Linear system and base points

algebraic-geometry

I'm using Chapter IV, Lemma 4.2 of Hartshorne's AG to better understand the concept of linear systems and base points.

This lemma states for any two points $P,Q$ (not necessarily distinct) in an elliptic curve $X$ (over an algebraically closed field), we can find an automorphism $\sigma$ of $X$ such that $\sigma$ sends $P$ to $Q$.

The first line of the proof talks about the linear system $|P+Q|$ having dimension $1$ and is base point-free. But the definition of a linear system is a subset of a complete linear system $|D_0|$ of all effective divisors linearly equivalent to some divisor $D_0$. In this case is it referring to the complete linear system $|P+Q|$ and the subset in this case is just the whole of $|P+Q|$?

Also, why does $|P+Q|$ have dimension $1$? I know that the dimension is given to be the dimension of $$V=\{s \in \Gamma(X, \mathcal{L}(P+Q)):(s)_0 \in |P+Q|\}\cup \{0\}$$ minus $1$. Does the dimension of $V$ somehow coincide with the degree of the divisor $P+Q$, which in this case is $2$?

Finally, why is $|P+Q|$ base point-free? The definition of base point given says that $S \in X$ is a base point of the linear system $|P+Q|$ if $S \in \mathrm{Supp}\,D$ for all $D \in |P+Q|$. $\mathrm{Supp} \, D$ refers to the union of the prime divisors of $D$. I cannot make any sense of a prime divisor of an element in a linear system, so I'm confused by this definition.

Any explanations would be appreciated.

Best Answer

  1. Yes, the complete linear system.

  2. Because $l(P+Q)=2$ by Riemann-Roch, and $|D|$ is the projectivization of the vector space of global sections.

  3. The term "linear system" is really an abbreviation of "linear system of divisors". Given $s\in V$ as in your post, the divisor associated to this is $(s)_0$ (this has some different conventions associated to it - Hartshorne takes poles instead of zeros, for instance). Since it's a specific divisor, you can ask whether some point is in the support or not.

  4. Suppose $|P+Q|$ had a base point - that is, there was some point $A$ so that every divisor in $|P+Q|$ could be written as $A+B$ for $B$ variable. Then given two distinct points $B_1,B_2$ so that $A+B_1$ and $A+B_2$ are in $|P+Q|$ (which must exist as $\dim |P+Q|=1$), the ratio of a global section with zeroes at $A+B_1$ and a global section with zeroes at $A+B_2$ is a global section with a single zero and a single pole, therefore defining an isomorphism of your elliptic curve with $\Bbb P^1$. But that's absurd.

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