I'm going through Example II.3.3 from Silverman.
Assume $\text{char}(K)\neq 2$. Let $e_1,e_2,e_3\in \overline{K}$ be distinct, and consider the curve $$ C:y^2 = (x-e_1)(x-e_2)(x-e_3) $$. Let $P_i = (e_i,0)$ and $P_\infty = [0:1:0]$ be points in $C(\overline{K})$. Then $$ \text{div}(x-e_i) = 2(P_i) – 2(P_\infty) $$ $$ \text{div}(y)=(P_1)+(P_2)+(P_3)-3(P_\infty) $$
I can easily show that $\text{ord}_{P_i}(x-e_i) = 2$ and $\text{ord}_{P_i}(y)=1$ for all $i=1,2,3$. I am stuck when trying to show that $\text{ord}_{P_\infty}(x-e_i)=-2$. I want to do this calculation directly using uniformizers, and not using the Proposition 3.1(b) stating that $\text{deg}(\text{div}(f))=0$.
I first change the elliptic curve into the chart containing $P_\infty$.
$$ 0=(x-e_1z)(x-e_2z)(x-e_3z)-z=f(x,z) $$
Let $R=\overline{K}[x,z]/(f)$ be the coordinate ring and $M_{P_{\infty}} = (x,z)$ be the maximal ideal corresponding to $(x,z)=(0,0)$. Now, $\text{ord}_{P_\infty}(x-e_i)$ becomes $\text{ord}_{P_\infty}(x-e_iz)$ and it is clear that $$ x-e_iz\in M_{P_\infty} – M_{P_\infty}^2 $$ since it is homogeneous of degree one, and hence $\text{ord}_{P_\infty}(x-e_iz) = 1$. But we were expecting $\text{ord}_{P_\infty}(x-e_iz) = -2$ instead. Where did I go wrong?
Best Answer
You really shouldn't call the function $x$ in your chart at infinity. In terms of homogeneous coordinates, we have $C: Y^2 Z = (X - e_1 Z)(X - e_2 Z)(X - e_3 Z)$, so $x = X/Z$ and $y = Y/Z$. At infinity, where $Y \neq 0$, we instead take $w = X/Y$ (which you have again called $x$) and $z = Z/Y$ as affine coordinates, so $C$ is given by $$ C: z = (w - e_1 z)(w - e_2 z)(w - e_3 z) \, . $$ In fact, $w - e_i z$ does indeed vanish to order $1$ at $P_\infty$, but this is not the same function as $$ x - e_i = \frac{X}{Z} - e_i = \frac{X/Y}{Z/Y} - e_i = \frac{w}{z} - e_i \, . $$ Can you show that this function has a double pole at $(w,z) = (0,0)$? Note that $$ z = (w - e_1 z)(w - e_2 z)(w - e_3 z) \implies \frac{w}{z} - e_1 = \frac{1}{(w - e_2 z)(w - e_3 z)} $$ as was more or less suggested in the comments.