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!
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! 


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. 

