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In Silverman's Book on elliptic curves, in example 4.6, he says:

Let $C:y^2=(x-e_1)(x-e_2)(x-e_3)$ then $div(dx) = div(d(x-e_i)) = div(-x^2d(\frac{1}{x}))$. He concludes that $div(dx) = (P_1) + (P_2) + (P_3) - 3(P_{\infty})$ where $P_i$ corresponds to the point $(e_i,0)$

Now here's what I don't understand. $x-e_i$ should be a uniformizer for $P_i$. Hence, $ord_{P_i}(dx) = ord_{P_i}(\frac{dx}{d(x-e_i)}) = ord_{P_i}(\frac{dx}{dx})$ This feels like it should definetly be $0$. Further, everything I've been led to beleive up to this point would tell me that $ord_\infty(dx) = -2$. Even in his previous example over the projective line this was the case.

Now, logically - since $e_1, e_2, e_3$ appear once each as a factor of the RHS of our curve, I'm inclined to believe Silverman. Also, since the genus is 1, its to be expected that the divisor is 0.

What am I missing?

gradintherockies
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    The curve has vertical tangents at $P_1,P_2,P_3$, so $x-e_i$ is NOT a uniformizer at $P_i$. At a smooth point $P=(x_0,y_0)$ of a curve the difference $x-x_0$ is a uniformizer, unless the tangent is vertical, and $y-y_0$ is a uniformizer unless the tangent is horizontal. – Jyrki Lahtonen Apr 10 '20 at 04:07
  • This might help https://math.stackexchange.com/questions/511950/calculating-the-divisors-of-the-coordinate-functions-on-an-elliptic-curve?rq=1 – Álvaro Lozano-Robledo Apr 10 '20 at 17:50
  • Related: https://mathoverflow.net/questions/24501, https://math.stackexchange.com/questions/3246218/, https://math.stackexchange.com/questions/660158/ – Watson Nov 23 '20 at 08:41

1 Answers1

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$x-e_1$ is not a uniformiser at $P_1$ as it has a double zero there: note that $$y^2=(x-e_1)(x-e_2)(x-e_3)=x^3+Ax^2+BX+C\tag1$$ so that $$\text{ord}_{P_1}(x-e_1)=2\text{ord}_{P_1}(y)-\text{ord}_{P_1}(x-e_3)-\text{ord}_{P_1}(x-e_3)=2\text{ord}_{P_1}(y).$$ Actually $y$ is a uniformiser.

Differentiating $(1)$ gives $$2y\,dy=(3x^2-2Ax+B)\,dx.$$ Then $$\frac{dx}{dy}=\frac{2y}{3x^2-2Ax+B}$$ which has a simple zero at $P_1$ ($y$ has a simple zero and the denonomiator is nonzero). So $\text{ord}_{P_1}(dx)=1$.

I've assumed that $2\ne0$.

Angina Seng
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