TITLE: $f(x) $ be the minimal polynomial of $a$ (algebraic element) over $\mathbb Q$ , let $b=f'(a) \in \mathbb Q(a)$ , then is $\mathbb Q(a)=\mathbb Q(b)$?
QUESTION [30 upvotes]: Let $a \in \mathbb C$ be algebraic over $\mathbb Q$ , let $f(x) \in \mathbb Q[x]$ be the minimal polynomial of $a$ over $\mathbb Q$ , let $b=f'(a) \in \mathbb Q(a)=\mathbb Q[a]$ , then is it true that $\mathbb Q(a)=\mathbb Q(b)$ ?

REPLY [12 votes]: Here is a hand calculation that can be used to replace
the Sage calculation in Plamondon's answer.
Since $a$ is a primitive 24th root of unity,
the number $a^5$ is also a primitive 24th root of unity
and there is an automorphism $\sigma$ of $\mathbb Q[a]$
such that $\sigma(a) = a^5\;(\neq a)$. Since $b = 8a^7-4a^3$ we have
$\sigma(b) = 8\sigma(a)^7-4\sigma(a)^3 = 8a^{35}-4a^{15} =
{\mathbf{8a^{11}-4a^{15}}}$.
Since $a$ is a solution to $x^8-x^4+1=0$, we have
$a^8=a^4-1$, which can help reduce expressions
involving powers of $a$. Using this we find
$$
\begin{array}{rl}
\sigma(b)&=8a^{11}-4a^{15}\\
&=(4a^{11}+4a^{11})-4a^{15}\\
&=(4a^3(a^8)+4a^{11})-4a^7(a^8)\\
&=(4a^3(a^4-1) + 4a^{11})-4a^7(a^4-1)\\
&=8a^7-4a^3=b.
\end{array}
$$
The automorphism $\sigma$ fixes $b$, hence also fixes $\mathbb Q[b]$,
but $\sigma$ moves $a$. This proves that $a\notin \mathbb Q[b]$.