$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

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 $\mathbb Q(a)=\mathbb Q(b)$ ?

Answer 1

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.

Answer 2

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 $a\notin \mathbb Q[b]$.

Answer 3

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.