Algebraic numbers abhorrent to cyclotomic fields Consider an algebraic number $\alpha$, which can be taken to be an
integer. With $\deg\alpha$ a prime number, one can easily  arrange that
to be such that all powers $\alpha^n$ to be of the same degree as $\alpha$.
I would like to know how to get numbers with the same property without any restriction on its degree.  If $\deg\alpha=n$,  my guess for a condition to guarantee this behaviour is  $\mathbf{Q}[\alpha]$ be linearly disjoint with the $n$-th cyclotomic extension.
My second question is:
Is it possible for a number field to be linearly disjoint with every cyclotomic extension?
 A: The second question is rather easy. 
One can construct Galois extensions $F/\mathbb Q$ with Galois groups nonabelian and simple. Then they are certainly linearly disjoint from all cyclotomic extensions.
For the first question, let $F = \mathbb Q(\sqrt[15]{2})$. The element $\alpha = \sqrt[15]{2}$ has its minimal polynomial over $\mathbb Q$ be $x^{15} - 2$. For any cyclotomic field $\mathbb Q(\mu_n)$ with $15\mid n$, if $\alpha^3\in \mathbb Q(\mu_n)$, then $\mathbb Q(\sqrt[5]{2}, \mu_5)$ would be abelian over $\mathbb Q$, which is impossible. Thus $\sqrt[5]2\notin \mathbb Q(\mu_n)$, and similarly $\sqrt[3]2\notin \mathbb Q(\mu_n)$. This shows that $x^{15} - 2$ remains irreducible over $\mathbb Q(\mu_n)$, and that $F$ is linearly disjoint from $\mathbb Q(\mu_n)$. Indeed, if $[F(\mu_n), \mathbb Q(\mu_n)] = d < 15$, then there is a prime $p\mid 15$ but $p\nmid d$. Then $2^d = N_{F(\mu_n)/\mathbb Q(\mu_n)}(2) = N_{F(\mu_n)/\mathbb Q(\mu_n)}\left(\sqrt[p]{2}\right)^p\in \mathbb Q(\mu_n)^{\times p}$. Since $p\nmid d$, this implies $\sqrt[p]2\in \mathbb Q(\mu_n)$, contradiction.
However it is clear that $\mathbb Q(\alpha^3)\subsetneq F$, not as you expected.
A: This type of problem arises in Dobrowolski's famous result [1] on Lehmer's conjecture. Here is a result from Dobrowolski's paper: 
Lemma 3. Let $\alpha$ be an algebraic number of degree $n$. Then
$$
\#\{\text{primes }p : \deg(\alpha^p) < n \} \le \frac{\log n}{\log 2}.
$$
So although this doesn't answer your first question, it gives, in some sense, a uniform bound for the number of exceptions (as well as an interesting application). 
[1] E. Dobrowolski, On a question of Lehmer and the number of irreducible factors of a polynomial, Acta Arithmetica 34(4), 1977, pages 391-401.
