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.

  • 1
    $\begingroup$ 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 $O=[0:1:0]$. $\endgroup$ Sep 6, 2020 at 17:51

2 Answers 2



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:

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


  • $v_\theta(f(X)) = 3 + v_\theta(t) - v_\theta(u) = 1$ (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


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


as desired.


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