Lemma 15.124.10. 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 15.124.8 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 $1$. $\square$

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