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I'm interested in necessary and sufficient conditions for a nonreal algebraic integer $\alpha$ to satisfy the equality above. I know that if $\alpha$ has prime degree then the equality holds, but I don't know whether this condition is also necessary.

Thanks in advance.

Oliver Miller
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Not sure how useful this is, but a translation of this to the language of groups via Galois correspondence would go as follows.

Let $K$ be the splitting field of the minimal polynomial $m(x)$ of $\alpha$. IOW the normal closure of $\Bbb{Q}[\alpha]$. Let $G$ be the Galois group $Gal(K/\Bbb{Q})$, and $G_\alpha$ the stabilizer of $\alpha$ when $G$ is viewed as a group of permutations of the zeros of $m(x)$. Let $\sigma$ be the restriction of complex conjugation to $K$.

Galois correspondence then tells us that $\Bbb{Q}[\alpha]\cap\Bbb{R}$ is the fixed field of the group $H=\langle G_\alpha,\sigma\rangle$. Thus $$ \Bbb{Q}[\alpha]\cap\Bbb{R}=\Bbb{Q}\Leftrightarrow H=G. $$

So for example if $\alpha$ is non-real and $G_\alpha$ is a maximal subgroup of $G$, then the condition is satisfied. This is because $\sigma\notin G_\alpha$, and there are no subgroups properly between $G_\alpha$ and $G$. This happens for example when $\alpha$ is a non-real root of an irreducible quartic with Galois group $S_4$, because in that case $G_\alpha\cong S_3$ is a maximal subgroup.

If $[Q[\alpha]:\Bbb{Q}]$ is a prime, then $G_\alpha$ is automatically a maximal subgroup of $G$ also. So your observation fits into this as well.

Jyrki Lahtonen
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  • OP actually has $\mathbb{Q}[\alpha]$, though from context perhaps they mean $\mathbb{Q}(\alpha)$? – Steven Stadnicki Feb 05 '15 at 19:29
  • Makes no difference given that $\alpha$ is algebraic $\implies \Bbb{Q}(\alpha)=\Bbb{Q}[\alpha]$. – Jyrki Lahtonen Feb 05 '15 at 19:30
  • Oh, of course. :/ Mea culpa on my part. – Steven Stadnicki Feb 05 '15 at 19:38
  • If I remember the basics of permutation groups correctly a point stabilizer is maximal, iff the action is primitive. This suggests that finding examples can be done by picking your favorite primitive permutation action together with a bit of inverse Galois theory. – Jyrki Lahtonen Feb 05 '15 at 19:53
  • Still thinking about locating an example where $G_\alpha$ is non-maximal. – Jyrki Lahtonen Feb 05 '15 at 19:53
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    By galois correspondence $G_\alpha$ is a maximal subgroup of the galois group of the splitting field iff $\mathbb Q(\alpha)/\mathbb Q$ does not admit an intermediate field.

    Let $\alpha$ be a root of $X^4+2$. $G_\alpha=Gal(\mathbb Q(\alpha,i)/\mathbb Q(\alpha))$ is not maximal, because $\mathbb Q(\alpha)/\mathbb Q$ admits the intermediate field $\mathbb Q(\alpha^2)$. Furthermore this is the unique intermediate field of $\mathbb Q(\alpha)/\mathbb Q$. In particular $\mathbb Q(\alpha) \cap \mathbb R = \mathbb Q$.

     – MooS Feb 05 '15 at 20:48
  • @MooS: That comment could be promoted to an answer. – Jyrki Lahtonen Feb 06 '15 at 08:25