Let $n\in\mathbb N$ and $p$ be any prime. Denote by $\mathbb Q\_p$ the $p$-adic numbers. For a field extension $L/K$ denote by $Tr\_{L/K}$ the corresponding trace function.


Let $\zeta\_n$ be a primitve $n$-th root of unity and $\xi\_n$ any $n$-th root of unity. I would like to know the following:


What is $Tr\_{\mathbb Q\_p(\zeta\_n)/\mathbb Q\_p}(\xi\_n)$?


In particular: Is $Tr\_{\mathbb Q\_p(\zeta\_n)/\mathbb Q\_p}(\xi\_n)\in\mathbb Q$?


I know that the trace is a coefficient of the minimal polynomial of $\xi\_n$ over $\mathbb Q\_p$. So if $n=p^l$ the minimal polynomial of $\xi\_n$ is the same as over $\mathbb Q$, so it is a cyclotomic polynomial and thus has integer coefficients and above trace is an integer.


But what about the case $n\neq p^l$?



Trace of n-th root of unity in cyclotomic extension of p-adic rationals
Certainly not in general: If $p \equiv 1 \bmod n$ (but $n \neq \pm 1$) then $\mathbb{Q}\_p(\zeta\_n) = \mathbb{Q}\_p$) so $Tr(\zeta\_n) = \zeta\_n$ which is not in $\mathbb{Q}$. To give another example, take $n=5$ and $p \equiv 4 \bmod 5$. Then $\mathbb{Q}\_p$ contains a square root of $5$ and $x^4+x^3+x^2+x+1 = (x^2 + (1-\sqrt{5})/2 x +1) (x^2 + (1+\sqrt{5})/2 x +1)$ so the trace of a primitive $5$th root of unity will be $(-1 \pm \sqrt{5})/2$.


I am tempted to guess that the only time that this happens is when the $n$-th cyclotomic polynomial is irreducible in $\mathbb{Q}\_p$, which is equivalent to asking that $(\mathbb{Z}/n \mathbb{Z})^{\ast}$ is cyclic and $p$ is a generator. But I don't see an easy proof of that. Guess is wrong, see below. But it is right when $n$ is square free!


