# 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?

-

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.
For the first question avoiding nth roots elements of the base field is an obvious necessary condition. So $1+\sqrt[15]{2}$ would be a better primitive element with the property. –  P Vanchinathan Oct 24 '13 at 16:38