Lemma 15.121.10. Let $R$ be a Bézout domain. Let $n \geq 1$ and $f_1, \ldots , f_ n \in R$ generate the unit ideal. There exists an invertible $n \times n$ matrix in $R$ whose first row is $f_1 \ldots f_ n$.

Proof. This follows from Lemma 15.121.8 but we can also prove it directly as follows. By induction on $n$. The result holds for $n = 1$. Assume $n > 1$. We may assume $f_1 \not= 0$ after renumbering. Choose $f \in R$ such that $(f) = (f_1, \ldots , f_{n - 1})$. Let $A$ be an $(n - 1) \times (n - 1)$ matrix whose first row is $f_1/f, \ldots , f_{n - 1}/f$. Choose $a, b \in R$ such that $af - bf_ n = 1$ which is possible because $1 \in (f_1, \ldots , f_ n) = (f, f_ n)$. Then a solution is the matrix

$\left( \begin{matrix} f & 0 & \ldots & 0 & f_ n \\ 0 & 1 & \ldots & 0 & 0 \\ & & \ldots \\ 0 & 0 & \ldots & 1 & 0 \\ b & 0 & \ldots & 0 & a \end{matrix} \right) \left( \begin{matrix} & & & 0 \\ & A \\ & & & 0 \\ 0 & \ldots & 0 & 1 \end{matrix} \right)$

Observe that the left matrix is invertible because it has determinant $1$. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0ASW. Beware of the difference between the letter 'O' and the digit '0'.