Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
Thanks, Second question has such a simple answer. –  P Vanchinathan Oct 24 '13 at 15:43
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.