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I'm trying to understand the proof of the Theorem at page 163 from Mumford, Abelian Varieties, and I have a question about one step. This is the situation:

$X$ is an abelian variety (hence there's the correspondence between line bundles, Cartier divisors and Weil divisors), $\mathcal{L}$ is an ample line bundle on $X$ and Mumford shows that we can take an effective divisor $D$ without multiple components such that $\mathcal{L}\cong\mathcal{O}_X(D)$. We want to show that $\mathcal{L}^3$ is very ample, hence that:

(1) Given $x_0, x_1\in X$, $x_0\neq x_1$, there exists $D' \in |3D|$ such that $x_0\in Supp D'$, $x_1\notin Supp D'$.

(2) Given any tangent vector $t$ to $X$ at $x_0$, there exists $D'\in|3D|$ such that $x_0\in Supp D'$ and $t$ is not tangential to $D'$.

In the proof of the point (1) above, we suppose that (1) is not true (without loss of generality for $x_0=0$) so we want to arrive to a contradiciton. With this hypothesis, one has that $t_{x_1}^*(D)=D$ where $t_{x_1}:X\rightarrow X, y\mapsto y+x_1$ is the translation morphism. Hence $x_1\in K(\mathcal{L})=\{ x\in X: t_x^*\mathcal{L}\cong \mathcal{L}\}$ which is a finite group for $\mathcal{L}$ ample. Hence $x_1$ generates a finite subgroup of $X$, which we call $F$. We consider then the étale morphism $\pi:X\rightarrow X/F$ (see costruction of $(X/F, \pi)$ from page 66 to page 69). Now, $D_1=\pi(Supp F)$ is a closed subset of pure codimension one in $X/F$, so we can see it as a divisor with all components of multiplicity one; the same holds for $\pi^*(D)$ and Mumford shows that $D=\pi^*(D_1)$, so $\mathcal{L}\cong \pi^*(\mathcal{O}_{X/F}(D_1))$, and that $\dim\Gamma(\mathcal{L})>\dim\Gamma(\mathcal{O}_{X/F}(D_1))$. Then, Mumford says:

Since the set of all divisors $D_1$ such that $\mathcal{L}\cong\pi^*(\mathcal{O}_{X/F}(D_1))$ fall into a finite set of equivalence classes, this proves that all sections $s\in\Gamma(L)$ either define multiple divisors, or lie into one of a finite number of lower-dimensional subspaces $\pi^*\Gamma(\mathcal{O}_{X/F}(D_1))$. This is a contradiction.

I don't understand this conclusion. Can someone explain it to me?

Thanks!

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  • $\begingroup$ Do you agree that the set of $D_1$ is finite in this way? $\endgroup$ – Hoot Jun 12 '16 at 19:57
  • $\begingroup$ I can't see it. But I don't understand the rest anyway. $\endgroup$ – Poecilia Jun 12 '16 at 20:48
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    $\begingroup$ Well maybe we should start there. Is there a relationship between the Picard groups of $X$ and $X/F$? $\endgroup$ – Hoot Jun 12 '16 at 20:50
  • $\begingroup$ The only thing I found is here: citeseerx.ist.psu.edu/viewdoc/… in section 4, Picard Group of quotients. $\endgroup$ – Poecilia Jun 13 '16 at 15:57
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    $\begingroup$ I would say that $\sum n_iD_i$ is multiple if some $n_i \neq 0,1$. $\endgroup$ – Hoot Jun 13 '16 at 20:33

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