Rational functions on elliptic curves

Recall that an elliptic curve over a field $$k$$ i.e a proper smooth connected curve of genus $$1$$ equipped with a distinguished $$k$$-rational point, I'll be really grateful for any help in understanding the following part of our course

Let $$(E,0)$$ be an elliptic curve, using Riemann-Roch we construct an isomorphism into $$\operatorname{Proj}\,k[X,Y,Z]/Y^2Z+a_1XYZ+a_3YZ^2-X^3-a_2X^2Z-a_4XZ^2-a_6Z^3$$ that can be written informally as $$P\rightarrow [x(P):y(P):1(P)]$$, where $$x$$ and $$y$$ are rational functions such that $$v_0(x)=-2$$ and $$v_0(y)=-3$$.

Why does $$0$$ map to the infinity point $$O=[0:1:0]$$? According to Hartshorne it is because both $$x$$ and $$y$$ have poles in $$0$$ but I can't see why.

• In this affine chart we have $P\mapsto [X/Y:1:Z/Y]$. Here $X/Y=(X/Z)/Y/Z)=x/y$. As $y$ has a higher order pole at $O$, it follows that $x/y=0$ at $O$. Because $y=Y/Z$ has a pole of order three there, it follows that $Z/Y$ has a zero. Hence $O=[0:1:0]$. Sep 6, 2020 at 17:51

Prelude

Throughout, I will refer to the distinguished point of $$E$$ as $$\theta$$ instead of $$0$$.

It is important to understand that your rational map $$p$$ is simply not defined at $$\theta$$, which is what you likely mean by saying that the morphism $$\varphi:E\to\Bbb P^2$$ can "informally" be written as follows: $$P \mapsto [x(P):y(P):1]$$ I personally think of this scenario in two ways:

• A continuous extension of this rational map to $$\theta$$. This is a more geometric approach and my below explanation is not rigorous (due to the field being completely general here), but I think it is helpful.
• For a rigorous approach, you have to use a different rational representation of $$\varphi$$ around $$\theta\in E$$ and prove that it:
• has the desired property and
• is compatible with the one that you use everywhere else on the curve.

Geometric Intuition

Let us think of $$k$$ as a continuous and friendly field, like $$\Bbb C$$. Let's imagine $$E$$ as being embedded into some projective space $$\Bbb P(V)$$. For a point $$p\in V$$, we use the notation $$[p]\in E$$ if the projectivization of this point lies on the curve. With this notation, you have $$[\lambda p]=[p]$$ for any $$\lambda\in k^\times$$. Let $$p\in V$$ be such that $$\theta=[p]$$ is the distinguished point of $$E$$. Note that: \begin{align*} \varphi([p]) &= [x([p]):y([p]):1] \\ & = [x([\lambda p]):y([\lambda p]):1] \\ &= [\lambda^{-2}\cdot x([p]): \lambda^{-3}\cdot y([p]) : 1] \\ &= [\lambda x([p]): y([p]) : \lambda^3] \end{align*} Because $$x$$ and $$y$$ have the aforementioned order at $$[p]$$. Now if you let $$\lambda$$ approach zero in this expression on the right (which is constant!), the point approaches $$[0:1:0]$$. This can be turned into a formal proof if your field is actually continuous, but I will not spend a lot of time on it because we don't have or need this assumption here.

Rigorous Approach

To prove this properly, we need to understand the morphism globally, and provide a different rational representation around $$\theta$$. The morphism $$\varphi$$ corresponds to a morphism of (function) fields $$\varphi^\ast:k(X,Y,Z)\to k(E)$$. The distinguished point $$\theta$$ is a maximal ideal in $$k[E]$$ and we can write

• $$\varphi^\ast(X) = x = \frac ab$$
• $$\varphi^\ast(Y) = y = \frac cd$$

While $$\zeta := \varphi^\ast(Z)$$ is some element that is not in $$\theta$$, we have $$a,b,c,d\in\theta$$ and as per the assumption:

• $$v_\theta(a) - v_\theta(b) = -2$$
• $$v_\theta(c) - v_\theta(d) = -3$$

Now let's define an a-priori different rational map $$f:k(X,Y,Z)\to k(E)$$ as the composition of $$\varphi^\ast$$ with the multiplication by $$\lambda^3$$ where $$\lambda$$ is a function with $$v_\theta(\lambda)=1$$. In other words:

• $$f(X) := \frac{\lambda^3 a}b$$
• $$f(Y) := \frac{\lambda^3 c}d$$
• $$f(Z) := \lambda^3\zeta$$

Hence:

• $$v_\theta(f(X)) = 3 + v_\theta(t) - v_\theta(u) = 1$$ (first coordinate vanishes)
• $$v_\theta(f(Y)) = 3 + v_\theta(v) - v_\theta(w) = 0$$ (second coordinate doesn't)
• $$v_\theta(f(Z)) = 3$$ (third coordinate vanishes)

With $$\psi:=f^\ast$$, this implies $$\psi(\theta)=[0:1:0]$$. Now we are left to verify that the rational functions $$\psi$$ and $$\varphi$$ are the same, which would prove that $$\psi$$ is an extension of $$\varphi$$ to $$\theta$$ which proves what we desire.

This is fairly straightforward: We only need to check that they agree on an open subset. For this subset, simply choose one where all of the functions $$\zeta,\lambda,a,b,c,d$$ are nonzero and you will get that the projective coordinates of $$\psi(P)$$ and $$\varphi(P)$$ differ by the nonzero scalar factor $$\lambda^3(P)$$, and so they are identical.

I think I explained this to you in this post. The point is that if you have an $$(n+1)$$-tuple $$S:=(s_0,\ldots,s_n)$$ of globally generating global sections of a line bundle $$\mathscr{L}$$ on a $$k$$-scheme globally generating $$\mathscr{L}$$ then one gets a map $$F_S:X\to\mathbb{P}^n_k$$ given by $$F_S(x):=[s_0(x),\ldots,s_n(x)]$$

What does this mean concretely at the level of $$k$$-points? Note that one has an isomorphism

$$\mathscr{L}_x\cong \mathcal{O}_{X,x}$$

as an $$\mathcal{O}_{X,x}$$-module and this isomorphism is actually unique up to scaling by $$\mathcal{O}_{X,x}^\times$$. One then gets an induced isomorphism

$$\mathscr{L}_x/\mathfrak{m}_x\mathscr{L}\cong \mathcal{O}_{X,x}/\mathfrak{m}_x\mathcal{O}_{X,x}=k(x)$$

where $$k(x)$$ is the residue field. Let's pretend that $$x$$ is a $$k$$-point so that $$k(x)=k$$. Note that this isomorphism is well-defined up to multiplication by $$k^\times$$. Thus, from $$s_0,\ldots,s_n\in\mathscr{L}(X)$$ one obtains an element

$$(s_0(x),\ldots,s_n(x))\in k^{n+1}$$

where $$s_i(x)$$ is shorthand for the image of $$s_i$$ under the composition

$$\mathscr{L}(X)\to \mathscr{L}_x/\mathfrak{m}_x\mathscr{L}_x\cong k$$

Note that this map is only well-defined up to scalar multiplication and so

$$(s_0(x),\ldots,s_n(x))\in k^{n+1}$$

is only well-defined up to scalar multiplication. Moreover, this tuple is not zero for one (equivalently) any choice of isomorphism by the assumption that $$S$$ is globally generating. Thus,

$$(s_0(x),\ldots,s_n(x))\in k^{n+1}$$

defines an element in $$\mathbb{P}^n_k(k)$$ independent of the choice of isomorphism. This is what we're denoting by $$F_S(x)$$.

Let us now assume that $$X$$ is some smooth curve and define for any element $$s\in\mathscr{L}(x)$$ its valuation $$v_{X,\mathscr{L}}(s)$$ as follows. Consider the composition

$$\mathscr{L}(X)\to \mathscr{L}_x\to\mathcal{O}_{X,x}$$

Then, the image of $$s$$ under this map is not well-defined but it is well-defined up to $$\mathcal{O}_{X,x}^\times$$ which, in particular, means that it has a well-defined valuation since $$\mathcal{O}_{X,x}$$ is a DVR. Note then that if we have a collection $$s_i(x)=0$$ iff $$v_{x,\mathscr{L}}(s_i)>0$$.

Now, for the case of $$X=E$$ an elliptic curve you are considering the line bundle $$\mathscr{L}=\mathcal{O}(3p)$$. What then is the isomorphism of $$\mathscr{O}_{X,x}$$-modules

$$\mathscr{L}_p\to \mathcal{O}_{E,p}$$

but the map which multiplies an element of $$\mathscr{L}_p$$ by $$\pi^3$$ where $$\pi$$ is uniformizer of $$\mathcal{O}_{E,p}$$. Note then that if you're thinking of $$1,x,y\in\mathcal{O}(3p)(E)$$ as such that $$v_p(1)=0$$, $$v_p(x)=-2$$, and $$v_p(y)=-3$$ IN THE SENSE OF RATIONAL FUNCTIONS then under our isomorphism multiplication-by-$$\pi^3$$ they have valuation $$3$$, $$1$$, and $$0$$ respectively. This means that $$v_{p,\mathcal{O}(3p)}(1)=3$$, $$v_{p,\mathcal{O}(3p)}(x)=1$$, and $$v_{p,\mathcal{O}(3p)}(y)=0$$. From this, we see that under the map

$$F_S:E\to \mathbb{P}^2_k$$

with $$S=(x,y,1)$$ we have that

$$F_S(p)=[x(p):y(p):1(p)]=[0:c:0]$$

where $$c\ne 0$$. But, this then means that

$$F_S(p)=[0:1:0]$$

as desired.