# Hartshorne detail in definition of $j$-invariant: pull back of hyperplane from morphism $X\to\mathbb{P}^n$ defined by a linear system

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)?

## 1 Answer

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 $$\mathfrak{d}$$.