Consider a 2x2 matrix $A$ with entries from $\mathbb{C}[x,y]$. Assume that $\mathrm{det} A$ is a square. Is it true that then $A$ can be represented as a noncomuting product $A=A\_1 A\_2 … A\_{2n}$, in which any given matrix $A\_k$ occurs an even number of times up to transposition? (Formally, the latter means that for each k=1,…,2n there is an even number of indices j such that either $A\_j=A\_k$ or $A\_j=A\_k^T$). 


If this is not true, is there a nice description of 2x2 polynomial matrices whose determinant is a square?


In fact the following particular case is of most interest for me:




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Let $A=A(x,y)$ be a *biquadratic* polynomial with quaternionic coefficients. Assume that $|A|^2$ is a square of a real polynomial. Is it true that then $A$ can be represented as a noncomuting product $A=\lambda A\_1 A\_2 … A\_{2n}$, in which any given quaternionic polynomial $A\_k$ occurs an even number of times up to conjugation, and $\lambda$ is a real polynomial? 




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This question is a continuation of the following one:
[Pythagorean 5-tuples](https://mathoverflow.net/questions/62820/pythagorean-5-tuples)



When the determinant of a 2x2 polynomial matrix is a square?
I can show that there is no universal formula. More precisely, let $k$ be an algebraically closed field and let $R = k[w,x,y,z,\Delta]/(\Delta^2-wz+xy)$. I claim that there do not exist $2 \times 2$ matrices $A\_1$, $A\_2$, ..., $A\_{2t}$ with entries in $R$, such that every matrix occurs an even number of times in the list up to transpose, and $\left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right) = A\_1 A\_2 \cdots A\_{2t}$. In fact, we are going to show something much stronger: It is impossible to write $\left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right)$ as $MN$ for two noninvertible $2 \times 2$ matrices with entries in $R$.


Since $R$ is a graded integral domain, $\det M$ and $\det N$ would have to be of pure degree. If $\det M$ has degree $0$ then $M$ is invertible. So we must have $\det M$ and $\det N$ of degree $1$.


Write $M=M\_0 + M\_1 + \cdots$ where $M\_k$ is pure of degree $k$, and likewise for $N$. So $$\left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right) = M\_0 N\_1 + M\_1 N\_0.$$ 
Since $\det M$ has no constant term, we have $\det M\_0 =0$ and, similarly, $\det N\_0 =0$. Let $\left( \begin{smallmatrix} u\_1 & u\_2 \end{smallmatrix} \right) \in k^2$ be in the left kernel of $M\_0$ and let $\left( \begin{smallmatrix} v\_1 \\ v\_2 \end{smallmatrix} \right) \in k^2$ be in the right kernel of $N\_0$. So the above equation shows that $\left( \begin{smallmatrix} u\_1 & u\_2 \end{smallmatrix} \right) \left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right) \left( \begin{smallmatrix} v\_1 \\ v\_2 \end{smallmatrix} \right) =0$. 


But there are no nonzero vectors in $k^2$ for which $\left( \begin{smallmatrix} u\_1 & u\_2 \end{smallmatrix} \right) \left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right) \left( \begin{smallmatrix} v\_1 \\ v\_2 \end{smallmatrix} \right) =0$, a contradiction.


**UPDATE:** Pushing this argument further, I can show that $\begin{pmatrix} 0 & u & v & w \\ -u & 0 & x & y \\ -v & -x & 0 & z \\ -w & -y & -z & 0 \end{pmatrix}$ cannot be expressed as $MN$ with $\det M$ and $\det N$ nonconstant. **Proof:** Abbreviate the above matrix to $S$. As before, deduce that we can write $M\_0 N\_0 =0$ and $M\_0 N\_1 + M\_1 N\_0 = S$ with $\det M\_0 = \det N\_0 = 0$. So $\mathrm{rank} M\_0 + \mathrm{rank} N\_0 \leq 4$. Without loss of generality, suppose that $M\_0$ has nullity at least $2$ and $N\_0$ has nullity at least $1$. So, as before, we can find linearly independent constant vectors $p$ and $q$ with $p^T M\_0=q^T M\_0 =0$ and a nonzero vector $r$ with $N\_0 r=0$. We have $p^T S r = q^T S r = 0$ as before. 


Let $p=(p\_1, p\_2, p\_3, p\_4)$ and $r=(r\_1, r\_2, r\_3, r\_4)$. Looking at the coefficient of $u$ in $p^T S r$, we get $p\_1 r\_2 = p\_2 r\_1$. Looking at $v$, $w$, $x$, $y$ and $z$, we see that $p\_i r\_ = p\_j r\_i$ for all $(i,j)$, so $p$ and $r$ are proportional. Similarly, $q$ and $r$ are proportional, so $p$ and $q$ are proportional, contradicting that we choose them linearly independent. QED.


I have no $2 \times 2$ counter-example.


