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Abelian varieties are projective algebraic varieties endowed with an Abelian group structure. Over the complex numbers, they can be described as quotients of a vector space by a lattice of full rank. They are analogs in higher dimensions of elliptic curves, and play an important role in algebraic geometry and number theory.
2
votes
Accepted
Ramification in Division field of Abelian Varieties
What if $m=pq$ with $\mathfrak p \mid p$ and $p\ne q$ and $k(A[p])=k$? Then the $p$-torsion doesn't cause ramification since its defined over $k$, and the $q$-torsion won't cause $\mathfrak p$ ramific …
3
votes
Curve through the 16 singular points of a Kummer surface
To answer your second question, no, $C$ does not have to be isomorphic to its image. First, if the image is singular, you may have the common situation of a map $C\to C'$ that is bijective on points, …
5
votes
Accepted
morphism of abelian variety
This is Theorem 4 on page 73 of Mumford's Abelian Varieties, where you will also find a proof. Here's the statement: Let $X$ be an abeian variety. There is a 1-1 correspondence between the two sets of …
5
votes
Intersection multiplicity in abelian varieties
Can't you always find a geometric point $T\in A$ such that the intersection of $X$ and the translation $Y+T$ is either empty or 0-dimensional? Since $X\cdot Y$ is numerically equivalent to $X\cdot(Y+T …
3
votes
Picard number of abelian variety
Do you know about the embedding of $\hbox{NS}(A)\otimes\mathbb{R}$ into $\hbox{End}(A)\otimes\mathbb{R}$ once you choose a polarization? The image is the subset (actually, a Jordan algebra) fixed by t …
8
votes
n-th root of unity in n-th division field of abelian variety?
Let $\hat A/K$ be the dual of $A/K$. Then $K(A[n],\hat A[n])$ contains a primitive $n$'th root of unity, since it contains the image of the Weil pairing $e_n:A[n]\times\hat A[n]\to\mu_n$, which is non …
5
votes
Faltings height of a CM abelian variety
Fix an $A$ with CM by $K$, and for each $D$, let $A_D$ be the quadratic twist of $A$ by $K(\sqrt{D})/K$. Also let $$h_{sf}(D)=\min(h(Dd^2):d\in K^*)$$ denote the "square-free height" of $D$. Then
$$
…
6
votes
What is the motivation for defining the conductor of an abelian variety?
The reference to Serre is good, but for a somewhat more elementary introduction (for elliptic curves), you could look at Chapter IV Section 10 of my book Advanced Topics in the Arithmetic of Elliptic …
5
votes
Conductor CM abelian variety
The short answer is no, since if we take the quadratic twist of $A$ by $\sqrt{D}$, then the conductor of $A$ will more-or-less acquire divisibility by the primes dividing $D$. So we can make the condu …
3
votes
Accepted
Potential good reduction of abelian varieties
The given conditions ensure that $L(A_m)$ is an unramified extension of $L$, since $A$ has good reduction over $L$ and $m$ is prime to the residue characteristic. If one were to merely assume that $K$ …
7
votes
Accepted
Another question related to the isogeny theorem for elliptic curves
I think that maybe what is meant is that there is no functorial way to define the bijection in the category of algebraic geometry. So suppose that we let $\hbox{Ell}(R)$ denote the set of elliptic cur …
4
votes
Accepted
Triviality of torsors after a field extension of bounded degree
This is not even true for elliptic curves over $\mathbb Q$ or $\mathbb Q_p$. For example, by Tate duality the group $H^1(\mathbb Q_p,E)$ is dual to $E(\mathbb Q_p)$, which is an infinite group. This s …
7
votes
Accepted
Generalization of $j(E) \in \overline { \Bbb{Z}}$ to abelian varieties of arbitrary dimension
The $j$ invariant gives an isomorphism over $\mathbb Z$,
$$j:\mathcal A_1\to\mathbb A^1,$$
of the moduli space of elliptic curves. So $j(E)\in\overline{\mathbb Z}$ can be interpreted as saying that $\ …
5
votes
Torsion group of the following elliptic curve
An alternative to the "reduction mod $p$ and Dirichlet's theorem" approach described by Jeremy Rouse is to do a height computation. It is not that hard to find explicit absolute constants $c_1>0$ and …
23
votes
Accepted
When $k = \mathbb{F}_q$ finite field, $X$ always has $k$-rational point, and so $A \simeq X$?
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 …