0
$\begingroup$

In page 317 of Hartshorne's Algebraic Geometry, the author takes a point $P_0$ of an elliptic curve $X$ (over an algebraically closed field $k$ of characteristic $\neq2$) and from it, defines a morphism $f:X\to \mathbb{P}^1$ from the linear system $|2P_0|$. By Riemann-Roch, this linear sytem has dimension 1 and it can be shown that it has no base point. Riemann-Hurwitz tells us that this morphism has four ramification points.

Problem: Hartshorne says that $P_0$ is one of these ramification points.

Idea: I see that for any point $Q$ of $\mathbb{P}^1$ one has $f^*(\mathscr{L}(Q))=f^*(\mathscr{O}_{\mathbb{P}^1}(1))=\mathscr{L}(2P_0)$ (from II.7.1 (b)) so we should have $f^*Q\sim2P_0$ but I don't see why there should exist a point $Q'$ with $f^*(Q')=2P_0$.

More generally: I see somewhere that when a linear system $\mathfrak{d}$ defines a morphism $\varphi:X\to\mathbb{P}^n$ then the hyperplanes of $\mathbb{P}^n$ are pulled back in effective divisors of $\mathfrak{d}$. Is this true (and why)?

$\endgroup$

1 Answer 1

0
$\begingroup$

Recall theorem II.7.1 and the general setup of morphism to $\Bbb P^1$ coming from line bundles:

Theorem. Let $A$ be a ring, and let $X$ be a scheme over $A$.

(a) If $\varphi:X\to\Bbb P^n_A$ is an $A$-morphism, then $\varphi^*(\mathcal{O}_{\Bbb P^n_A}(1))$ is an invertible sheaf on $X$, which is generated by the global sections $s_i=\varphi^*(x_i)$, $i=0,1,\cdots,n$.
(b) Conversely, if $\mathcal{L}$ is an invertible sheaf on $X$, and if $s_0,\cdots,s_n\in\Gamma(X,\mathcal{L})$ are global sections which generate $\mathcal{L}$, then there exists a unique $A$-morphism $\varphi:X\to\Bbb P^n_A$ such that $\mathcal{L}\cong\varphi^*(\mathcal{O}_{\Bbb P^n_A}(1))$ and $s_i=\varphi^*(x_i)$ under this isomorphism.

Since $f:X\to\Bbb P^1$ is given by the complete linear system $|2P_0|$, we pick a basis of $\Gamma(X,\mathcal{O}_X(2P_0))$ as the $s_i$ and therefore for any divisor in $|2P_0|$ (represented as the divisor associated to some global section) there must be some global section of $\mathcal{O}_{\Bbb P^1}(1)$ which pulls back that global section. The zero locus of this global section of $\mathcal{O}_{\Bbb P^1}(1)$ is the point you seek.

This also answers your more general question in the affirmative: the global sections of $\mathcal{O}(1)$ pull back to global sections of $\mathfrak{d}$, which determine the effective divisors in $\mathfrak{d}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .