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Let $C$ be an elliptic curve defined by $y^2=(x-e_1)(x-e_2)(x-e_3)$.

My question is how to determine the order of the differential $dx$ at infinity, $ord_{\infty}(dx)$.

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The same way you find the order of any differential: use a local coordinate. This should be a fairly standard exercise; what textbook are you using? – Qiaochu Yuan May 13 '10 at 13:36
Sorry, I'm voting to close. See Silverman's Arithmetic of Elliptic Curves, page 33. The following link may not work:… – S. Carnahan May 13 '10 at 14:19
I'm reading Silverman's book on my own, I can't understand why it can be manipulated this way, and find no one to consult. So I pose it here. – Yinbang Lin May 13 '10 at 16:48
$x$ has order $-2$ at $\infty$, so if $t$ is a uniformizer there, $x= a_{-2} t^{-2} + a_{-1} t^{-1}+\cdots$ where $a_{-2}\ne0$. So what does dx look like in terms of $t$ and $dt$? – Robin Chapman May 13 '10 at 17:34

When I saw that example in Silverman for the first, it didn't understand much out of it either. I'm still not good at it, but I'm trying to solve it using theorems and examples in Silverman. However, It is possible that I don't see some trivial stuff and try to prove them. On the bright side, in the begining, not assuming any thing, has some benefits.

I want to use Proposition II.4.3.d. which says:

$\mathrm{ord}_p(fdx) = \mathrm{ord}_p(f) + \mathrm{ord}_p(x) - 1$

To compute $\mathrm{div}(x)$ We need to check all $P \in C$ and compute $\mathrm{ord}_p(dx)$. If $P = (x_0, y_0) \neq (e_i, 0)$ then $M_P = (x - x_0, y - y_0)$. However, on can see that $x - x_0$ is a uniformizer for $P$, because $x - x_0$ can generate the whole ideal. To see that, it is enough to show that it can generate $y - y_0$. We have:

$(x-x_0)f(x) = (x - e_1)(x - e_2)(x - e_3) - (x_0 - e_1)(x_0 - e_2)(x - e_3)$

For some $f(x) \in K[E]_{(x_0,y_0)}$, because $x_0$ is a root for the right hand side. Now using the curve equation we can write the right hand side as:

$(x-x_0)f(x) = y^2 - y_0^2$

Now because $y + y_0 \not \in (y - y_0, x-x_0)$ so $1/(y+y_0) \in K[E]_{(x_0,y_0)}$, so we can multiply both side by $1/(y+y_0)$ and we get:

$(x-x_0)\frac{f(x)}{y + y_0} = y - y_0$, hence $M_{(x_0, y_0)} = (x-x_0)$ and therefore $\mathrm{ord}_{(x_0, y_0)}(dx) = \mathrm{ord}_{(x_0, y_0)}(d(x-x_0)) = \mathrm{ord}_{(x_0, y_0)}(x-x_0) - 1 = 0$

Using Proposition II.4.3.d.

So we only need to compute $\mathrm{ord}_p(dx)$ for $p = (e_i, 0)$ and $p = \infty$.

We know that $M_{(e_i, 0)} = (y, x - e_i)$ However, clearly $y^2 = (x - e_i)f(x)$ for some $f(x) \in K[E]_{(e_i,0)}^*$. So, $y$ is a unifromizer for $M_{(e_i, 0)}$ and $\mathrm{ord}_{(e_i,0)}(x - e_i) = 2$. Using the same proposition we have: $\mathrm{ord}_{(e_i, 0)}(dx) = \mathrm{ord}_{(e_i, 0)}(d(x-e_i)) = \mathrm{ord}_{(e_i, 0)}(x-e_i) - 1 = 2 -1 =1$

For the point at infinity, We can use $dx = -x^2(d(1/x))$. Now from example II.3.3, We know that $ord_{\infty}(x) = -2$. So we can easily use the proposition again:

$ord_{\infty}(dx) = ord_{\infty}(-x^2d(1/x)) =$ $2ord_{\infty}(x) + ord_{\infty}(1/x) - 1 = 2 (-2) + (2) - 1 = -3$.

Therefore $\mathrm{div}(dx) = (e_1, 0) + (e_2, 0) + (e_3, 0) - 3\infty$

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