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Consider the elliptic curve $y^2=(x-\alpha)(x^2+ax+b)=x^3+(a-\alpha)x^2+(b-a\alpha)x-\alpha b$ over the field $K$ with $\text{char}\ K\not= 2$.

The questions I am doing asks for a formula for the $x$-coordinate of $(x_1,y_1)+(\alpha ,0)$ in the form $\dfrac{t_{11} x+t_{12}}{t_{21}x+t_{22}}$ where $t_{ij}$ are expressed in terms of $a,b,\alpha$.

By brute force I find the $x$-coordinate to be $$\left(\dfrac{y_1}{x_1-\alpha}\right)^2-a-x_1=x_1^4+2ax_1^3+(a^2+2b)x_1^2+(2ab-1)x_1+b^2-a$$

but I have no idea how to arrive at a form of a Möbius transformation. Any ideas?

Andrew
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1 Answers1

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The $x$ coordinate is a branched map $E \to P^1$ and makes $P^1$ into the quotient of $E$ by its automorphism $(P \mapsto -P)$.

If you know the $x$ coordinate of $P$, this is the same thing as knowing $\{P, -P\}$. Let $Q = (\alpha,0)$. Since $Q= -Q$, $\{P+Q,-P+Q\} = \{P+Q,-(P+Q)\}$ so the addition by $Q$ induces a map $f : P^1 \to P^1$ taking the $x$ coordinate of $\pm P$ and returning the $x$ coordinate of $\pm(P+Q)$.

Furthermore, that map is obviously an involution, so it is in $Aut(P^1)$, which means it is a Mobius transformation.

Now we can look at what $f$ is on the points of order $2$ (i.e. $x = \infty,\alpha$, and the roots of $x^2+ax+b$). Clearly, $f$ switches $\infty$ with $\alpha$ and it switches the other two roots.

Since $f(\alpha) = \infty$ and $f(\infty) = \alpha$, $f$ must be of the form $x \mapsto (\alpha x + t) / (x - \alpha)$ for some expression $t$ depending on $\alpha,a,b$.

Let $\beta_1,\beta_2$ be the other two roots. Then $-a = \beta_2 + \beta_1 = f(\beta_1)+f(\beta_2) = \frac{2\alpha\beta_1\beta_2+(t-\alpha^2)(\beta_1+\beta_2)-2\alpha t}{\alpha^2-\alpha(\beta_1+\beta_2)+\beta_1\beta_2} = \frac{2\alpha b+\alpha^2a-t(a+2\alpha)}{\alpha^2+a\alpha+b}$

Hence $t = \frac{2a\alpha^2+a^2\alpha+ab+2\alpha b}{a+2\alpha} = a\alpha+b$

mercio
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