0
$\begingroup$

I'm having some trouble with this problem, and I wanted to know if someone could help me out. Let $K=\mathbb{Q}(\theta)$ be an algebraic number field of degree $n$. Let $\theta_1=\theta,\ldots,\theta_n$ be the conjugates of $\theta$ over $\mathbb{Q}$. Suppose there are exactly $m$ distinct fields among $\mathbb{Q}(\theta_1),\ldots,\mathbb{Q}(\theta_n)$ prove that $m\mid n$ and each field occurs $n/m$ times.

Any advice?

$\endgroup$
4
  • $\begingroup$ I can't even think of any examples! $\endgroup$
    – Unit
    Dec 8, 2014 at 20:55
  • $\begingroup$ Well, a simple example is $\mathbb Q(i)$. $i$'s conjugates are $\pm i$, but there is only one distinct field extension. $\endgroup$
    – Nishant
    Dec 8, 2014 at 20:56
  • $\begingroup$ I meant of an extension where two conjugates give rise to different fields! $\endgroup$
    – Unit
    Dec 8, 2014 at 21:03
  • $\begingroup$ $\mathbb Q(\sqrt[3]{2})$? $\endgroup$
    – Nishant
    Dec 8, 2014 at 21:26

1 Answer 1

2
$\begingroup$

Say $S$ is the set of $\theta_i$'s that are in $\mathbb Q(\theta_1)$. Let $\sigma_j$ be an automorphism that sends $\theta_1\mapsto\theta_j$. Then the set $\sigma_j(S)$ is exactly the set of $\theta_i$'s in $\mathbb Q(\theta_j)$. Thus, each of the distinct fields among them has the same number of $\theta_i$'s, which means that the $n$ of them are partitioned into $m$ same-size sets of $n/m$ elements.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .