For an arbitrary cyclotomic polynomial, is this true? Let $F$ be a field and $F_n$ be the splitting field for $X^n-1$ over $F$. Let $\zeta$ be a primitive $n$-th root of unity and $f$ be the minimal polynomial for $\zeta$ over $F$.
Then, for a prime $p$ not dividing $n$, is $\zeta^p$ a root of $f$?
This is true when $F=\mathbb{Q}$, but I'm not sure whether this is true for an arbitrary field $F$.
 A: If $\zeta$ is a primitive $n$-th root of unity then $\zeta^n=1$, and hence $$(\zeta^p)^n=\zeta^{pn}=(\zeta^n)^p=1^p=1,$$
so $\zeta^p$ is an $n$-th root of unity. If $m$ is another integer such that $(\zeta^p)^m=1$ then $\zeta^{pm}=1$, so because $\zeta$ is a primitive $n$-th root of unity it follows that $pm$ is a multiple of $n$. Because $p$ is prime and does not divide $n$ it follows that $m$ is a multiple of $n$. This shows that $\zeta^p$ is also a primitive $n$-th root of unity.
The field $F$ either contains all primitive $n$-th roots of unity or none of them. If it contains all of them then the minimal polynomials of $\zeta$ and $\zeta^p$ are $X-\zeta$ and $X-\zeta^p$, respectively, which are the same if and only if $p\equiv1\pmod{n}$. If it contains none of them then the minimal polynomial of $\zeta$ and $\zeta^p$ is the $n$-th cyclotomic polynomial.
A: Note that this is false when $F_n=F$, i.e. when $F$ already contains the $n$th roots of unity. In that case, the minimal polynomial of $\zeta$ is just $x-\zeta$, which $\zeta^p$ is not a root of (unless $p\equiv 1\bmod n$).
More generally, this is false when the $n$th cyclotomic polynomial $$\Phi_n(x)=\prod_{\substack{1\leq d\leq n\\\gcd(d,n)=1}}(x-\zeta^d)$$
is not irreducible over $F$, which can happen even when $F$ does not contain the $n$th roots of unity.
For example, $\Phi_8(x)=x^4+1$ factors into $(x^2-i)=(x-\zeta)(x-\zeta^5)$ and $(x^2+i)=(x-\zeta^3)(x-\zeta^7)$ over $F=\mathbb{Q}(i)$, so that $\zeta^p$ is a conjugate of $\zeta$ (i.e., a root of the same minimal polynomial over $F$) only when $p\equiv 1\bmod 4$.
