This is from example II.4.6 of Joseph Silverman's Arithmetic of Elliptic curves. Let $e_1,e_2$ and $e_3$ be distinct points and let $C$ be the curve $$y^2=(x-e_1)(x-e_2)(x-e_3)$$
Then Silverman claims that div$(dx)$ = $(P_1)+(P_2)+(P_3)-3(P_\infty)$, where $P_i=(e_i,0)$ and $P_\infty$ is the point at infinity.
I have not been able to verify this. I know that $y$ is a uniformizer at $P_i$ and $\frac{1}{y}$ is a uniformizer at $P_\infty$. To calculate the $Ord_{P_i}$ we must first find an expression for $dx$ in the form
$$dx = g\ dy,\ \ g\in K(C)$$
Then, by definition, $Ord_{P_i}(dx)=Ord_{P_i}(g)$. However, the expression I obtain to $g$ is
$$g= \frac{2}{(x-e_1(x-e_2)+(x-e_1)(x-e_3)+(x-e_2)(x-e_3)} $$
whose ordinal at each $P_i$ should be 0 and at $P_\infty$ is 2.
I'm also unsure about computing div $d(x/y)$.
Any help in understanding this example is greatly appreciated
Best Answer
I’m assuming that $6$ is invertible in the base field. Let $P(x)=(x-e_1)(x-e_2)(x-e_3)$.
It’s easy to see that the correct value of $g$ (as pointed out by Ted) has vanishing order $1$ at each $P_i$, so the vanishing order of $dx$ at these points is $1$. At $P_{\infty}$, the correct uniformizer is $x/y$ and $d(x/y)=\frac{ydx-xdy}{P(x)}=dx \times \frac{y-x/g}{P(x)}:= fdx$.
It’s not hard to see that the vanishing orders of $x,y$ at infinity are $-2,-3$ respectively, and that the vanishing order of $g$ at infinity is that of $\frac{y}{x^2}$ so it’s $1$. So the vanishing order of $x/(g)$ at infinity is $-3$, exactly that of $y$, so we need to show that the vanishing order of $y-xg^{-1}$ at infinity is still $-3$.
In other words, we need to show that the vanishing order at infinity of $yg-x$ is $-2$, but $yg-x=\frac{2y^2-xP’(x)}{P’(x)}=\frac{x^3+O(x^2)}{P’(x)}$, which concludes.