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Let $a,b\in\mathbb{C}$, and $\sigma$ be an automorphism of $\mathbb{C}$ such that $b=\sigma(a)$. My question is: Why if the set $$\{\sigma'(a)\mid \sigma' \text{ is an automorphism of }\mathbb{C}\}$$ has at most $n$ elements then $b$ is an algebraic number of degree at most $n$, i.e. $[\mathbb{Q}(b):\mathbb{Q}]\leq n$?. I know that every automorphism permutes the roots of a polynomial, but my question is like ''if $$\{\sigma(a)\mid \sigma \text{ is an automorphism of }\mathbb{C}\}=\{\alpha_{1},\ldots,\alpha_{m}\}$$ then $\alpha_{1}$, $\ldots$, $\alpha_{m}$ are roots of a polynomial over $\mathbb{Q}$ with degree at most $n$?'' and I don't know if this is true.

Obviously $b$ is a root of $(x-\alpha_{1})\ldots(x-\alpha_{m})$ but I don't know if this polynomial has rational coefficients.

Any help will be appreciated.

Nah
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  • Given $a,b\in K$ algebraically closed and $F\subset K$ there is $\sigma\in Aut(K/F)$ such that $\sigma(a)=b$ iff $F(a)\cong F(b)$. If $a$ is not algebraic over $F$ then $F(a)\cong F(a^n)$ for all $n$ (thus the orbit of $a$ is infinite). The proof is a bit abstract (axiom of choice..) as shown in the transcendental basis of Keenan Kidwell's answer. – reuns Jun 09 '20 at 23:59

2 Answers2

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Suppose that $a$ is transcendental over $\mathbf{Q}$. Choose a transcendence basis $S$ for $\mathbf{C}$ over $\mathbf{Q}$ containing $a$. The extension $\mathbf{C}/\mathbf{Q}(S)$ is then algebraic, and because $\mathbf{C}$ is algebraically closed, $\mathbf{C}$ is necessarily an algebraic closure of $\mathbf{Q}(S)$. For each $b\in S-\{a\}$, let $\varphi_b:S\to S$ be the bijection that is the identity on $S-\{a,b\}$ and interchanges $a$ and $b$. Then $\varphi_b$ extends to an automorphism $\overline{\sigma}_b$ of $\mathbf{Q}(S)$. Finally, since $\mathbf{C}$ is an algebraic closure of $\mathbf{Q}(S)$, $\overline{\sigma}_b$ extends to an automorphism $\sigma_b$ of $\mathbf{C}$. The set $S$ is infinite (if $S$ is finite, $\mathbf{Q}(S)$ is countable, which implies that $\mathbf{C}$ itself is countable because $\mathbf{C}$ is algebraic over $\mathbf{Q}(S)$). Thus $S-\{a\}=\{\sigma_b(a):b\in S-\{a\}\}$ is an infinite subset of $\{\sigma(a):\sigma\in\mathrm{Aut}(\mathbf{C})\}$, and the latter set is infinite as well.

So, given $a\in\mathbf{C}$, if $\{\sigma(a):a\in\mathrm{Aut}(\mathbf{C})\}$ is finite, then $a$ is algebraic over $\mathbf{Q}$. If $a^\prime$ is a Galois conjugate of $a$ over $\mathbf{Q}$, i.e., another root of the minimal polynomial $f$ of $a$ over $\mathbf{Q}$, then there is an automorphism of $\overline{\mathbf{Q}}$ sending $a$ to $a^\prime$, and any such automorphism extends to an automorphism of $\mathbf{C}$ (use a transcendence basis for $\mathbf{C}$ over $\overline{\mathbf{Q}}$). This means that the set $\{\sigma(a):a\in\mathrm{Aut}(\mathbf{C})\}$ contains all the roots of $f$, and conversely, since any automorphism of $\mathbf{C}$ fixes $\mathbf{Q}$, $\sigma(a)$ is a root of $f$ for every $\sigma\in\mathrm{Aut}(\mathbf{C})$. Thus $\{\sigma(a):\sigma\in\mathrm{Aut}(\mathbf{C})\}$ is exactly the set of roots of $f$ in $\mathbf{C}$. There are $\deg(f)$ such roots, which means that, for any element $b$ of the set in question, we have

$$[\mathbf{Q}(b):\mathbf{Q}]=\deg(f)=|\{\sigma(a):\sigma\in\mathrm{Aut}(\mathbf{C})\}|\text{.}$$

Keenan Kidwell
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  • It seems like a pretty tricky argument, but I was told that is was ''standard Field Theory''. Maybe there is a simpler proof? – Nah Jun 10 '20 at 00:06
  • I am not sure if there is a simpler proof. But everything I am using is reasonably classified as "standard field theory." – Keenan Kidwell Jun 10 '20 at 00:23
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Hint: Here is a possible roadmap:

  • If $a \in \mathbb C$ is fixed by every automorphism of $\mathbb C$, then $a \in \mathbb Q$.

  • The elementary symmetric functions of $\alpha_{1},\ldots,\alpha_{m}$ are fixed by every automorphism of $\mathbb C$.

  • $(x-\alpha_{1})\cdots(x-\alpha_{m})$ has rational coefficients.

The hard part is the first one.

lhf
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  • See https://math.stackexchange.com/questions/412010/wild-automorphisms-of-the-complex-numbers – lhf Jun 12 '20 at 00:35