# Can a product of Cohn matrices over the Eisenstein integers with non-zero, non-unit coefficients be a Cohn matrix?

For $$k > 1$$, is it possible that $$\begin{pmatrix} a_1 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} a_2 & 1 \\ -1 & 0 \end{pmatrix}\ldots \begin{pmatrix} a_k & 1 \\ -1 & 0 \end{pmatrix} = \pm \begin{pmatrix} b & 1 \\ -1 & 0 \end{pmatrix}$$ if $$a_1,a_2,\ldots a_k,b$$ are Eisenstein integers and $$|a_i| > 1$$ for $$i=1,2,\ldots k$$?

If the Eisenstein integers are replaced with the Gaussian integers, this is possible. $$\begin{pmatrix} 3 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 - i & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 + i & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 - i & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} -2 & 1 \\ -1 & 0 \end{pmatrix} = -\begin{pmatrix} i & 1 \\ -1 & 0 \end{pmatrix}$$

This problem came up in my research; I am primarily interested in the case where $$b = (\pm 1 \pm \sqrt{-3})/2$$, but I suspect that there might not be a solution for any choice of $$b$$.

I originally asked this question on math.stackexchange (https://math.stackexchange.com/q/3903567/202799), but it seems to be more difficult than I first thought.

• You probably know this already: if there is a counter-example, it must involve some Cohn matrix $\operatorname{E}(a_i)$ with $1 < | a_i | < 2$ because of the inequalities 5.6 of "On the structure of the $\text{GL}_2$ of a ring" by P. Cohn,1966. (If $|a_i| \ge 2$ for every $i$, then $|b| \ge 2$.) Nov 15, 2020 at 6:54
• @LucGuyot I am embarrassed to say that I just realized I never read Cohn's original paper; just papers citing it. Nov 15, 2020 at 23:16

To my surprise, not only is there a solution for some $$b$$, there is actually a very simple infinite family of solutions for every $$b$$. Let $$\omega = \frac{1 + \sqrt{-3}}{2}$$. Then
$$\begin{pmatrix} a_0 + a_1\omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2 -\omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 + \omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2 - \omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 + \omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2 - \omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} b_0 + b_1\omega & 1 \\ -1 & 0 \end{pmatrix} = -\begin{pmatrix} a_0 + b_0 - 1 + (a_1 + b_1 - 1)\omega & 1 \\ -1 & 0 \end{pmatrix}$$
I wish I could say that there was some clever trick to finding this solution, but I cannot: it was found by a brute force search through $$\approx 5$$ million possibilities using Python after I realized that the bottom right coordinate of the product of $$n$$ Cohn matrices only depends on the inner $$n - 2$$ matrices.
I do not know if there are shorter solutions. One can prove that there are no solutions with $$k < 5$$, but there could be solutions with $$k = 5$$ or $$k = 6$$. If there are, however, they have to involve some elements with squared norm at least $$4$$, as I checked everything smaller.