5,066 reputation
21438
bio website iazd.uni-hannover.de/…
location Hanover, Germany
age 29
visits member for 5 years
seen 7 hours ago

I am a postdoc at the Leibniz Universität 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
comment 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
comment 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