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Section 7.7 Raynaud 3) example?

This is just me (Alex) experimenting with the feasibility of doing an example of the case 3) of Raynauds proof.

As explained in my notes for raynaud2 the group \(D_{2\ell}\) for prime \(\ell\) is quasi-\(2\) and satisfies \(G(S) \ne G\) and has no normal \(2\)-subgroup. So it lands in the third case of Raynaud's proof.

The first step is to find tuple of generators for the group whose product is one but we defer this because in Introduction to Branched Galois Covers Hiroo Tokunang http://www.math.ac.vn/publications/vjm/VJM_33/Pdf_files_DB_2005/Bai7_Tokunaga.pdf the following model is given for a \(D_{2n}\) cover in characteristic 0

\begin{equation*} \underbrace{s_0 t_1^{2 n}-2 s_1 t_1^n t_0^n+s_1 t_0^{2 n}}_F=0 \subseteq \PP^1 \times \PP^1 \end{equation*}

this maps to \(\PP^1\) via projection onto the the first factor.

Letting \(x = s_1/s_0\text{,}\) \(y = t_0/t_1\) this becomes

\begin{equation*} 1-2 x y^n+x y^{2 n} \end{equation*}

or

\begin{equation*} 1=x(2 y^n-y^{2 n}) \end{equation*}
\begin{equation*} x ( 2y^n - 1) = y^{2n} \end{equation*}

so

\begin{equation*} x = \frac{y^{2n}}{2y^n - 1} = \frac{y^{2n}}{2y^n - 1} \end{equation*}

so the partial derivative is

\begin{equation*} ny^{n-1}(2- y^n)-ny^ny^{n-1}= 2ny^{n-1}(1 -y^{n}) \end{equation*}

this has zeroes when \(y = 0\) or \(y^{n} = 1\text{.}\) In the first case \(x= 0\text{.}\) In the second \(x= 1\text{.}\)

The branch points are

\begin{equation*} (1:1),(-1:1), (0:1) \end{equation*}

as we can take the partial derivative w.r.t \(t_0\) and \(t_1\) giving

\begin{equation*} \partder[f]{t_0} = -2n s_1 t_1^n t_0^{n-1}+2n s_1 t_0^{2 n - 1} = 2ns_1(t_0^{n} -t_1^n) t_0^{n-1} \end{equation*}
\begin{equation*} \partder[f]{t_1} = 2n s_0 t_1^{2 n - 1}-2n s_1 t_1^{n-1} t_0^{n} =2n(s_0t_1^{n} - s_1 t_0^n) t_1^{n-1} \end{equation*}

I don't understand these equations, but I do understand this one

\begin{equation*} x = y^n + \frac{1}{y^n} \end{equation*}

as this clearly has a \(D_{2n}\) worth of automorphisms, from \(y \leftrightarrow 1/y\) and \(y \mapsto \zeta _n y\text{.}\)

This can be rewritten as

\begin{equation*} xy^n = y^{2n} + 1 \end{equation*}

but for the purposes of the ramification locus take the first equation and take partials.

\begin{equation*} ny^{n-1} - n y^{-n-1} = n\frac{y^{2n} - 1}{y^{-n-1}} \end{equation*}

which is ramified for \(y^{2n} = 1\) so \(y^n = \pm 1\text{.}\) Hence

\begin{equation*} x = 1 + 1\text{ or } x = -1 - 1 \end{equation*}

giving ramification when \(x = \pm 2\text{,}\) or infinity.

\begin{equation*} xy^n = y^{2n} + 1 \end{equation*}
\begin{equation*} nxy^{n-1}- 2n y^{2n-1} = ny^{n-1}(x- 2 y^{n}) \end{equation*}

ramified if \(y = 0 \) or \(x = 2y^n\text{.}\) First case \(x = 0 + 1/0 = \infty\text{,}\) second \(x = x/2 + 2/x \implies x / 2= 2/x,\, x = \pm 2\text{.}\)