I'm currently studying Silverman's Arithmetic in Elliptic Curves book. I hope someone can help me with example 2.4.6. Let $C$ be the curve $$C:y^2=(x-e_1)(x-e_2)(x-e_3)$$ with char$(K)\neq 2$, $e_1,e_2,e_3\in\bar{K}$ distinct. Silverman claims that $\text{div}(dx) = (P_1)+(P_2)+(P_3)-3(P_\infty)$, where $P_i=(e_i,0)$. He uses the fact that $dx=d(x-e_i)=-x^2d(1/x)$. However, I do not see where these values come from . Why is the order at $P_i$ of $d(x-e_i)$ equal to 1? and why at infinity equal to -3?

  • $\begingroup$ Do you understand why $\text{div}(d(x-e_i)) = (P_i)-(P_\infty)$? $\endgroup$
    – Somos
    May 31, 2019 at 12:15
  • $\begingroup$ @Somos actually not. Can you help me? $\endgroup$
    – jbuser430
    May 31, 2019 at 14:44

1 Answer 1


It's easier to see this using the short Weierstrass form. Write $C : y^2 = x^3 + ax + b$. Then, by calculus, $$\frac{dy}{dx} = \frac{3x^2+a}{2y}$$ By definition, $$\operatorname{div} dx = \sum_{P \in C} \operatorname{ord}_P(dx)(P)$$ and furthermore, by definition $$\operatorname{ord}_P(dx) = \operatorname{ord}_P\left(\frac{dx}{dt}\right)$$ for any choice of uniformizer $t$ at $P$. Note crucially that $t$ may depend on $P$: a particular function which qualifies as a uniformizer at some point $P$ may not qualify as a uniformizer at some other point $P'$.

The function $y-y_P$ qualifies as a uniformizer on every point $P$ of $C$ other than $\infty$ and the set of points $S \subset C$ where $\frac{dy}{dx} = 0$. Hence $$ \operatorname{div}{dx} = \left(\sum_{P \in C\setminus (\{\infty\} \cup S)} \operatorname{ord}_P \left(\frac{dx}{d(y - y_P)}\right) (P)\right) + \left(\sum_{P \in \{\infty\} \cup S} \operatorname{ord}_P (dx) (P)\right) $$ For all points $P \in C\setminus (\{\infty\} \cup S)$, we have $$ \frac{dx}{d(y-y_P)} = \frac{dx}{dy} = \frac{2y}{3x^2+a} $$ where $y$ is always non-infinite, and $3x^2+a$ is always non-infinite and nonzero. The only zeros of $\frac{dx}{dy}$ are the three points $P_1, P_2, P_3$ satisfying $y=0$, and each of these zeros is a simple zero. This explains the $(P_1)+(P_2)+(P_3)$ term in $\operatorname{div} (dx)$.

Any point $Q \in S$ has tangent line parallel to the $x$-axis. Therefore $x-x_Q$ is a uniformizer at $Q$, and $\frac{dx}{d(x-x_Q)} = \frac{dx}{dx} = 1$, so $\operatorname{ord}_{Q} (dx) = 0$.

It remains to deal with the point $\infty$. Note that $\frac{x}{y}$ is a uniformizer at $\infty$ since $$\operatorname{ord}_\infty \left(\frac{x}{y}\right) = \operatorname{ord}_\infty (x) - \operatorname{ord}_\infty (y) = (-2) - (-3) = 1.$$ By calculus, we have $$ \frac{d(\frac{x}{y})}{dx} = \frac{\frac{y\ dx - x\ dy}{y^2}}{dx} = \frac{1}{y} - \frac{x}{y^2} \cdot \frac{dy}{dx} = \frac{1}{y} - \frac{x}{y^2} \cdot \frac{3x^2+a}{2y} = \frac{2y^2 - 3x^3 - ax}{2y^3} = \frac{2b+ax-x^3}{2y^3} $$ (where the last equality uses the substitution $y^2 \mapsto x^3 + ax + b$ in the numerator). Hence $$ \frac{dx}{d(\frac{x}{y})} = \frac{2y^3}{2b+ax-x^3} $$ and $$\operatorname{ord}_\infty (dx) = \operatorname{ord}_\infty (y^3) - \operatorname{ord}_\infty (x^3) = 3 (\operatorname{ord}_\infty (y) - \operatorname{ord}_\infty (x)) = 3((-3) - (-2)) = -3.$$ QED

  • $\begingroup$ I have seen several answers of yours about computing differentials on curves. All of them are really wonderful. $\endgroup$
    – Yang Pei
    Dec 11, 2020 at 22:16
  • 1
    $\begingroup$ For the point $\infty$, I think we can directly know the order is $-3$, since the sum of these coefficients are $0$ for a meromorphic differential. $\endgroup$
    – Yang Pei
    Dec 11, 2020 at 22:26

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