# Completing unimodular vectors with $3$ entries in $F_2[t]$ to a $3$ by $3$ matrix with determinant equal to $1.$

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 $12.$

-
How does this relate to what used to be called Serre's Conjecture, which was settled by Quillen and Suslin, and later by Vaserstein and Suslin? – Gerry Myerson Feb 16 '12 at 22:39
@OP: You're working over a principal ideal domain, aren't you? Whenever you have $(a,b,c)\in \mathbb F_2[t]^{1\times 3}$ with $gcd(a,b,c)=1$, you will be able to (explicitly) complete this to a matrix in $GL_3(\mathbb F_2[t])$ (which in this case is the same as $SL_3$). Imposing any further conditions seems unnecessary. The slides you link to do this for an arbitrary ring, and only there does the problem become difficult. – Florian Eisele Feb 17 '12 at 13:57
nice observation florian; probably you can post it as an answer ?; what is the explicit matrix on terms of $a,b,c$ ? – Luis H Gallardo Feb 17 '12 at 19:11

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.

-

Let $R$ be an arbitrary ring, and $(x,y,z)$ a unimodular row. View this row as a linear map from $R$ to $R^3$. Then the unimodularity implies that this map splits, so its cokernel (call it $M$) is projective.

Now suppose that $M$ is free, and that free modules over $R$ have well-defined ranks (e.g. if $R$ is commutative). Then $M$ is isomorphic to $R^2$. A complementary splitting $R^2\rightarrow R^3$ tells you how to complete your matrix.

In particular, this works whenever all projective $R$-modules are free, which is certainly the case for any PID and in particular for $F_2[t]$.

-