Given $a\_1,b\_1,c\_1$ in $F\_2[t]$ with $\gcd(a\_1,b\_1,c\_1)=1$ it is known that there exists an element
$g$ of $SL(3, F\_2[t])$ (by explicit construction) such that $g$ has first line
$$
[a\_1^2, b\_1,c\_1],
$$
in particular this works when $a\_1^2$ is in the ideal $(b\_1,c\_1)$ generated by $a\_1$ and $b\_1$ in $F\_2[t].$


Question: It is possible to extend in the above manner the following line $L$ to an element of $SL(3, F\_2[t])$


$$
L =[1+a,b,c]
$$
where $a,b,c \in F\_2[t]$ with $\gcd(1+a,b,c)=1$ and with
$$
a^2 \in (b,c).
$$


The case when $\gcd(b,c)=1$ is trivial since $1+a = M^2+ t N^2$ and $tN^2 \in (b,c)$ in these case.


The question (for a general characteristic $2$ ring R) appears in page $14$ of


www.math.psu.edu/oldColloquium/Ravi2.pdf


while the explicit construction described above appears in page $12.$



Completing unimodular vectors  with $3$ entries in $F_2[t]$  to a $3$ by $3$ matrix with determinant equal to $1.$
Okay, suppose $K$ is a field and you have three elements $a,b,c\in K[t]$ with $gcd(a,b,c)=1$.
I will show that the row $(a,b,c)$ can always be extended to a matrix in $SL(3,K[t])$.
Suppose that at least one of $b$ and $c$ is nonzero (otherwise $a\in K$ and extending the row is trivial).


Let $g:=gcd(b,c)$ and choose $x,y\in K[x]$ such that $xb+yc=g$. Since $gcd(a,g)$ must be one, we find $r,s\in K[x]$ such that $ra+sg=1$. Now the following $3\times 3$-matrix has determinant $1$ and extends the row $(a,b,c)$:
$$
\left[\begin{array}{ccc}a&b&c\newline -s& \frac{rb}{g} & \frac{rc}{g} \newline 0 & -y & x\\ \end{array}\right]
$$
It is an easy computation that this matrix has determinant one (unless I mistyped something). Note that the occurring fractions lie in $K[x]$, since $g$ divides both $b$ and $c$ by definition.


