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How might I find

$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ where $a,b,c,d \in \mathbb Z[x]$ such that there does not exist $B, C \in M_2(\mathbb Z[x])$ such that $B^{-1}AC$ is diagonal?

Added conditions: $B,C$ are invertible -- such that $B^{-1},C^{-1}\in M_2(\mathbb Z[x])$.

Anyone?

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    @J.D.: I am not sure how eigenvectors and stuff work in $\mathbb Z[x]$? If so, wouldn't \begin{pmatrix} 0&1\\0&0\end{pmatrix} be a solution? But I don't think this is what they are looking for?2012-03-16

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A necessary (and sufficient) condition for matrices $B,C$ with entries in $\mathbb{Z}[x]$ to be invertible is that the respective determinant be a unit of $\mathbb{Z}[x]$, i.e 1 or -1.

Consider $A = \begin{pmatrix} x & 2 \\ 0 & 0 \end{pmatrix}$.

If matrix $A$ were equivalent to diagonal matrix $D$:

$B^{-1} A C = D$

then by taking determinants we see that $D$ must have a zero on its diagonal:

$D = \begin{pmatrix} p & 0 \\ 0 & 0 \end{pmatrix}$

where $p$ is a nonzero element of $\mathbb{Z}[x]$. Rewriting the equivalence condition:

$\begin{pmatrix} x & 2 \\ 0 & 0 \end{pmatrix} C = B \begin{pmatrix} p & 0 \\ 0 & 0 \end{pmatrix}$

we see that $p B_{21} = 0$, so $B_{21} = 0$, and that $x C_{12} + 2 C_{22} = 0$. Since $x,2$ are different primes (not associates), this implies for some polynomial $q$, $C_{12} = 2q$ and $C_{22} = -xq$.

Now invertibility of $B$ requires $|B_{11}| = 1$, and without loss of generality any sign of $B_{11}$ may be combined with $p$ in our calculations, so that $x C_{11} + 2 C_{21} = p$.

But invertibility of $C$ requires $C_{11} C_{22} - C_{12} C_{21}$ be 1 or -1. Substituting $C_{22} = -xq$ and $C_{12} = 2q$ from above gives us that $-q (x C_{11} + 2 C_{21})$ is a unit, and thus both factors of that must be units.

This is a contradiction because factor $x C_{11} + 2 C_{21}$ cannot be a unit. The ideal generated by $x,2$ in $\mathbb{Z}[x]$ is proper.

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    Equivalent matrices have the same determinantal divisors. This leads to $(2,x)$ is a principal ideal, contradiction.2015-05-14