations. For instance, $$X^2-Y^2-Z^2-W^2=(X+Ya+Zb+Wc)(X-Ya-Zb-Wc)$$ where $a,b,c$ are the Pauli matrices $$\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \end{pmatrix},\qquad \begin{pmatrix} 0 & -i \\\\ i & 0 \end{pmatrix},\qquad \begin{pmatrix} 1 & 0 \\\\ 0 & -1 \end{pmatrix}.$$ This gives a factorisation of the wave operator $\partial_t^2-\partial_x^2-\partial_y^2-\partial_z^2$ into first-order operators. Im am not sure whether there is a characterization of those hyperbolic operators that can be split into first-order factors.

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Your example is included in my $\dim 2$ case, because $X^2-Y^2-Z^2-W^2=\det\left(\begin{array}{cc} X-Y&Z+iW\\ Z-iW&X+Y\end{array}\right)$. ;) –  darij grinberg Nov 10 '10 at 11:52
Right! At least, my answer could be useful to those who didn't make a connection we differential operators. –  Denis Serre Nov 10 '10 at 12:25
Yes, there seems to be more literature on factoring PDOs. –  darij grinberg Nov 10 '10 at 13:14