bio | website | www2.imperial.ac.uk/~rpanneko |
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location | London, United Kingdom | |
age | ||
visits | member for | 3 years, 3 months |
seen | 58 mins ago | |
stats | profile views | 1,082 |
I am a postdoc at Imperial College. My interests are rational points on varieties, in particular K3 surfaces and abelian varieties.
Dec 3 |
revised |
rational points of a hyperelliptic curve
added latex |
Dec 3 |
suggested | approved edit on rational points of a hyperelliptic curve |
Nov 30 |
suggested | rejected edit on p-power roots of unity in local fields |
Nov 25 |
comment |
Fermat's last theorem over larger fields
I think the answer is yes, since a non-hyperelliptic genus 3 curve is a plane quartic, and every quartic polynomial (in one variable) has solvable Galois group. Nice answer/comment, by the way! |
Nov 25 |
comment |
Fermat's last theorem over larger fields
Ah, of course, thanks! |
Nov 25 |
comment |
Fermat's last theorem over larger fields
Just curious, what is the reason for asking about the $n=5$ case specifically? I thought about the $n=4$ case and I couldn't answer that either -- but perhaps you can? |
Nov 25 |
comment |
Fermat's last theorem over larger fields
(OK, except the trivial rational solutions, that is.) |
Nov 25 |
comment |
Fermat's last theorem over larger fields
I'm really curious about how you want to apply HIT here! I would think that, on the contrary, no preimage of a rational point will give a Galois point, since you're extracting a fifth root, so you're missing a primitive fifth root of unity in your residue field! |
Oct 24 |
awarded | Disciplined |
Oct 23 |
comment |
Does every smooth, projective morphism to $\mathbb{C}P^1$ admit a section?
Those are never smooth, I think? |
Oct 17 |
comment |
Equidistribution of rational points on an algebraic variety
You're absolutely right. I just wanted to show the answer is "no" even if one restricts attention to those elements of $X(\mathbb{F}_q)$ that are in the image of the reduction map. |
Oct 17 |
comment |
Equidistribution of rational points on an algebraic variety
Probably yes. Don't have time to think about a proof though. |
Oct 17 |
revised |
Equidistribution of rational points on an algebraic variety
edited body |
Oct 17 |
answered | Equidistribution of rational points on an algebraic variety |
Oct 10 |
comment |
Are there noncommutative extensions of $\alpha_p$ by $\mathbb{G}_m$?
Actually, I know of no serious source that uses that phrase to mean an extension of the form $0 \rightarrow A \rightarrow E \rightarrow B \rightarrow 0$, but I would love to be shown an example. |
Sep 18 |
awarded | Yearling |
Sep 14 |
comment |
Is there such thing as the Gorensteinification of a one-dimensional local ring?
+1 for "Gorensteinification". |
Sep 10 |
comment |
Squarefree numbers with all digits equal 1
When $p=3$ the condition $9 \mid k$ is necessary and sufficient. So I guess the real question is whether there are infinitely many squarefree numbers of this form, given how many are definitely not squarefree. |
Sep 10 |
answered | Hasse principle and twists of $\mathbb{P}^n$ |
Aug 25 |
comment |
Rational points techniques on curves not using their Jacobian
I am wondering whether $J(\mathbb{Q})$ can indeed be dense in $J(\mathbb{Q}_p)$ for all $p$, as you say is true for your $C$. I actually think this can't ever happen, using an argument very much analogous to the one presented in this question: mathoverflow.net/questions/113968/… (to which you supplied the winning answer, incidentally). I am of course also curious what your $C$ is, if you'd care to disclose such information. :) |