Isomorphism between complex tori and complex elliptic curves

Question

I'm trying to understand the following proof in Joseph H. Silverman's book "The Arithmetic of Elliptic Curves". The author is showing that the map $\phi:\mathbb{C}/\Lambda\to E(\mathbb{C})$ defined as $z\longmapsto[\wp(z),\wp'(z),1]$ is a complex analytic isomorphism of Lie groups. The part that's tripping me up is showing that the map actually is a group homomorphism. Here is how his argument goes.

Using previous propositions, he shows there exists a function $f(z)\in\mathbb{C}(\Lambda)$ such that $\text{div}(f)=(z_1+z_2)-(z_1)-(z_2)+(0)$. Since $\mathbb{C}(\Lambda)=\mathbb{C}(\wp,\wp')$, we have that there exists an $F\in\mathbb{C}(x,y)$ where $\mathbb{C}(x,y)$ is viewed as $\mathbb{C}(E)$. We then have $\text{div}(F)=(\phi(z_1+z_2)) - (\phi(z_1)) - (\phi(z_2)) + (\phi(0))$. He then invokes Corollary 3.5 which tells us that since $\text{div}(F)$ is principal, $\phi(z_1+z_2)-\phi(z_1)-\phi(z_2)+\phi(0)=O_E$, where $O_E$ is the identity on our elliptic curves. Re-arranging the terms, I get $$\phi(z_1+z_2)+\phi(0)=\phi(z_1)+\phi(z_2)$$ so we're almost done but I don't know how to show $\phi(0)=0$. I have shown that $\phi$ is bijective but without the morphism property, I don't know how to conclude. Can anyone help shed some light on this proof?

Answer 1

It helps to remember that $E$ is sitting in the projective space $\mathbb{P}^2$. You've indicated that by writing the map $\phi$ using homogeneous coordinates $\phi(z)=[\wp(z),\wp'(z),1]$. So just as when working with algebraic maps, you need to take local coordinates that make sense at your point before evaluating. So as a map on $\mathbb{C}/\Lambda$, if you want to know what $\phi$ looks like near $0\bmod\Lambda$, you can't just plug in $z=0$ since, as Angina Seng indicated, both $\wp$ and $\wp'$ have poles at $z=0$. But since $\phi$ is written using homogeneous coordinates, it is also given by the formula $$ \phi(z)=[z^3\wp(z),z^3\wp'(z),z^3] $$ at all points of $\mathbb{C}/\Lambda$ where these three coordinate functions are (1) well-defined and (2) don't all vanish. Then using $$ z^3\wp(z)\Big|_{z=0}=0, \quad z^3\wp'(z)\Big|_{z=0}=-1,\quad z^3\Big|_{z=0}=0, $$ we get $\phi(0)=[0,1,0]=\mathcal{O}$.

Answer 2

$\wp(z)\sim 1/z^2$ as $z\to0$ and $\wp'(z)\sim -2/z^3$. For small nonzero $z$, $$\phi(z)=[\wp(z),\wp'(z),1]=[z^3\wp(z),z^3\wp'(z),z^3]\to[0,-2,0]=[0,1,0]$$ as $z\to0$. So we define $\phi$ at the poles of $\wp$ by defining $\phi(0)=[0,1,0]$ in the projective plane, which is the zero point $O_E$ of the elliptic curve $E$.