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Results tagged with abelian-varieties
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user 11926
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.
3
votes
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Effective version for Silverman’s specialization theorem
All of the height estimates used to prove Theorem B can in principal be made effective, in terms of the equations defining your family of abelian varieties and the equations defining the section. The …
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4
votes
Is the set of points on an abelian surface which project to rational points on the Kummer su...
The answer to your second question is no. If $x\notin J(\mathbb Q)$, then the condition $\pi(x)\in X(\mathbb Q)$ implies that $\mathbb Q(x)/\mathbb Q$ is a quadratic extension. If you take a point $y$ …
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7
votes
Accepted
Néron model, torsion and ramification
If, for example, $A[n]$ is already contained in $A(K)$, then there will be no ramification, regardless of whether or not $A'$ is an abelian scheme (although that may well force the special fiber to be …
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7
votes
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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 $\ …
- 47.4k
7
votes
What integer value can be the conductor of a $g$-dimensional abelian variety over $\mathbb Q$?
You seem to be asking about which small integers cannot occur as the conductor $N$ of an abelian variety over $\mathbb Q$ of dimension $g$, but possibly you'll also be interested in other constraints …
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4
votes
Singular abelian surfaces that can be defined over $\mathbb Q$
I think that what you want is contained in the theory of $\mathbb Q$-curves, which are elliptic curves defined over $\overline {\mathbb Q}$ that are isogenous to all of their Galois conjugates. Dick G …
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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 …
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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 …
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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 …
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3
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How to construct an abelian variety with CM by a given CM field?
A reasonably good answer to your question of finding a number field $L$ is given in the comments to Is there an excplicit number field of definition for an Abelian Variety $A/\mathbb{C}$ with CM? . Ho …
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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, …
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8
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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 …
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5
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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 …
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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 $K$ …
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5
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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 …
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