Let $k \subsetneq L$ be a finite separable field extension, and let $a \in L-k$ satisfy: For every $n \geq 1$, $k(a^n)=L$.

In other words, all the non-zero powers of the primitive element $a$ are also primitive elements.

Is there something interesting to say about such an extension? Should it be Galois?

Partial answer: According to this question, if the extension is of prime degree, and if there exist infinitely many $m$'s such that $k(b^m) \neq L=k(b)$, then $k \subsetneq L$ is not Galois.

But I am asking about the opposite direction, namely, if only finitely many (= more precisely, zero) $2 \leq m$'s are such that $k(b^m) \subsetneq k(b)=L$, then $k \subseteq L$ is Galois? Can we reverse the argument in that question?

Special case $k=\mathbb{Q}$: Can one find $a \in \bar{\mathbb{Q}}-\mathbb{Q}$ satisfying $\mathbb{Q}(a)=\mathbb{Q}(a^2)=\mathbb{Q}(a^3)=\ldots$?

The following are two non-examples for $k=\mathbb{Q}$:

(1) In the spirit of the comment for this question, notice that here taking $a=p^{\frac{1}{m}}$ for some (positive) prime $p \geq 2$ and fixed $m \geq 2$ will not help, since $a^m=(p^{\frac{1}{m}})^m=p$ so $\mathbb{Q}(a^m)=\mathbb{Q}(p)=\mathbb{Q} \subsetneq \mathbb{Q}(a)$.

(2) $a=\frac{-1+\sqrt{3}i}{2}$ will not help; indeed, $a^2+a+1=0$, so $a^2=-a-1$, then $a^3=-a^2-a=-(-a-1)-a=1$, so $\mathbb{Q}(a^3)=\mathbb{Q}(1)=\mathbb{Q} \subsetneq \mathbb{Q}(a)$.

Any hints are welcome!

  • 1
    $\begingroup$ A simple example take $a=1+\sqrt{2}$ $\endgroup$ – mouthetics Jan 3 at 16:26
  • $\begingroup$ Oh, of course.. thank you. Is it true that also any $a=1+p^{\frac{1}{m}}$ will work, where $p \geq 2$ is prime and $m \geq 2$? (Perhaps one has to be careful, and not all such $p$ and $m$ will work). $\endgroup$ – user237522 Jan 3 at 16:36
  • $\begingroup$ No It won't. As you explained in you question. $\endgroup$ – mouthetics Jan 3 at 16:41
  • $\begingroup$ @mouthetics, please, is it possible to find $b \in \mathbb{Q}(\sqrt{2})$ such that: $\mathbb{Q}(b^n)=\mathbb{Q}(\sqrt{2})$ (for all $n \geq 2$), and also $b-a \notin \mathbb{Q}$? $\endgroup$ – user237522 Jan 3 at 16:43
  • $\begingroup$ Well, $b=1+2\sqrt2$ =) $\endgroup$ – Kenny Lau Jan 3 at 18:15

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