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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
3
votes
Accepted
Integers in residue classes $\mathcal{O}_K/\mathfrak{p}$
Let $p$ be the prime number that satisfies $p\mathbb{Z} = \mathbb{Z} \cap \mathfrak{p}$.
Then your claim is equivalent to the inclusion $\mathbb{F}_p\subset \mathcal{O}_K/\mathfrak{p}$ being an equali …
3
votes
Accepted
Selmer groups and fppf cohomology
The following paper of Kestutis Cesnavicius answer these and related questions completely: https://arxiv.org/abs/1301.4724
8
votes
If it quacks like an abelian variety over a finite field
One possible answer to this could be Lang's theorem: it says that if $G/\mathbb{F}_q$ is a smooth connected algebraic group, then $H^1(\mathbb{F}_q,G)$ is trivial, or otherwise put every $G$-torsor ha …
4
votes
Accepted
Generalization of torsion points on Jacobian of genus 2 over finite fields (with respect to ...
The set $J(C)_{\Theta}[n]$ has the structure of a smooth irreducible algebraic curve, and the restriction of $J(C)\xrightarrow{\times n } J(C)$ to $C$ defines a morphism $J(C)_{\Theta}[n]\rightarrow C …
5
votes
1
answer
311
views
Conductor at 2 of abelian surfaces with real multiplication
Let $A/\mathbb{Q}$ be an abelian surface such that $\text{End}_{\mathbb{Q}}(A)\otimes \mathbb{Q}$ is a real quadratic field $E$. I am interested in bounding the conductor of $A$ at $2$.
Let $\mathfrak …
10
votes
0
answers
228
views
If $H$ is a quotient of $G$, does there exist an $H$-extension of $\mathbb{Q}$ not contained...
Let $\phi\colon G\rightarrow H$ be a surjective homomorphism between finite groups. Assume that $\phi$ is not split, in other words there exists no homomorphism $\sigma\colon H\rightarrow G$ such that …