Irrational numbers in a cyclotomic extension $\mathbb{Q}(\zeta_N)$ Let $\zeta_N$ be a primitive root of unity so that $\mathbb{Q}(\zeta_N)$
is an $n$-th cyclotomic extension. I am interested in irrational (real) numbers $\alpha$ in $\mathbb{Q}(\zeta_N)$ such that 
$\alpha^m \in \mathbb{Z}$ for some $m \in \mathbb{N}$.
Does such $\alpha$ have to be of the form $\alpha = A^{1/m}$ where $A$ only consists of prime factors dividing $N$? Thank you very much!
 A: As you've stated the question, there are some trivial counterexamples. For example, if $N=5$, then $\alpha = 2\sqrt5\in \mathbb Q(\zeta_5)$, and $\alpha^2 = 20\in \mathbb Z$ but $2\mid 20$ and $2\nmid 5$. 
However we can salvage things:

Let $K = \mathbb Q(\zeta_N)$. If $\alpha\in  K$ is a non-unit such that $a=\alpha^n$ is a squarefree integer for some $n\ge 2$ and $p$ is a prime dividing $a$, then $p\mid N$.

Let's rephrase this question a bit.  Since $\alpha^n = a$ and $\alpha$ is a non-unit, as ideals
$$a\mathcal O_K = (\alpha\mathcal O_K)^n=(\mathfrak p_1\cdots\mathfrak p_m)^n$$
for some (not necessarily distinct) primes $\mathfrak p_i\subset\mathcal O_K$. Hence if $p\mid a$, then $p\mathcal O_K\mid a\mathcal O_K$, so
$$p\mathcal O_K = \prod_{i=1}^m\mathfrak p_i^{a_i}$$
for some non-negative integers $a_i\le n$.
Now for the crucial fact: since $a$ is squarefree, the $a_i$ cannot all be $\le 1$ - otherwise $p^2\mathcal O_K$ would also divide $a\mathcal O_K$; restricting to $\mathbb Z$, this would mean that $p^2\mid a$, which can't happen.
Hence $p$ must be ramified in $\mathcal O_K$. And it is a general theorem that the only primes which ramify in $\mathbb Q(\zeta_N)$ are those dividing $N$.
