$f(x) $ be the minimal polynomial of $a$ (algebraic element) over $\mathbb Q$ , let $b=f'(a) \in \mathbb Q(a)$ , then is $\mathbb Q(a)=\mathbb Q(b)$? 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 $\mathbb Q(a)=\mathbb Q(b)$ ?
 A: I've been thinking about this for some time and while this is only a partial answer, perhaps someone can take this further.
Since $b=f'(a)$, we have that $\big(\mathbb{Q}(b)\big)(a)=\mathbb{Q}(a)$, which implies that the chain of inclusions $\mathbb{Q} \subset \mathbb{Q}(b) \subset \mathbb{Q}(a)$ is also a tower. Therefore, by the tower formula (in the obvious notation):
$$\deg(a)=\deg(b)\cdot \deg(a:b)$$
Now, suppose $a$ were prime. There are two possibilities:


*

*$\deg(b)=1$


In this case, $b \in \mathbb{Q}$, so $p(x)=f'(x) - b \in \mathbb{Q}(x)$. But then $f(x)$ is not minimal for $a$ over $\mathbb{Q}$, because $p(x)$ is also monic and its degree is less than $f$ (assuming that $\deg(f)>1$; otherwise the question is trivial anyway). This is a contradiction.


*

*$\deg(a:b)=1$


In this case, $a$ is a rational multiple of $b$, so indeed $\mathbb{Q}(a)=\mathbb{Q}(b)$.

Hence, if there is a counterexample, then $a$ may not have prime degree.

A: The answer seems to be no.
Cyclotomic fields are great (counter-)examples.  Take $a$ to be a primitive $24$th root of unity.  Its minimal polynomial is the cyclotomic polynomial $\Phi_{24}=x^8-x^4+1$.  
Now $b=8a^7-4a^3$.  Sage helpfully tells us that the minimal polynomial of $b$ is $x^4+2304$.   Hence $\mathbb{Q}(a)$ and $\mathbb{Q}(b)$ don't have the same degree over $\mathbb{Q}$, and cannot be equal.
A: 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 $a\notin \mathbb Q[b]$.
