# 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 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. Nov 17, 2021 at 14:50
• Does this answer your question? Definition and example of "order of a function at a point of a curve" Nov 17, 2021 at 14:54

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

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.

• Thanks! May I ask you to clarify the meaning of $v_P$ (is it the valuation?) and the nonarchimedean triangle inequality? Nov 17, 2021 at 17:55
• 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 $v_P(a)\neq v_P(b)$. Nov 17, 2021 at 18:46