# If $a \in L-k$ satisfies $k(a^n)=L$ (for all $n \geq 1$), then $L/k$ is Galois?

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!

• A simple example take $a=1+\sqrt{2}$ – mouthetics Jan 3 at 16:26
• 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). – user237522 Jan 3 at 16:36
• No It won't. As you explained in you question. – mouthetics Jan 3 at 16:41
• @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 $b-a \notin \mathbb{Q}$? – user237522 Jan 3 at 16:43
• Well, $b=1+2\sqrt2$ =) – Kenny Lau Jan 3 at 18:15