# Proving that the differential on an elliptic curve $E$ given by $\omega=\frac{dx}{y}$ is translation invariant

I'm taking a course on elliptic curves and I'm stuck on a line in a proof.

We're assuming we're in an algebraically closed field $K$ and char($K)\not=2$. We have our elliptic curve

$$E:y^2=(x-e_1)(x-e_2)(x-e_3)$$

where the $e_i$ are distinct.

We've proved that $\omega=\frac{dx}{y}$ is a regular differential on $E$.

Now for $P \in E$ we define

$\begin{array}{llll}\tau_P:&E &\rightarrow &E \\ & Q & \mapsto & P \bigoplus Q.\end{array}$

Since $\tau_P^*\omega$ is a regular differential on $E$, we have $\tau_P^*\omega=\lambda_P \omega$ for some $\lambda_P \in K^\times$. Then it is claimed that

The map $$\begin{array}{lll}E &\rightarrow & \mathbb{P}^1 \\ P & \mapsto & \lambda_P\end{array}$$ is a morphism of algebraic curves.

Now I'm not sure why this is true. If I could show it was a rational map then I'd be done, since rational maps from smooth projective curves are morphisms.

Well, so long as $P\not=0_E$ and we're not where the formulae for adding points degenerate, we have (say $P=(x_P,y_P)$)

$\tau_P^*\omega=\frac{d(x \circ \tau_P)}{y\circ \tau_P}=\left(\mbox{rational function of }x, y, x_P\mbox{ and }y_p\right) \frac{dx}{y}$

but I don't see why that rational function doesn't actually have any $x$ or $y$ dependence. Because $\omega$ is a $K$-basis for the space of regular differentials we know this rational function is in $K$. Is that it?

• If I am not mistaken, the definition of $x_P$ (resp. $y_P$) as a coordinate function is exactly the composition $x\circ\tau_P$ (resp. $y\circ\tau_P$). So the dependence on $x$ (resp. $y$) is "included" in the one on $x_P$ (resp. $y_P$). – Brenin Feb 12 '13 at 12:14
• I'm not sure what you mean, atricolf. In my post I said $P=(x_P,y_P)$. I don't want it to mean $x \circ \tau_P$ here, since I want to show that the coefficient of $\frac{dx}{y}$ is exactly a rational function of the coordinates of $P$. – porkramen Feb 12 '13 at 13:26
• I think I've got it now. For any $f(x,y)\in K(E)$ we have $f(x,y)\omega=k_{f(x,y)}\omega$ for some $k_{f(x,y)} \in K^\times$, so we can get rid of all the $x$, $y$ bits this way. Is this right? – porkramen Feb 12 '13 at 13:28
• Yes, this should work. The coefficient of $\omega$ actually is in $K$. – Brenin Feb 12 '13 at 14:40

The map $P\mapsto (\lambda_P:1)$ is a rational map because it is given by rational functions (the $\lambda_P$'s). As you pointed out in the question, it is indeed a morphism of algebraic curves. Now, since it is a morphism between nonsingular integral projective curves, it is either surjective or constant. But because $(1:0)$ is not hit, it must be constant. And you can compute, for instance, the value $\lambda_O$, which is $1$. (I named $O$ the origin $(0:1:0)\in E$.) This proves, in passing, that $\tau_P^\ast\omega=\omega$ for every $P\in E$.
• I don't know what an integral curve is. I thought you had to have a morphism for the "it is either surjective or constant" conclusion to hold. In my post I intended to ask why $P \mapsto (\lambda_P:1)$ was indeed a morphism. – porkramen Feb 12 '13 at 16:48