bio | website | www2.imperial.ac.uk/~rpanneko |
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location | London, United Kingdom | |
age | ||
visits | member for | 3 years, 7 months |
seen | 9 hours ago | |
stats | profile views | 1,196 |
I am a postdoc at Imperial College. My interests are rational points on varieties, in particular K3 surfaces and abelian varieties.
Apr 7 |
comment |
Algebraic integer with conjugates on the unit circle
Yes; see Mark Sapir's answer to this question: mathoverflow.net/questions/38680/… |
Mar 25 |
revised |
Can an abelian variety/Q have no non-trivial points over Q_sol?
title asked for something different than the main body |
Mar 24 |
suggested | approved edit on Can an abelian variety/Q have no non-trivial points over Q_sol? |
Mar 16 |
accepted | Singular models of K3 surfaces |
Mar 16 |
asked | Singular models of K3 surfaces |
Mar 6 |
comment |
Maryam Mirzakhani's works
@Neil Strickland: it's only mathematical Farsi, so you'll pick it up in no time. |
Feb 9 |
awarded | Nice Answer |
Feb 9 |
revised |
Why does inconstructibility of $\sqrt[3]{2}$ imply impossibility of cube doubling?
deleted 9 characters in body |
Feb 9 |
answered | Why does inconstructibility of $\sqrt[3]{2}$ imply impossibility of cube doubling? |
Feb 8 |
comment |
Why does inconstructibility of $\sqrt[3]{2}$ imply impossibility of cube doubling?
Wait, shouldn't $K$ be the union of all Galois extensions of $2$-power degree over $\mathbb{Q}$? I don't think the set of all algebraic numbers of $2$-power degree forms a field (e.g. any two degree-$4$ subextensions of an $S_4$-extension give rise to a degree-$12$ compositum...). |
Jan 19 |
revised |
The exponent of Ш of y^2 = x^3 + px, where p is a Fermat prime
change from "\colon" to actual colon |
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 |