# Example II.4.6 from Joseph Silverman's Elliptic Curves

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

• In $g$, the numerator should be $2y$ not 2.
– Ted
Oct 29, 2021 at 16:52
• Thank you, this silly mistake being illuminated for me makes this example clear. Nov 2, 2021 at 13:53

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.