Containment of $\mathbb Q(a)$

Question

Let $a$ be a complex number which is algebraic over $\mathbb Q$. Let $r$ be rational number. Can we say $\mathbb Q(a)$ is contained in $\mathbb Q(a^r)$.

My opinion: Since $\mathbb Q(a^r)$ is a field then we can say positive powers of $a^r$ are in itself. Is it true that $a$ is contained in $\mathbb Q(a^r)$?

Answer 1

Here is a simple counterexample. Let $a=i$ and $r=2$. Then $\mathbb Q(a)$ is the set of "Gaussian rationals" ($a+bi$ with $a$ and $b$ rational) but $\mathbb Q(a^r)=\mathbb Q$ as $a^r=-1\in\mathbb Q$, so $\mathbb Q(a)$ is not contained in $\mathbb Q(a^r)$.