Exercise 1.2.40 ($x_{i-1} y_i - x_i y_{i-1} = (-1)^i$)

Question

With the $x_i$ and $y_i$ determined as in Problem 39, show that $x_{i-1} y_i - x_i y_{i-1} = (-1)^i$ for $i = 0, 1,2,\ldots,j+1$. Deduce that $(x_i,y_i) = 1$ for $i = -1, 0,1,\ldots, j+1$.

Richard Ganaye · Accepted Answer

\begin{proof} The relations between $x_i$ given in Problem 39 are, for $-1 \leq i \leq j-1$,
\begin{align}
x_{i+2} &= x_i - q_{i+2} x_{i+1},\
y_{i+2} &= y_i - q_{i+2} y_{i+1}.
\end{align}
This gives the equivalent equality between matrices
$$
\begin{pmatrix} x_{i+2} & x _{i+1}\ y_{i+2}& y_{i+1} \end{pmatrix}
=
\begin{pmatrix} x_{i+1} & x _{i}\ y_{i+1}& y_{i} \end{pmatrix}
\begin{pmatrix} -q_{i+2} & 1\ 1& 0 \end{pmatrix}
$$
Taking the determinants, we obtain
$$
\begin{vmatrix} x_{i+2} & x _{i+1}\ y_{i+2}& y_{i+1} \end{vmatrix}
=
\begin{vmatrix} x_{i+1} & x _{i}\ y_{i+1}& y_{i} \end{vmatrix}
\begin{vmatrix} -q_{i+2} & 1\ 1& 0 \end{vmatrix}
= - \begin{vmatrix} x_{i+1} & x _{i}\ y_{i+1}& y_{i} \end{vmatrix}.
$$
If $d_i = x_{i+1} y_i - y_{i+1} x_i = \begin{vmatrix} x_{i+1} & x _{i}\ y_{i+1}& y_{i} \end{vmatrix}$, then $d_{i+1} = - d_i$ for $-1 \leq i \leq j$.

Moreover $d_{-1} = \begin{vmatrix} x_{0} & x _{-1}\ y_{0}& y_{-1} \end{vmatrix} =  \begin{vmatrix} 0 & 1\ 1& 0 \end{vmatrix} = -1$ (see Problem 39). An easy induction gives 
$$d_k = (-1)^{k+1} d_{-1} = (-1)^k$$ for all $k \in[\![-1,j]\!]$.

In particular,
$$x_{i-1} y_i - x_i y_{i-1} = -d_{i-1} = - (-1)^{i-1} = (-1)^i, \qquad i = 0,1,2,\ldots,j+1$$
This gives a B\'ezout's relation between $x_i$ and $y_i$, thus $x_i \wedge y_i = 1$.
\end{proof}

Exercise 1.2.40 ($x_{i-1} y_i - x_i y_{i-1} = (-1)^i$)

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Exercise 1.2.40 ($x_{i-1} y_i - x_i y_{i-1} = (-1)^i$)

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