Exercise 2.9.13

Question

Show that the automorphism group of the form $(1,0,-5)$ is infinite.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

\begin{proof}
By Theorem 2.5.5, the group $\mathrm{Aut}(f)$, where $f = (1,0,-5)$ is in bijective correspondence with the set (group) of solutions of the Pell equation
\begin{align}
x^2 - 20 y^2 = \pm 4.
\end{align}
Every solution of this equation is such that $x$ is even. If we write $X = x/2, Y = y$, then $(X,Y)$ is a solution of

\begin{align}
X^2 - 5 Y^2 = \pm 1,
\end{align}
and the sets of solutions of these two equations are in bijective correspondence.

The solution $(2,1)$ of equation (2) corresponds to the solution $(4,1)$ of (1).

If $(X,Y)$ is a solution of (2), so is $(2X + 5 Y, X + 2 Y)$, since 
$$(2X+5Y)^2 - 5(X+2Y)^2 = -(X^2 - 5 Y^2).$$
Define $(X_0,Y_0) = (1,0)$, and $(X_{n+1},Y_{n+1}) = (2X_n + 5 Y_n, X_n + 2 Y_n)$ for all $n \in \N$.

Then $X_n > 0, Y_n >0$ for all $n>0$, and $X_{n+1} > 2 X_n$, thus all these solutions are distinct. Therefore the sets of solutions of (2) or (1) are infinite, and so  $\mathrm{Aut}(f)$ is infinite.

The first positive solutions of (2) are 
$$(1,0), (2,1), (9,4), (38, 17) , (161, 72), \ldots$$
corresponding to the solutions
$$(2, 0), (4, 1), (18, 4), (76, 17), (322, 72),\ldots.$$
This gives the automorphisms
$$I_2, U = \begin{pmatrix} 2 & 5\ 1 & 2 \end{pmatrix}, U^2 = \begin{pmatrix} 9 & 20\ 4 & 9 \end{pmatrix}, U^3 = \begin{pmatrix} 38 & 85\ 17 & 38 \end{pmatrix}, U^4 =  \begin{pmatrix} 161 & 360\ 72 & 161 \end{pmatrix},\ldots$$
with $$U^{-1} =  \begin{pmatrix} 2 & -5\ -1 & 2 \end{pmatrix}, U^{-2}, \cdots$$
The matrices $U^n$ with $n$ even are in $\mathrm{Aut}^+(f)$, which is also infinite.
\end{proof}

Exercise 2.9.13

Answers

Comments

Exercise 2.9.13

Answers

Comments

Add answer