# $f(x)$ 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 $\mathbb Q(a)=\mathbb Q(b)$ ?

• @Transcendental : The smallest " subfield " .. but since $a$ is algebraic over $\mathbb Q$ , the smallest subring and subfield are the same anyway – user228168 Aug 26 '16 at 6:46
• @Trancendental $\Bbb Q(a)$ is the smallest subfield, $\Bbb Q[a]$ is the smallest subring, but the notation makes sense even if $a\notin \Bbb C$. – Arthur Aug 26 '16 at 6:54
• @Arthur But as $\mathbb{C}$ is algebraically closed, any algebraic element $a$ of $\mathbb{Q}$ can be identified with an element of $\mathbb{C}$, no? – Daniel Robert-Nicoud Aug 26 '16 at 7:18
• @DanielRobert-Nicoud Anything algebraic over $\mathbb{Q}$, sure. But you can also use this notation for something like a variable $x$ so that $\mathbb{Q}[x]$ is the polynomial ring and $\mathbb{Q}(x)$ is the field of rational functions. I think this is what Arthur was getting at. Correct me if I'm wrong, Arthur. – Josh Keneda Aug 26 '16 at 7:35
• @JoshKeneda Ok, that makes sense (even though the particular case of $\mathbb{Q}[x]$ can be obtained as the extension by an arbitrary transcendental number...) – Daniel Robert-Nicoud Aug 26 '16 at 7:37

## 3 Answers

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.

• I know my first thought after I read any question is "Well, does it work for the 24th roots of unity"? :) – user98602 Aug 30 '16 at 17:21

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

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.

• It might be worth pointing out that your reasoning for $deg(b)\neq 1$ applies no matter what $a$ is. That is, $deg(b)>1$ whether or not $a$ has prime degree. – Josh Keneda Aug 27 '16 at 5:50