Inspired by $$ $$ http://mathoverflow.net/questions/65738/when-matrix-multiplication-commutes  $$ $$ and $$ $$ http://www.imdb.com/title/tt0293702/ $$ $$
is it true that, when $$ A \in SL_n(\mathbf Z),$$
then all integral matrices that commute with $A$ are an integral (or at least rational) polynomial in $A$? I dimly recall proving this for a specific 3 by 3 $A$ that was all 0's and 1's, so calculations were easy. The use of the unit determinant is that $A^{-1}$ is an integral polynomial in $A$ by Cayley-Hamilton. The degree of the polynomial need be no larger than $n-1,$ also by Cayley-Hamilton.

EDIT: as both David Speyer and  Tommaso Centeleghe point out in comments below, the statement is true if all eigenvalues are distinct, false otherwise. People are smart. And quick.

With or without commutativity, I once made a "multiplicative" function  out of
$$ f(x_0, x_1, \ldots, x_{n-1}) = \det (x_0 I + x_1 A + x_2 A^2 + \cdots + x_{n-1} A^{n-1}),   $$
amounting to a kind of fake norm form. The guy I asked about it laughed at me but said that's what I had.


I asked Manjul Bhargava about this: take the matrix $A,$ 3 by 3, to have rows
<0,1,0; 0,0,1; 1,1,1> which I think may actually have had determinant -1, never mind. Then the prime values I got from my fake norm form were all nonresidues mod 11 and all $x^2 + 11 y^2.$ I'm not sure about 2 itself. No proof but presumably a known sort of problem.