# $P$ is a branching point of the elliptic curve $(E, P)$

In section 19.9.5 of Vakil, he says an elliptic curve $E$ with a point $P$ admits a double cover of $\mathbb{P}^1$ given by the line bundle $\mathscr{O}_E(2P)$. Then he says $P$ is a branch point because one of the sections of $\mathscr{O}_E(2P)$ vanishes at $P$ to order 2, so there is a point of $\mathbb{P}^1$ whose preimage consists of $P$ with multiplicity 2.

Here's my understanding: let $s$ be a global section of $\mathscr{O}(2P)$ that only has a zero of order 2 at $P$, and let $t$ be a global section of $\mathscr{O}_{\mathbb{P}^1}(1)$ that pulls back to $s$. Then the locus of where $s$ vanishes should be the pullback of the locus of where $t$ vanishes, which means there must be a point in $\mathbb{P}^1$ whose preimage is just $P$. Is that right?

But the definition of the global sections of $\mathscr{O}(2P)$ is $$\{t \in K(E)^{\times}: div \, t + 2P \geq 0\} \cup \{0\}$$ which means rational functions that can only have poles at $P$ of order less than equal to 2. Then can't we have a global section that only has a zero of order 2 at $Q$ for any point $Q$?

## 1 Answer

The confusion might stem from the fact that, taking the zero locus of a global section $s\in H^0(X,\mathscr{O}(D))$ is different from taking the divisor of zeros of the corresponding rational function.

As an example, consider $X=\mathbb{P}^1$ and $\mathscr{O}(p)$ for some point $p$. This is isomorphic to $\mathscr{O}(1)$, which has global sections degree 1 homogeneous polynomials. In particular, they vanish at a single point (except for the zero section). On the other hand, you can cook up rational functions $t$ such that $div t + p \ge 0$, where $t$ has a lot of zeros!

Anyway, one way to resolve your problem is to consider the Cartier divisor associated to $2P$ where you can consider $U_1 = E-P$ with a function $1$, $U_2$ to be a open containing $P$ with a function defined on $U_2$ which vanishes at $P$ to order $2$. Now if you consider the global section $1$, then it vanishes at $P$ of order $2$.

• I guess I've been having a misunderstanding there for a while... why is there a difference between the zero locus of a global section and the divisor of zeros of rational function? In Vakil, the definition of of $\mathscr{O}(p)$ is $\Gamma(U,\mathscr{O}(p)) = \{t \in K(X)^{\times} : div \, t |_U+p\geq 0\} \cup \{0\}$. Then if I take a $t$ satisfying $div \, t + p \geq 0$, $t$ should be a global section of $\mathscr{O}(p)$, right? And $div \, t$ at a point $q$ should be $v_q(t)$, which tells you whether $t$ has a zero/pole at $q$? What did I get wrong? – Aaron Johnson Jan 26 '18 at 0:25
• And about the last paragraph you wrote, is this what's going on - when you restrict the divisor $2P$ to the stalk at $P$, you still get $2P$, and this is principal since $\mathscr{O}_P$ is a UFD, so it's equivalent to $div \, f$ for some $f \in K(E)^{\times}$. $f$ should only have a zero of order 2 at $P$. On $U_2$ the Cartier divisor is generated by $1/f$. On $U_1$ it's generated by 1. Since 1 is clearly in $U_2$, these glue to a global section 1. Is this right? I still can't really understand why 1 vanishes at $P$ of order 2 though... Would be great if you could explain this more, thanks! – Aaron Johnson Jan 26 '18 at 0:32
• To answer your first question, when you talk about the zero locus of a global section of a vector bundle (locally free sheaf), you do that by taking a trivialisation (where your section now is equivalent to a map to $\mathbb{A}^n$ and then consider the corresponding zero locus). So you want to do the same here, take trivialisation and then consider the zero locus - which in general is different from simply taking the divisor of the rational function. – loch Jan 26 '18 at 1:06
• So to your second question, the answer is still the same -- take trivialisation! 1 is in both $U_1$ and $U_2$. Your sections on $U_2$ is generated by $1/f$, and the trivialisation is given by $\mathscr{O}(U_2) \rightarrow \mathscr{O}_E(2p)$ sending $1\mapsto 1/f$. The inverse map sends $1$ to $f\in \mathscr{O}(U_2)$, and $f$ vanishes at $P$ to order $2$. – loch Jan 26 '18 at 1:12
• So I was right about what the global section is, but wrong about their zeros/poles? Here we look at the global section corresponding to the rational function 1, whose restriction to both $\Gamma(U_1, \mathscr{O}_E(2p))$ and $\Gamma(U_2, \mathscr{O}_E(2p))$ is 1, and via the trivialization it corresponds to $f \in \mathscr{O}_E(U_2)$ and $1 \in \mathscr{O}_E(U_1)$, so it has a zero at $p$ of order 2. Is it true that in general, given a divisor $D$ and a rational function $t$ satisfying $div t+ D \geq 0$, this $t$ corresponds to the global section that is $t$ in each trivialization of $D$? – Aaron Johnson Jan 26 '18 at 6:27