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!



Is there a known example of a curve X of genus > 1 over Q such that we know the number of points of X over the n-th cyclotomic field, for every n?
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.


