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gcd of polynomials is 1. There is an nxn matrix with determinant...

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Let $F$ be any field. Let $p_1,\ldots, p_n\in F[x]$. Assume that $\gcd(p_1,\ldots,p_n)=1$. Show that there is an $n\times n$ matrix over $F[x]$ of determinant $1$ whose first row is $p_1,\ldots,p_n$.

When $n=2$ this is easy since then there exist $a_1,a_2\in F[x]$ such that $p_1a_1+p_2a_2=1$. So the required matrix has first row $p_1,p_2$ and the second row $-a_2,a_1$.

I am stuck when $n>2$.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Re: gcd of polynomials is 1. there is an nxn matrix with determinant...

Let $F$ be any field. Let $p_1,\ldots, p_n\in F[x]$. Assume that $\gcd(p_1,\ldots,p_n)=1$. Show that there is an $n\times n$ matrix over $F[x]$ of determinant $1$ whose first row is $p_1,\ldots,p_n$.

When $n=2$ this is easy since then there exist $a_1,a_2\in F[x]$ such that $p_1a_1+p_2a_2=1$. So the required matrix has first row $p_1,p_2$ and the second row $-a_2,a_1$.

I am stuck when $n>2$.
For $n=3$ there exist $a_1,a_2,a_3\in F[x]$ such that $p_1a_1+p_2a_2+p_3a_3=1$.
So a matrix that has a matching determinant will do the trick.

For instance:
$$\begin{bmatrix}p_1 & p_2 & p_3 \\ 1 & -a_1/a_2 & 0 \\ 0 & a_3 & -a_2 \end{bmatrix}$$
Next step is to try and construct such a matrix for any n.
Put for instance 1 below $p_1$ and zeroes below that.
Then complete the proof with full induction.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Re: gcd of polynomials is 1. there is an nxn matrix with determinant...

For $n=3$ there exist $a_1,a_2,a_3\in F[x]$ such that $p_1a_1+p_2a_2+p_3a_3=1$.
So a matrix that has a matching determinant will do the trick.

For instance:
$$\begin{bmatrix}p_1 & p_2 & p_3 \\ 1 & -a_1/a_2 & 0 \\ 0 & a_3 & -a_2 \end{bmatrix}$$
Next step is to try and construct such a matrix for any n.
Put for instance 1 below $p_1$ and zeroes below that.
Then complete the proof with full induction.
This looks promising but there seems to be a small problem with this construction that $a_1/a_2$ might not necessarily lie in $F[x]$. Can we somehow still make it work?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Re: gcd of polynomials is 1. there is an nxn matrix with determinant...

This looks promising but there seems to be a small problem with this construction that $a_1/a_2$ might not necessarily lie in $F[x]$. Can we somehow still make it work?
Good point. :eek:

You already know that $\begin{bmatrix}p_2 & p_3 \\ -a_3' & a_2' \end{bmatrix}$ is a solution for n=2.

So we're looking for a solution of the following form for n=3:
$$\begin{bmatrix}p_1 & p_2 & p_3 \\ 1 & x & y \\ 0 & a_3 & -a_2 \end{bmatrix}$$
In other words:
$$p_1(-a_2 x - a_3 y) = p_1a_1$$
$$a_2 x + a_3 y = -a_1$$
This works if $\gcd(a_2, a_3)$ divides $a_1$.
Not yet sure how and if we can proof that though...
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Let $F$ be any field. Let $p_1,\ldots, p_n\in F[x]$. Assume that $\gcd(p_1,\ldots,p_n)=1$. Show that there is an $n\times n$ matrix over $F[x]$ of determinant $1$ whose first row is $p_1,\ldots,p_n$.

When $n=2$ this is easy since then there exist $a_1,a_2\in F[x]$ such that $p_1a_1+p_2a_2=1$. So the required matrix has first row $p_1,p_2$ and the second row $-a_2,a_1$.

I am stuck when $n>2$.
For $n=3$ there exist $a_1,a_2,a_3\in F[x]$ such that $p_1a_1+p_2a_2+p_3a_3=1$. Let $d = \text{gcd}(a_2,a_3)$ and let $a_2 = db_2$ and $a_3 = db_3$, where $\text{gcd}(b_2,b_3) = 1.$ Then there exist $c_2$ and $c_3$ such that $a_1 = c_2b_2+c_3b_3.$ Consequently $$\begin{vmatrix}p_1&p_2&p_3 \\ 0&b_3&-b_2 \\ -d&c_2&c_3 \end{vmatrix} = p_1(b_3c_3+b_2c_2) -p_2(-b_2d) + p_3(b_3d) = p_1a_1 + p_2a_2+p_3a_3=1.$$

Where do we go from there? I don't have time at present to think about how to deal with $n>3$, but here are a couple of comments. First, this problem does not really appear to be about $F[x]$, it looks as though the result should apply to elements of any euclidean domain. Second, the upper right corner $\begin{array}{cc}p_2&p_3 \\b_3&-b_2 \end{array}$ of the above determinant has the same general appearance as the determinant that solves the $n=2$ case. That might perhaps mean that there is scope for some sort of inductive procedure here.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
For $n=3$ there exist $a_1,a_2,a_3\in F[x]$ such that $p_1a_1+p_2a_2+p_3a_3=1$. Let $d = \text{gcd}(a_2,a_3)$ and let $a_2 = db_2$ and $a_3 = db_3$, where $\text{gcd}(b_2,b_3) = 1.$ Then there exist $c_2$ and $c_3$ such that $a_1 = c_2b_2+c_3b_3.$ Consequently $$\begin{vmatrix}p_1&p_2&p_3 \\ 0&b_3&-b_2 \\ -d&c_2&c_3 \end{vmatrix} = p_1(b_3c_3+b_2c_2) -p_2(-b_2d) + p_3(b_3d) = p_1a_1 + p_2a_2+p_3a_3=1.$$

Where do we go from there? I don't have time at present to think about how to deal with $n>3$, but here are a couple of comments. First, this problem does not really appear to be about $F[x]$, it looks as though the result should apply to elements of any euclidean domain. Second, the upper right corner $\begin{array}{cc}p_2&p_3 \\b_3&-b_2 \end{array}$ of the above determinant has the same general appearance as the determinant that solves the $n=2$ case. That might perhaps mean that there is scope for some sort of inductive procedure here.
Thanks a ton!