Take the 2-minute tour ×

By Falting's theorem, these numbers are of course finite. Is there an example where we can explicitly compute them for every $n$?

Thank you!

share|improve this question

1 Answer 1

up vote 28 down vote accepted

Let $n > 1$ be odd. The curve:

$$X: \ x^n + 2 y^n + 4 z^n = 0$$

does not have any points over $K/\mathbf{Q}_2$ unless the ramification index $e(K/\mathbf{Q}_2)$ is divisible by $n$. Proof: At least two of the terms $x^n$, $2 y^n$, $4 z^n$ must have the same $2$-adic valuation. On the other hand, the ramification index $e$ of any abelian extension $K/\mathbf{Q}_2$ is a power of two. So $X$ has no points over any Galois extension of $\mathbf{Q}$ whose decomposition group at any prime above $2$ is abelian. In particular, it has no point over any cyclotomic extension. It has genus $>1$ if $n>3$.

You can ask whether any smooth projective curve $X/\mathbf{Q}$ has at least one rational point over any solvable extension. Since local Galois groups are solvable, there are no longer any local obstructions. This is an open problem, and a positive answer would have various nice consequences including (generalizations of) Serre's conjecture, etc. There's no particular reason why it should be true, however.

share|improve this answer
Concerning solvable points on projective curves: there are some results of Ciperiani and Wiles "Solvable points on genus one curves" and of Pal as well "Solvable points on projective algebraic curves", "Solvable points on genus one curves over local fields" and "Curves which do not become semi-stable after any solvable extension". –  Laurent Berger Jun 24 '13 at 7:07

Your Answer


