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Recall that any smooth projective variety of dimension $g$ embeds into $\mathbf{P}^{2g+1}$. Consider now an abelian variety $A$ of dimension $g$ which embeds into $\mathbf{P}^{2g}$. Van de Ven proves (essentially by applying the self-intersection formula to the normal bundle of $A$ in $\mathbf{P}^{2g}$) that the degree of $A$ in $\mathbf{P}^{2g}$ is given by ...

23

This is a theorem of Lang's from 1956. Here's an online document giving a proof (in the form $H^1(A,k)=0$):
Lecture 14: Galois Cohomology of Abelian Varieties over Finite Fields,
William Stein. http://wstein.org/edu/2010/582e/lectures/582e-2010-02-12/582e-2010-02-12.pdf
Stein notes that there is a "more modern proof" in the first few sections of
Chapter ...

answered Oct 4 '16 at 17:10

Joe Silverman

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Weil introduced the term "polarization" in connection with his study of abelian varieties with complex multiplication. His definition is slightly different from what one sees today; one might call it a polarization up to isogeny instead of a polarization. One can find a discussion in Weil's article "On the theory of complex multiplication" ([1955d] in volume ...

21

Really this is mostly just consolidating what has been (implicitly) said in the comments and cleaning it up a bit (e.g. using the Chow ring instead of singular cohomology), but might as well make it an answer...
Let $A$ be an abelian variety of dimension $d$. We prove that $A$ cannot embed in $\mathbb{P}^{2d-1}$ and can only embed in $\mathbb{P}^{2d}$ if $d=...

21

I don't have any contribution for the intuition beyond the fact that, I can't construct something outside the image of (1) so I hope it's surjective.
Here is a sketch of the central idea of Tate's proof. Consider $A = A' \times A''$ and try to find an endomorphism of $A$ from an endomorphism of its Tate module. The endomorphism of the Tate module gives you ...

20

Here is Ribet's proof (expanding on Ulrich's comment):
Let $G_K:=Gal(\bar{K} / K)$ and $V_l:=T_l(E)\otimes \mathbb{Q}_l$.
The image $\rho_{l,E}(G)$ is a closed subgroup of the $l$-adic Lie group $\text{Aut}(V_l(E)) \cong \text{GL}_{2}(\mathbb{Q}_l)$ and is therefore a Lie subgroup of $\text{Aut}(V_l(E))$. Its Lie algebra $\mathfrak{g}_l$ is a subalgebra of ...

18

I am not aware of anybody seriously considering hyperelliptic curves for actual real-world usage, beyond toys, and I would be rather surprised to hear differently from anyone.
As you say, hyperelliptic provide comparatively few (if any!) advantages over elliptic curves but have the huge disadvantage that virtually nobody has well-tested, battle-hardened ...

17

Let me answer your last question "How can I think geometrically (in the lattice) about fixing a polarization?". I will follow the treatment given in [Birkenhake-Lange, Complex Abelian Varieties, Chapter 3].
Let $X = V / \Lambda$ be a complex torus of dimension $g$ and $L$ a line bundle on $X$ with first Chern class $H$. Then $H$ is an Hermitian form on $V$, ...

answered Jul 28 '14 at 9:06

Francesco Polizzi

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17

Embed the dual abelian variety into projective space. Take a smooth hyperplane section and interate until it's one-dimensional, obtaining a smooth curve $C$. By Lefschetz $C$ is irreducible, and the natural map $H_1(C, \mathbb Z) \to H_1(A^\vee, \mathbb Z)$ is surjective. Because $H_1(C, \mathbb Z)= H_1(J(C), \mathbb Z)$, the natural map $H_1(J(C), \mathbb Z)...

17

Welcome new contributor. Yes, that is true. Let $k$ be any field, let $A$ be an Abelian variety over $k$, and let $U\subset A$ be a dense open affine. Denote by $D\subset A$ the complementary divisor with its induced reduced structure. Denote the invertible sheaf of this divisor by $\mathcal{L}:=\mathcal{O}_A(D).$ Denote by $s$ the global section of $\...

16

In fact, it is no for completely elementary reasons. If $A$ is simple and
$B\subset A^n$ is an abelian variety with $\dim B < g$, then $Hom(B,A^n)=Hom(B,A)^n$ is necessarily zero. So $B=0$.

16

No.
It is always difficult to "prove" that something is "not known", but this may do: I claim it is not even known when $K=\mathbb Q$, $A$ is an elliptic curve $E$. In fact in this case, the result you ask for is not even known when $E$ has CM. In fact, even in this very special case, it is not known that the $l$-primary torsion subgroup of the Sha is ...

16

This follows from the following well-known lemma.
Lemma. Let $A$ be an abelian variety over $k$. Then any map $f \colon \mathbb P^1 \to A$ is constant.
Proof 1. The map $f$ induces a map on the Albanese $f_* \colon \operatorname{Alb}_{\mathbb P^1} \to \operatorname{Alb}_A$ sitting in a commutative diagram
$$\begin{array}{ccc}\mathbb P^1 & \stackrel{f}\...

answered Feb 20 '18 at 16:11

R. van Dobben de Bruyn

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Let me put it this way, Tate's conjecture for abelian varieties is known to imply the Hodge conjecture for abelian varieties, and the last is very much open for this class. For the implication, see the article by Deligne and Milne on "Hodge cycles on abelian varieties" (you can get a copy off of Milne's website).
There are lot's of interesting cases known ...

15

Actually, this is an exercise in Serre's Lectures on Mordell--Weil Theorem:
$K(A[n])$ always contains $\mu_n$ if $char(K)$ does not divide $n$ and $A$ is an abelian variety of positive dimension over $K$. (You don't need to assume that $K$ is a number field)
Here is a solution. First, it suffices to check the case when $n=\ell^m$ is a power of a prime $\...

14

Abelian varieties over the rationals are modular if and only if they are of "$GL_2$"-type, which is a notion introduced by Ribet who proved that this statement is a consequence of Serre's conjecture which, as you know, has since been proved. Here is a link to Ribet's paper:
http://math.berkeley.edu/~ribet/Articles/korea.pdf
Generalizing the statement of ...

answered May 23 '14 at 14:53

Felipe Voloch

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A bit of overkill, but it follows from the Weil conjectures. The structure of cohomology ($H^i = \wedge^i H^1$) is computed over the algebraic closure and it follows that the number of points is $\prod(\alpha_i-1)$ where the $\alpha_i$ are the eigenvalues of Frobenius on $H^1$ so $|\alpha_i| = q^{1/2}$ and the product is therefore not zero.

13

Let $C$ be a non hyperelliptic curve of genus 3, let $f\colon C'\to C$ be an e'tale double cover and let $A$ be the Prym variety of $f$. Then $A$ is a principally polarized surface and the Abel-Prym map embeds $C'$ in $A$ as an element of $|2\Theta|$. Multiplication by $-1$ on $A$ restricts on $C'$ to the involution induced by $f$, hence the image of $C'$ ...

13

Another proof that $L = \,\overline{\bf \!Q\!}\,$:
Clearly $L$ is contained in $\,\overline{\bf \!Q\!}\,$,
so we need only show $L$ contains every algebraic number $x \notin \bf Q$.
Let $P(X)$ be the minimal polynomial of $x$. If $\deg P$ is odd,
then the class of $((x,0)) - (\infty)$ is a $2$-torsion point on
the Jacobian of the elliptic or hyperelliptic ...

ag.algebraic-geometry nt.number-theory arithmetic-geometry algebraic-number-theory abelian-varieties

answered Dec 3 '14 at 20:56

Noam D. Elkies

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For a prime $p$ and an elliptic curve $E/\mathbb{Q}$, we have the exact sequence
$$\displaystyle 0 \rightarrow E(\mathbb{Q})/p E(\mathbb{Q}) \rightarrow S_p(E) \rightarrow \text{Sha}_E[p]\rightarrow 0,$$
where $S_p(E)$ is the $p$-Selmer group of $E$ and $\text{Sha}_E$ is the Tate-Shafarevich group of $E$ and $\text{Sha}_E[p]$ is the $p$-part of it. This ...

answered Mar 14 '18 at 0:22

Stanley Yao Xiao

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13

Let me give an answer for $k = \mathbb{C}$.
By a theorem of Matsusaka, every abelian variety $A$ over an algebraic closed field $k$ is a quotient of a Jacobian.
Now just apply Matsusaka's theorem to $A^{\vee}$, and dualize. Since we are over $\mathbb{C}$, dualization sends surjective morphisms of Abelian varieties into injective ones, so we are done.
I ...

answered Jul 5 '18 at 12:18

Francesco Polizzi

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13

First, when defining the stack you will have the issue that there are formal deformations of abelian varieties which do not extend to families of abelian varieties over any reduced scheme. These are the deformations that do not respect any polarization. (In the complex analytic world these correspond to deformations of complex tori) So unless you have some ...

13

We give a uniform approach to $p \leq 61$ by applying
analytic discriminant bounds to the Hilbert class field.
To be sure this is not entirely "conceptual", but then
some computation is needed even to deal with $p < 36$ using Minkowski.
If $p = 4k+1$ is prime then $K = {\bf Q}(\sqrt{p})$ has odd class number $h$,
so either $h=1$ or $h \geq 3$. If $h \...

answered May 25 at 18:42

Noam D. Elkies

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12

The answer is no: an easy (not interesting) counterexample is provided by "constant" abelian varieties, i.e., when $A$ and $B$ are defined over an algebraically closed field $k$ of characteristic zero while $K$ is finitely generated over $k$.
So, let's assume that the $K/k$-traces of $A$ and $B$ are zero, i.e., both $A$ and $B$ do not contain constant ...

12

Yes, assuming you mean a way to go from the double cover to the curve, (rather than how to go from the Prym variety to the curve, which is just a constructive Torelli argument). This is based on the fact that curves of genus 3 and 4 are "trigonal", i.e. admit a degree 3 map to the projective line. The corresponding "trigonal construction" is attributed to ...

12

Theorem If $m$ is relatively prime to $30$, then $N=3$ works.
I'll use additive notation. The cyclic group of order $n$ is denoted $C(n)$, the generators of this cyclic group are denoted $U(n)$.
Lemma Let $p \geq 7$ be prime and let $X$, $Y$ and $Z$ be subsets of $C(p)$ of size at least $(p-1)/2$. Then $0 \in X+Y+Z$.
Proof By the Cauchy-Davenport theorem ...

answered Jul 10 '13 at 18:42

David E Speyer

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12

I think the following should give a counterexample. Let $\mathcal{O}$ be an order in an imaginary quadratic field $K$ and $\mathcal{O}_K$, the ring of integers. Then it's not too hard to find a (non-split) short exact sequence of $\mathcal{O}$-modules:
$$0 \to \mathcal{O}_K \to \mathcal{O} \oplus \mathcal{O} \to \mathcal{O}_K \to 0,$$
e.g. if $1, \omega$...

12

If $\lambda\in\overline{\mathbb{Q}}$, the elliptic curve
$$
E_\lambda\colon y^2=x(x-1)(x-\lambda)
$$
has $(\lambda,0)$ as $2$-torsion point and is defined over (a subfield of) $L=\mathbb{Q}(\lambda)$. Its Weil restriction $A_\lambda:=\operatorname{Res}_{L/\mathbb{Q}}(E_\lambda)$ is an abelian variety defined over $\mathbb{Q}$ and shares the same points of $...

ag.algebraic-geometry nt.number-theory arithmetic-geometry algebraic-number-theory abelian-varieties

answered Dec 2 '14 at 12:28

Filippo Alberto Edoardo

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12

The answer is affirmative (in Serre's formulation via principal homogeneous spaces) for proper, geometrically reduced, and geometrically connected schemes $X$ over any field $k$, giving a strong mapping property relative to working over all $k$-schemes. That is:
Theorem There exists a map $f:X \rightarrow E$ to a torsor for an abelian variety $A$ over $k$ ...

11

The answer to question (*) is yes. It is Theorem 1.2.2 in the following preprint.

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