5
$\begingroup$

Let $n \geq 1$ and $A = \mathbb{k}[x]$, where $\mathbb{k}$ is a field. Let $a_1, \dots, a_n \in A$ be such that

$$Aa_1 + \dots + Aa_n = A.$$

Does there exist an invertible matrix $\|r_{ij}\| \in M_n\left(A\right)$ such that $r_{1j} = a_j$ for all $j = 1, \dots, n$?

$\endgroup$
  • $\begingroup$ Your question appears to use only the ring $A$ of polynomials in $x$, and not mention rational functions at all. Did you forget to do anything with them? $\endgroup$ – Marc van Leeuwen Jul 14 '15 at 15:21
  • $\begingroup$ Whoops sorry, I deleted that part. $\endgroup$ – user254398 Jul 14 '15 at 15:24
  • $\begingroup$ Aren't the elements of $A$ polynomials, not scalars? The $a_i$ appear to belong to $\mathbb{k}$ $\endgroup$ – user237392 Jul 14 '15 at 16:47
2
$\begingroup$

As mentioned in the other answer, this result holds for all principal ideal domains $A$. You can prove by induction on $n$ that there exists $U\in {\rm GL}_n(A)$ such that $U\cdot (a_1,\dotsc,a_n)^t = (1,0,\dotsc,0)^t$:

If $n=2$, write $1 = ra_1 + sa_2$ for $r,s\in A$. Then $U = \begin{pmatrix} r & s\\ -a_2 & a_1\end{pmatrix}$ has determinant $1$ and satisfies $U\cdot (a_1, a_2)^t = (1,0)^t$.

Now assume that the result holds for $n-1$. Let $d$ be the gcd of $a_{n-1}$ and $a_n$. Write $d = ra_{n-1} + sa_n$ for some $d,r,s\in A$. Let $U_1 := \begin{pmatrix} E_{n-2} & 0\\ 0 & V\end{pmatrix}$ for $V = \begin{pmatrix}r & s\\ -a_n/d & a_{n-1}/d\end{pmatrix}$ (where $E_{n-2}$ is the $(n-2)\times(n-2)$ identity matrix). Then $U_1\in {\rm SL}_n(A) \subseteq {\rm GL}_n(A)$ and $$U_1 \cdot \begin{pmatrix}a_1\\ \vdots\\ a_n\end{pmatrix} = \begin{pmatrix} E_{n-2} & 0 & 0\\ 0 & r & s\\ 0 & -a_n/d & a_{n-1}/d \end{pmatrix}\cdot \begin{pmatrix}a_1\\ \vdots\\ a_n\end{pmatrix} = \begin{pmatrix} a_1\\ \vdots\\ a_{n-2}\\ d\\ 0\end{pmatrix}$$ reduces to the case $n-1$. So by induction hypothesis there exists $U'_2\in {\rm GL}_{n-1}(A)$ such that $U'_2\cdot (a_1,\dotsc,a_{n-2},d)^t = (1,0,\dotsc,0)^t$. So let $U_2 = \begin{pmatrix} U'_2 & 0\\ 0 & 1\end{pmatrix}$. By construction we have $U_2 \in {\rm GL}_n(A)$ and $U_2 \cdot (a_1,\dotsc,a_{n-2},d,0)^t = (1,0,\dotsc,0)^t$. Hence $U_2\cdot U_1 \cdot (a_1,\dotsc,a_n)^t = U_2 \cdot (a_1,\dotsc,a_{n-2},d,0)^t = (1,0,\dotsc,0)^t$ and $U_2 \cdot U_1\in {\rm GL}_n(A)$. This ends the induction.

Now, if you take $U\in {\rm GL}_n(A)$ such that $U\cdot (a_1,\dotsc,a_n)^t = (1,0,\dotsc,0)^t$, then $(a_1,\dotsc,a_n)^t$ is actually the first column in $U^{-1}$ (since otherwise, replacing the first column in $U^{-1}$ by $(a_1,\dotsc,a_n)^t$ yields another inverse for $U$, a contradiction to the uniqueness of the inverse). At last, $(U^{-1})^t$ has the desired form.

$\endgroup$
4
$\begingroup$

This property holds for any PID $A$ and $a_1,\dots,a_n\in A$ such that $(a_1,\dots,a_n)=A$.

The Smith Normal Form of the matrix $(a_1\dots a_n)$ is $(1\ 0 \dots 0)$ (why?). So, there is an invertible matrix $U\in M_n(A)$ such that $(a_1 \dots a_n)=(1\ 0 \dots 0)U$, and that's all.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy