Reduction of the Degree of a Curve by a Substitution Let $y^2=P_{2n}(x)$ be an (hyper)elliptic curve, where $P_{2n}$ is a polynomial of degree $2n.$
It is said that the substitution $$x=x_1^{-1}+\alpha\qquad \text{and} \qquad y=y_1x_1^{-n}$$ reduces the degree of the curve to some $y_1^2=P_{2n-1}(x_1).$ Here $\alpha$ is a zero of the polynomial $P_{2n}.$
I do not understand the rationale behind this substitution. Can somebody explain to me what is happening here? Is there a geometric meaning for this substitution?
 A: TL;DR: For an even model hyperelliptic curve, the line at infinity $Z=0$ intersects the curve in two distinct points. For an odd model, the line at infinity intersects the curve in just one point, and is tangent to the curve at this point. The substitution you've given maps a Weierstrass point $(\alpha, 0)$, at which the vertical line $x=\alpha$ is tangent to the curve, to infinity, and maps the vertical line $x = \alpha$ to the line at infinity. This ensures that we obtain an odd model.

First, the complete model of a hyperelliptic curve $C$ given by $y^2 = P_{2n}(x)$ or $y^2 = P_{2n-1}(x)$ naturally lives in the weighted projective space $\mathbb{P}(1,n,1) = \mathbb{P}(1,g+1,1)$, where $g$ is the genus of $C$. (If one simply tries to take the closure in $\mathbb{P}^2$, there is a singularity at infinity; see this chapter (Waybackup) by Galbraith for more on weighted projective models.) This already explains to some extent the factor of $1/x_1^n$ in the expression for $y$, as in $\mathbb{P}(1,n,1)$ we have $[X:Y:Z] = [\lambda X : \lambda^n Y : \lambda Z]$ for all $\lambda \neq 0$.
Given a curve with affine equation $y^2 = P_{2n}(x)$, where
$$
P_{2n}(x) = a_{2n} x^{2n} + a_{2n-1} x^{2n-1} + \cdots + a_2 x^2 + a_1 x + a_0 \, ,
$$
we can compute its projective closure in $\mathbb{P}(1,n,1)$ as follows. Writing $X,Y,Z$ for the weighted projective coordinates, then $x = X/Z$ and $y = Y/Z^n$ on the affine open where $Z \neq 0$, so the affine equation above becomes
\begin{align*}
\frac{Y^2}{Z^{2n}} = \left(\frac{Y}{Z^n}\right)^2 = a_{2n} \frac{X^{2n}}{Z^{2n}} + \cdots + a_2 \frac{X^2}{Z^2} + a_1 \frac{X}{Z} + a_0 \, .
\end{align*}
Multiplying both sides by $Z^{2n}$, we find that the closure of $C$ in $\mathbb{P}(1,n,1)$ is given by
$$
Y^2 = a_{2n} X^{2n} + a_{2n-1} X^{2n-1} Z + \cdots + a_2 X^2 Z^{2n-2} + a_1 X Z^{2n-1} + a_0 Z^{2n} \, .
$$
To compute the curve's intersection with the line at infinity $Z = 0$, we set $Z = 0$, which yields $Y^2 = a_{2n} X^{2n}$. Thus $Y = \pm \sqrt{a_{2n}} X^n$, so
$$
[X:Y:Z] = [X : \pm \sqrt{a_{2n}} X^n : 0] = [1 : \pm \sqrt{a_{2n}} : 0]
$$
and we find two points at infinity.
If instead we have an odd model given by $y^2 = P_{2n-1}(x)$, then the projective closure is given by
$$
Y^2 = a_{2n-1} X^{2n-1} Z + a_{2n-2} X^{2n-2} Z^2 + \cdots + a_2 X^2 Z^{2n-2} + a_1 X Z^{2n-1} + a_0 Z^{2n} \, .
$$
Since every term on the right has a factor of $Z$, setting $Z = 0$ yields $Y^2 = 0$, so $[X:Y:Z] = [1:0:0]$ is the only point at infinity.
Suppose that $C$ is given by $y^2 = f(x)$, where $f$ has degree $2n$. Replacing $x$ by $x+\alpha$, we may assume that $f(0) = 0$, so $(0,0)$ is a Weierstrass point of the curve. Then $f$ has the form
$$
f(x) = a_{2n} x^{2n} + a_{2n-1} x^{2n-1} + \cdots + a_2 x^2 + a_1 x \, .
$$
Note the lack of constant term!
As before, we find that the closure of $C$ in $\mathbb{P}(1,n,1)$ is given by
$$
Y^2 = a_{2n} X^{2n} + \cdots + a_2 X^2 Z^{2n-2} + a_1 X Z^{2n-1} \, .
$$
We now dehomogenize with respect to $X$, letting $u = Z/X$ and $v = Y/X^n$ on the affine open where $X \neq 0$. This amounts to applying the transformation $[X:Y:Z] \mapsto [Z:Y:X]$, which interchanges the line $X = 0$ and the line at infinity $Z = 0$ as described in the first paragraph. Dividing both sides of the above equation by $X^{2n}$ yields
\begin{align*}
v^2 &= \left(\frac{Y}{X^n}\right)^2 = a_{2n} + \cdots + a_2 \frac{Z^{2n-2}}{X^{2n-2}} + a_1 \frac{Z^{2n-1}}{X^{2n-1}} = a_{2n} + \cdots + a_2 u^{2n-2} + a_1 u^{2n-1}
\end{align*}
which is an odd, model, as desired. Note that
\begin{align*}
u &= \frac{Z}{X} = \frac{1}{x}\\
v &= \frac{Y}{X^n} = \frac{Y}{Z^n} \, \frac{Z^n}{X^n} = y \frac{1}{x^n}
\end{align*}
which is the substitution you gave in the question.
Here's an illustration for the curve $y^2 = x^6 - x$, for which the corresponding odd model is $v^2 = 1 - u^5$. The point $(0,0) = [0:0:1]$ is mapped to the point at infinity $[1:0:0]$ and red vertical line $X = 0$ is mapped to $Z = 0$.
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