What's the structure of the sequence of fields $\mathbb Q(\alpha^n)$?

Question

Given an algebraic number $x$ and a natural number $n>0$, we define the $n$th powerfield of $x$ as $\mathbb Q(x^n)$.

$$\mathbb Q(x),\; \mathbb Q(x^2),\; \mathbb Q(x^3),\; \mathbb Q(x^4),\; \mathbb Q(x^5),\; \cdots$$

Obviously these are subfields of $\mathbb Q(x)$. What more can be said? E.g. is the sequence eventually periodic?

$\mathbb Q(x^m) \subseteq \mathbb Q(x^n)$ when $m$ is a multiple of $n$. Hence, for arbitrary $m,n>0$, the intersection of $\mathbb Q(x^m)$ and $\mathbb Q(x^n)$ contains $\mathbb Q(x^{\operatorname{LCM}(m,n)})$, and the union (or compositum) of $\mathbb Q(x^m)$ and $\mathbb Q(x^n)$ is contained in $\mathbb Q(x^{\operatorname{GCD}(m,n)})$. Indeed, by Bezout, there exist $a,b$ such that $am+bn = \operatorname{GCD}(m,n)$, which shows that $\mathbb Q(x^{\operatorname{GCD}(m,n)}) \subseteq \mathbb Q(x^m,x^n)$, so these fields must be equal.

Given any powerfield $\mathbb F$, let $n>0$ be the smallest number such that $\mathbb Q(x^n) = \mathbb F$, and suppose $\mathbb Q(x^m) \subseteq \mathbb F$. Then $\mathbb Q(x^{\operatorname{GCD}(m,n)}) = \mathbb F$ by the previous paragraph. But since $\operatorname{GCD}(m,n) \leq n$, and $n$ is minimal, we must have $\operatorname{GCD}(m,n) = n$. That is, $m$ must be a multiple of $n$.

There are finitely many subfields of $\mathbb Q(x)$ over $\mathbb Q$. From this (and LCMs) it follows that the intersection of all powerfields of $x$ is itself a powerfield of $x$. How can we find this smallest powerfield? Is there a bound on the required exponent, in terms of the degree $d$ of the minimal polynomial of $x$? Note, for example, that $x=\tfrac12(3+\sqrt{-3})$ has degree $2$ but the required exponent is $6$, as $x^6 = -27$ but no lower power of $x$ is rational. Does the totient work, $\phi(n) \leq d$?

In fact we can prove periodicity rather easily. Let $\mathbb F$ be the smallest powerfield of $x$, and $n$ smallest such that $\mathbb Q(x^n) = \mathbb F$. Any other exponent can be written in the form $mn+k$, with $0 \leq k < n$. Clearly $\mathbb Q(x^{mn+0}) = \mathbb F$ (assuming $m > 0$), so suppose $k > 0$. Since $\mathbb F$ is contained in any powerfield, $\mathbb Q(x^k) \ni x^n \in \mathbb Q(x^{mn+k})$. Therefore, $x^k=x^{mn+k}/(x^n)^m \in \mathbb Q(x^{mn+k})$, and conversely $x^{mn+k}=x^k\cdot(x^n)^m \in \mathbb Q(x^k)$, which shows that $\mathbb Q(x^{mn+k}) = \mathbb Q(x^k)$. Also, by similar reasoning, $\mathbb Q(x^{n-k}) = \mathbb Q(x^k)$.

(So far, I haven't used any specific properties of the field $\mathbb Q$. If instead the base field is $\mathbb R$, and the extension field is $\mathbb R(x)=\mathbb C$, with degree $2$, then $n$ can be arbitrarily large before $x^n \in \mathbb R$. If the base field is $\mathbb F_p$, and the extension field $\mathbb F_p(x)$ has degree $d$, then the size of the field is $q=p^d$, so $x^{q-1} = 1 \in \mathbb F_p$.)

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Here's another example. Take $x=(1+\sqrt{-1})/\sqrt[3]2$, with minimal polynomial $x^6+2x^3+2=0$. (Let's denote $\sqrt{-1} = i$ and $\sqrt[3]2 = r$ to make the table a bit easier to read.)

$$\begin{array}{c|c|cccc|} n & x^n & & \mathbb Q(x^n) \\ \hline 1&\tfrac12(1+i)r^2&\mathbb Q(i,r) \\ 2&ir&\mathbb Q(i,r) \\ 3&(-1+i)&&&\mathbb Q(i) \\ 4&-r^2&&\mathbb Q(r) \\ 5&(-1-i)r&\mathbb Q(i,r) \\ 6&-2i&&&\mathbb Q(i) \\ 7&(1-i)r^2&\mathbb Q(i,r) \\ 8&2r&&\mathbb Q(r) \\ 9&2(1+i)&&&\mathbb Q(i) \\ 10&2ir^2&\mathbb Q(i,r) \\ 11&2(-1+i)r&\mathbb Q(i,r) \\ 12&-4&&&&\mathbb Q \end{array}$$

The splitting field is $\mathbb Q(i,r,\omega)$, of degree $12$, where $\omega$ is a primitive cube root of unity.

Answer 1

Not a complete answer. Here's a sketch of at least how the argument should start using the Galois correspondence. Let $L$ be the splitting field of $x$, with Galois group $G$. Then $\mathbb{Q}(x)$, as a subfield of $L$, is the fixed field of $H = \text{Stab}_G(x)$, the stabilizer of $x$ with respect to the Galois action. Similarly $\mathbb{Q}(x^n)$ is the fixed field of $H_n = \text{Stab}_G(x^n)$.

If there is some $n$ such that $\mathbb{Q}(x^n)$ is strictly smaller than $\mathbb{Q}(x)$ then $H_n$ is strictly larger than $H$, so there exists $g \in H_n \setminus H$, meaning $g(x^n) = (gx)^n = x^n$ but $gx \neq x$. This gives

$$\left( \frac{gx}{x} \right)^n = 1$$

so $\frac{gx}{x} \in L$ is a nontrivial $n^{th}$ root of unity. Write $H_{\infty} \subset G$ for the set of all elements of $G$ which act on $x$ by multiplication by some root of unity; this is a subgroup (this requires a small calculation) and is the union of the $H_n$, and the intersection of all powerfields is the fixed field of $H_{\infty}$. We also have $H_{\infty} = H_d$ for some $d$, and the smallest such $d$ is the $\text{lcm}$ of all $n$ such that some $g \in H_{\infty}$ acts by multiplication by a primitive $n^{th}$ root of unity.

There ought to be something much more specific to say about this $H_{\infty}$ that produces the desired bound but at the moment I'm not sure what the right way to think about it is. It doesn't seem to be the case that it necessarily normalizes $H$ and the map from $H_{\infty}$ to roots of unity is not a homomorphism.

Answer 2

Qiaochu Yuan's answer led me to a general bound on the exponent as desired; see the comments. Here I'll give more specific results for certain degrees.

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Suppose $x=a+b\sqrt d$ is a quadratic irrational number, and some power of $x$ is rational. We wish to prove that either $x^2$ is rational (so $a = 0$), or $x$ is a rational multiple of $(1+\zeta)$ where $\zeta$ is a primitive $3$rd or $4$th or $6$th root of unity. Hence, assume $a \neq 0$.

If $d > 0$ then the binomial expansion shows that $x^n$ is never rational. For example, with $n = 5$:

$$x^5 = (a^5 + 10a^3b^2d + 5ab^4d^2) + (5a^4b + 10a^2b^3d + b^5d^2) \sqrt d$$ $$= a(a^4 + 10a^2b^2d + 5b^4d^2) + b(5a^4 + 10a^2b^2d + b^4d^2) \sqrt d$$

The coefficient of $\sqrt d$ is non-zero because all the terms in parentheses are positive. Hence, assume $d < 0$. Then we can write $x$ as a complex number: $x = r e^{i\theta}\in\mathbb C$. By the assumption that some $x^n$ is rational and thus real, $e^{i\theta}$ must be a root of unity. Notice that $r=\sqrt{a^2+b^2|d|}$ is either rational or quadratic, so $e^{i\theta}=x/r$ is either rational or quadratic or quartic. Thus, considering totients, $e^{i\theta}$ must be a primitive $n$th root of unity for some $n\in\{1,2,3,4,5,6,8,10,12\}$. We can't have $n = 1$ or $2$ since $x$ is not real, and we can't have $n = 4$ since $a \neq 0$. Replacing $x$ with $-x$ if necessary, we can take $n$ to be even. Then $e^{i\theta n/2} = -1$, so $x^{n/2}=-r^{n/2}$ is real and thus rational (as $\mathbb Q(x)\cap\mathbb R = \mathbb Q$). Since we already knew that $r^2$ is rational, if $n/2$ is odd then $r^1$ must be rational. The case $n=6$ agrees with what we wanted to prove, and the case $n=10$ doesn't work because $x$ is supposed to be quadratic, not quartic. For $n = 8$ or $12$, we have $e^{i\theta} = \pm(1+\zeta)/\sqrt{n/4}$ where $\zeta$ is a primitive $n/2$ root of unity; thus $x = s(1+\zeta)$ for some real $s$. We know that the real part of $x$ is $a$, which is rational. On the other hand, for $n=8$ the real part of $x$ is $s$, and for $n=12$ the real part of $x$ is $\tfrac32s$. Therefore $s$ is rational.

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Now suppose $x$ has any odd degree $d$, and let $\mathbb F$ be the smallest powerfield of $x$. We wish to prove that $\mathbb F = \mathbb Q(x^d)$.

Let $n>0$ be the smallest exponent such that $\mathbb Q(x^n) = \mathbb F$, and let $z = x^n$, and denote by $M(T)$ the minimal polynomial of $z$. Suppose $p$ is any prime factor of $n$, and let $y = x^{n/p}$. Then $\mathbb Q(y) \neq \mathbb F$, and the degree $[\mathbb Q(y):\mathbb F] \neq 1$. Note that $[\mathbb Q(y):\mathbb F]$ is a factor of $[\mathbb Q(x):\mathbb F]$, which in turn is a factor of $[\mathbb Q(x):\mathbb Q]=d$, which is odd. So we can't have $p=2$; $n$ must be odd. Suppose there's some $u\in\mathbb F$ such that $u^p = z = y^p$. We can't have $y = u$ since $y \not\in \mathbb F$. The other possibility is $y = \zeta u$ where $\zeta$ is a primitive $p$th root of unity; but this is impossible since $\zeta$ has degree $p-1$, which is even, whereas anything in $\mathbb Q(x)$ such as $y/u$ has odd degree. Therefore $z$ is not any such $p$th power. It follows that the polynomial $M(T^n)\in\mathbb Q[T]$ is irreducible. Since $x$ is a root of $M(T^n)$, it must be the minimal polynomial, with degree $d=[\mathbb F:\mathbb Q]n$. Thus $x^d = z^{[\mathbb F:\mathbb Q]} \in \mathbb F$.