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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?

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    $\begingroup$ A bit of an easier argument: If $x \in M_P^3$ then so is $y^2$. But then $\mathrm{ord}_P(y^2) \geq 3$ which implies $\mathrm{ord}_P(y) \geq 3/2$, contradicting $\mathrm{ord}_P(y) = 1$. Similarly, if $2y^2 - x \in M_P^3$, then $2y^2 - x - x^3 = y^2\in M_P^3$ and you can use the same argument. $\endgroup$
    – Rushy
    Nov 17, 2021 at 14:50
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    $\begingroup$ Does this answer your question? Definition and example of "order of a function at a point of a curve" $\endgroup$ Nov 17, 2021 at 14:54

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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 $ord_P(2y^2-x)=2$.

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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.

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  • $\begingroup$ Thanks! May I ask you to clarify the meaning of $v_P$ (is it the valuation?) and the nonarchimedean triangle inequality? $\endgroup$ Nov 17, 2021 at 17:55
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    $\begingroup$ Yes, my $v_P$ is your $\operatorname{ord}_P$. Nonarchimedean triangle inequality says that $v_P(a+b) \geq \min(v_P(a), v_P(b))$, with equality if $v_P(a)\neq v_P(b)$. $\endgroup$
    – WhatsUp
    Nov 17, 2021 at 18:46

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