bio | website | iazd.uni-hannover.de/… |
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location | Hanover, Germany | |
age | 29 | |
visits | member for | 5 years |
seen | 7 hours ago | |
stats | profile views | 2,412 |
I am a postdoc at the Leibniz UniversitÃ¤t Hannover. I specialise in arithmetic geometry, algebraic number theory and analytic number theory.
Apr 6 |
answered | Dirichlet series without order term |
Apr 6 |
comment |
Dirichlet series without order term
Could you please clarify what you mean by "Dirichlet series without the order term"? Do you mean series of the shape $\sum_{n=-\infty}^\infty a_n/n^s$? |
Apr 5 |
comment |
Deciding a quadratic diophantine equation
@Turbo: Well I don't know the answer, I just had an idea how one could approach the problem. It seems that Aurel's suggested approach is better than mine however, namely this is a fairly explicit problem which one can probably handle using the classical theory of binary quadratic forms à la Gauss. |
Apr 5 |
comment |
Deciding a quadratic diophantine equation
@Aurel: Yes I see now, good point. |
Apr 5 |
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Deciding a quadratic diophantine equation
My naive guess however is that for $2$ variables, the Hasse principle for integral points holds. One approach to this would be to notice that the equation is a torsor for some norm one torus. Once one has this structure, its make studying the problem much easier as there are many tools for the Hasse principle for torsors under algebraic groups. I would not be surprised if it was already known that the Brauer-Manin obstruction is the only one to the existence of integral points; one then just has to show that the Brauer group is trivial to deduce that the Hasse principle holds. |
Apr 5 |
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Deciding a quadratic diophantine equation
This is an interesting problem. As Silverman points out, the natural approach is to check solubility in all $\mathbb{Z}_p$ first. A priori there is no guarantee however that this will imply there is a solution $\mathbb{Z}$ (note that the Hasse principle can fail here when you have 3 variables instead of 2: see the paper by Colliot-Thélène and Xu on the Brauer-Manin obstruction for integral points). |
Apr 2 |
awarded | Yearling |
Mar 30 |
awarded | Nice Question |
Mar 30 |
answered | Tauberian theorem with better error term |
Mar 29 |
awarded | Nice Answer |
Mar 29 |
comment |
Relative Picard functor for the Zariski topology
If you voted my question up, then please also do vote Martin's answer up. This is one of the best answers I have ever received on mathoverflow. |
Mar 29 |
accepted | Twists of projective automorphisms |
Mar 29 |
answered | Twists of projective automorphisms |
Mar 29 |
accepted | Relative Picard functor for the Zariski topology |
Mar 29 |
comment |
Relative Picard functor for the Zariski topology
Thanks very much Martin, this answer is great! It answers everything I wanted, and more. I have an application in mind where the base is regular, so this is perfect for me. |
Mar 29 |
answered | number theory which is close to analysis |
Mar 28 |
asked | Relative Picard functor for the Zariski topology |
Mar 23 |
revised |
Twists of projective automorphisms
deleted 9 characters in body |
Mar 20 |
comment |
Twists of projective automorphisms
I mean those $\sigma \in \mathrm{Aut} X$ such that $\sigma^*L \cong L$. Certainly I want $\mathrm{Aut}(\mathbb{P}^n, \mathcal{O}(1) )$ to be $\mathrm{PGL}_{n+1}$. |
Mar 20 |
asked | Twists of projective automorphisms |