TITLE: Why are most coefficients of these minimal polynomials divisible by $p$?
QUESTION [6 upvotes]: For an odd prime $p$, let $\zeta:=e^{\frac{\pi i}p}$ and choose odd $1<n<p$. Further let $q(x)$ and $r(x)$ be integer polynomials such that $r(x)$ has no common factor with $x^n+1$, and $\xi$ any root of unity (not necessarily related to $\zeta$). 

I have observed that the minimal polynomial of $$(1+\zeta^n)\frac{q(\xi)}{r(\xi)}$$ always seems to have all its coefficients divisible by $p$, except either the first or the last one. 

Likewise for the minimal polynomial of $(1+\zeta^n)a$, where $a$ is any real algebraic number. 
How to prove this?

REPLY [5 votes]: The second statement is false as stated. Let $p=3$, $n=1$, $a=1/\sqrt{3}$. The minimal polynomial of $(1+\zeta)a$ is $x^4-x^2+1$. Joe Silverman's argument is inapplicable since the $p$-adic value of $(1+\zeta)a$ is 0.