Help with example about divisors of differentials

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?

• Do you understand why $\text{div}(d(x-e_i)) = (P_i)-(P_\infty)$? – Somos May 31 '19 at 12:15
• @Somos actually not. Can you help me? – jbuser430 May 31 '19 at 14:44

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