Valuation of points on curves I am reading Silverman's The Arithmetic of Elliptic Curves. I understand the definition of (normalized) valuation of a smooth point $P\in C$, where $C$ is a curve. What I don't really get is its computation.
Consider example II.1.3: let $C$ be the curve defined by $y^2=x^3+x$ and $P=(0,0)\in C$. $P$ is smooth. Now to calculate $\text{ord}_P(y),\text{ord}_P(x)$ and $\text{ord}_P(2y^2-x)$.
We have $M_P=(x,y)$. So $y\in M_P$ but $y\not\in M_P^2$ as it is no linear combination of $\{x^2,xy,y^2\}$, hence $\text{ord}_P(y)=1$. Now, $x\in M_P$, but also $x\in M_P^2$ since $x=y^2-x^3$ but how can I be sure that $x\not\in M_P^3$? The same goes for $2y^3-x^3$ (I know it belongs to $M_P^2$, but how can I be sure that it doesn't belong to $M_P^3$). Further, is there a more intuitive way of thinking about this? A faster method perhaps?
Another question of mine is: considering an elliptic curve in the Weierstrass form
$$E:F(x,y)=y^2+a_1xy+a_3y-x^3-a_2x^2-a_4x-a_6=0,$$
how do I come up with the valuations of $x$ and $y$ at infinity $\infty=(0,1,0)$?
Moving to the projective space, I would write $x=X/Z$ and then
$$\text{ord}_\infty(x)=\text{ord}_\infty(X)-\text{ord}_\infty(Z).$$
Intuitively, I would say $\text{ord}_\infty(X)=1$ because $X=0$ intersects $\infty\in E$ one time and $\text{ord}_\infty(Z)=3$ because $Z=0$ intersects $\infty\in E$ three times. But how can I show this?
 A: First: Consider $ M_p^3=(x^3, x^2y, xy^2, y^3)\supseteq M_p^2=(x^2, xy, y^2)$.
If $x\in M_p^3$, then since $x-y^2=x^3$, we get $x^3\in M_P^3$.
Hence $M_P^3=M_p^2$.
Now recall the Nakayama lemma, which will force $M_p^2=0$, which is impossible.    Hence $x\notin M_p^3$. So $ord_P(x)=2.$
Second: As you have already observed $2y^2-x\in M_P^2.$
Now $2y^2-x=2(x^3+x)-x=2x^3+x$. If $2y^2-x\in M_P^3$, then $x\in M_P^3$, but as we have already observed that is impossible.
Hence $2y^2-x\in M_P^2$, but $2y^2-x\notin M_P^3$, so $ord_P(2y^2-x)=2$.
A: You can compare the valuation at $P$ to the $p$-adic valuation on rational numbers.
To work out the example:
We have $P = (0, 0)$ on the curve $C: F(x, y) = y^2 - (x^3 + x) = 0$. The partial derivative $\frac \partial{\partial x} F$ is nonzero at $P$, hence $y$ is a local uniformizer at $P$.
You may intuitively think it as "$x$ is locally a function of $y$", by implicit function theorem. A rigorous proof can be done as follows: we have $v_P(x) > 0$ and $v_P(y) > 0$, hence $v_P(x + x^3) = v_P(x)$ by nonarchimedean triangle inequality. Therefore $v_P(x) = v_P(y^2) = 2v_P(y)$. Since $x, y$ generate the maximal ideal at $P$, it is clear that $y$ is a generator (= uniformizer).
It follows that $v_P(y) = 1$ and $v_P(x) = 2$. For $2y^2 - x$, you can write $2y^2 - x = 2y^2 - (y^2 - x^3) = y^2 + x^3$. Since $v_P(y^2) = 2$ and $v_P(x^3) = 6$, by nonarchimedean triangle inequality, we know that $v_P(2y^2 - x) = 2$.
For the case of a point at infinity, it's the same principle. Usually one starts by moving the point to $(0, 0)$ via a projective linear transform, then finds a uniformizer, then calculates the valuations.
