1
$\begingroup$

I cannot understand the definition of the order function in "Joseph H. Silverman, The Arithmetic of Elliptic Curves", which is as follows:

$$\text{ord}_p: \bar K[C]_p \to \{ 0,1,\dots \} \cup \{ \infty \}$$

where $C$ is a curve and $$\text{ord}_p(f) = \sup \{ d \in \mathbb N : f \in M^{d} _p \} , \\ M_p = \{ g \in \bar K [C] : g(p) = 0 \} .$$

How can we compute the order of a function at a point, with the above definition? Give an example.

$\endgroup$
3
  • $\begingroup$ There is an example (Example 1.3, p. 18) immediately following this definition in Silverman's book. Why don't you take a look at it and then ask a question about that example if you have trouble understanding it? $\endgroup$ Nov 6, 2016 at 20:17
  • $\begingroup$ As I mentioned in comments, I didn't understand that example. How can I understand it when I have not any idea about the definition of $ord_{P}(f)$? $\endgroup$
    – user386269
    Nov 7, 2016 at 5:35
  • $\begingroup$ I just wanted some clarification on which part of the example confused you. Is it the definition of the local ring? The properties of a discrete valuation? How to find a uniformizer for the maximal ideal? Anyway, I've written a pretty extensive answer, so hopefully will be helpful. $\endgroup$ Nov 7, 2016 at 6:21

1 Answer 1

11
$\begingroup$

Okay, let's look at Example 1.3 from the text: consider the affine curve $E: y^2 = x^3 + x$. This curve is smooth: letting $f(x,y) = y^2 - (x^3 + x)$, one can show there is no point where $f$ and both partial derivatives $f_x$ and $f_y$ vanish simultaneously.

Recall that the local ring $\overline{K}[E]_P$ can be defined as all rational functions on $E$ that are defined at the point $P$, i.e., with denominators that don't vanish at $P$. This ring is local hence has a unique maximal ideal $\mathfrak{m}_P$, which consists of all rational functions on the curve that vanish at $P$.

Since the curve is smooth, then the local ring $\overline{K}[E]_P$ is a discrete valuation ring for each point $P \in E(\overline{K})$. This means that the unique maximal ideal $\mathfrak{m}_P$ of $\overline{K}[E]_P$ is generated by a single element $t$, called a uniformizer. Usually, we pick $t$ to be some line that is not tangent to the curve at $P$: we don't want the tangent line because it "vanishes to order 2" at $P$. Once we've found a uniformizer, our lives are easier: in order to determine $\text{ord}_P(f)$ of a function $f$, we must find the highest power $d$ such that $f \in \mathfrak{m}_P^d$. But this is just the number of factors of $t$ in $f$: if $\text{ord}_P(f) = d$, then we can write $f = u t^d$ for some unit $u \in \overline{K}[E]_P$.

Let's consider the point $P = (0,0)$ on $E$, so $\mathfrak{m}_P = (x,y)$. We want to find a uniformizer for $\mathfrak{m}_P$. At the point $P = (0,0)$, the tangent line to $E$ is the line $x = 0$: one can see this because it is the lowest order term appearing in the "Taylor expansion" $0 = x^3 - y^2 + x$, or because $x = y^2 - x^3 \in \mathfrak{m}_P^2 = (x^2, xy, y^2)$. So instead, we can take $y$ as our uniformizer and $\mathfrak{m}_P = (x,y) = (y)$. Since $y$ has only one factor of $y$, then $\text{ord}_P(y) = 1$. Here are two ways to compute $\text{ord}_P(x)$:

1) Since $y^2 = x^3 + x = (x^2 + 1)x$, then we can write $x = \frac{y^2}{1+x^2}$. Since $1 + x^2$ doesn't vanish at $(0,0)$, then it is a unit, hence has order $0$ at $P$. Then we have $x = u y^2$, where $u = \frac{1}{1+x^2}$, so $\text{ord}_P(x) = 2$.

2) We have $x = y^2 - x^3$, so $$ \text{ord}_P(x) = \text{ord}_P(y^2 - x^3) = \min\{\text{ord}_P(y^2), \text{ord}_P(x^3)\} = 2 $$ by the second property of the discrete valuation, since $\text{ord}_P(x^3) \geq 3$.

Now that you've seen these two examples, see if you can show $\text{ord}_P(2y^2 - x) = 2$ as stated in the text.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy