Automorphism exhanging primitive $n$-th roots

Question

Suppose $k$ is a field of characteristic $0$ with algebraic closure $L$. Let $\xi,\zeta\in L$ be primitive $n$-th roots of unity (say for $n\geq 3$), and let $k\subset K\subset L$ be the splitting field of $X^n-1$, so that $K=k(\xi)=k(\zeta)$.

Does there always exist a field automorphism of $K$ that sends $\xi$ to $\zeta$? In particular, does there always exist a field automorphism of $K$ that sends $\xi$ to $\xi^{-1}$?

If $\xi\notin k$, the answer is yes : since $\xi$ and and $\zeta$ are conjugate over $k$, and there exists a (unique) $k$-linear field isomorphism $K=k(\xi)\simeq k(\zeta)=K$ sending $\xi$ to $\zeta$. EDIT @MathChat raises an interesting point : while $\xi$ and $\zeta$ are conjugate over $\Bbb Q$, they may not be conjugate over $\Bbbk$! The example of $\Bbbk=\Bbb R$ gives counter-examples. However, $\xi$ and $\xi^{-1}$ will still be conjugate, so allow me to forget about $\zeta$ from now on, and only consider the case of $\xi$ and $\xi^{-1}$.

What if $\xi\in k$?

I may be overlooking an obvious argument ... I only have two leads : establishing the existence of some smaller field $k^0\subsetneq k$ such that $k=k^0(\xi)$, or somehow extending the automorphism $\Bbb Q[\xi]\simeq\Bbb Q[\zeta],\;\xi\mapsto\zeta$ to $k$, but my ignorance of field and Galois theory means I don't know how to tackle either problem.

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The motivation for this question comes from character theory which I'm having a look at for an exposé : say one works over a field of characteristic zero $\Bbbk$, and one studies the $\Bbbk$-linear representations of $G$, a finite group. There is a bilinear form on class functions $$B(\varphi,\psi)=\frac1{|G|}\sum_{g\in G}\varphi(g)\psi(g^{-1})$$ Over the complex numbers, and for class functions that arise as characters, and more generally $\Bbb R$-linear linear combinations of characters, this inner product is sesquilinear with respect to complex conjugation, i.e. hermitian : for any character $\chi_V$ associated to some complex representation $V$ of $G$, one has $\forall g\in G, \chi_V(g^{-1})=\overline{\chi_V(g)}$ so that $$B(\chi_V,\chi_W)=\frac1{|G|}\sum_{g\in G}\chi_V(g)\overline{\chi_W(g)}.$$ I thus wondered whether, over an arbitrary field $\Bbbk$ of characteristic zero, when restricted to characters, the form $B$ coincides with a sesquilinear one with respect to some field automorphism $\sigma:\Bbbk\to\Bbbk$.

Since representations of $G$ become diagonalizable over the splitting field of $X^{|G|}-1$, and has eigenvalues that are $|G|$-th roots of unity, I was expecting the property of exchanging (in some sense) $\xi$ with $\xi^{-1}$ to be crucial, but the first part of the discussion shows, I believe, that whenever $\xi\notin\Bbbk$, then for any $\Bbbk$-linear character, $\chi_V(g^{-1})=\chi_V(g)$. So the question becomes : what about if the $G$-action is diagonalizable over $\Bbbk$? This will always be the case if $\xi\in\Bbbk$.

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Résumé of what's been found :

1. While $\xi$ and $\zeta$ may be conjugate over $\Bbb Q$, as @MathChat points out, they aren't necessarily conjugate over $\Bbbk$; maybe the simplest example of this phenomenon happens when $\Bbbk=\Bbb R$ : two algebraic numbers are conjugate over $\Bbb R$ if and only if they are complex conjugate.
2. @EricWofsey provides a very nice counter example to the modified question of whether $\xi$ and $\zeta=\xi^{-1}$ may always be interchanged by a field automorphism of $\Bbbk$ if $\xi\in\Bbbk$.
3. It seems to me that the answer to the question becomes yes if we add a further hypothesis and demand, say, that the extension $\Bbb Q\subset \Bbbk$ be normal, in the sense that it be algebraic and contains all the $\Bbb Q$-conjugates of its elements. In this context there is the following theorem

Theorem. Suppose $k\subset K\;(\subset \overline{k})$ is a normal extension of $k$ (and $\overline{k}$ is an algebraic closure of $k$), then for every $\sigma\in\mathrm{Gal}(\overline{k}/k)$, $\sigma(K)=K$.

When applied to a normal extension $\Bbb Q\subset\Bbbk$ of $\Bbb Q$, and $\sigma$ being the restriction to $\overline{\Bbb Q}$ of complex conjugation, we get that $\sigma$ induces an automorphims of the extension exchanging $\xi$ and $\xi^{-1}$.

Answer 1

Here's a counterexample. Let $n=3$ and let $k=\mathbb{Q}(\alpha)$ where $\alpha\in\mathbb{C}$ has minimal polynomial $x^4+3$. Then $\xi=\frac{-1+\alpha^2}{2}$ and $\xi^{-1}=\frac{-1-\alpha^2}{2}$ are the two primitive cube roots of unity, so we have $K=k$. If $k$ had an automorphism $\sigma$ mapping $\xi$ to $\xi^{-1}$, then it would satisfy $\sigma(\alpha^2)=-\alpha^2$ and so $\sigma(\alpha)/\alpha=\pm i$. It would follow that $k$ contains $\mathbb{Q}(i,\sqrt{3})$, and thus is equal to $\mathbb{Q}(i,\sqrt{3})$ since $[k:\mathbb{Q}]=4$.

So to show no such $\sigma$ exists, it suffices to show that $\alpha\not\in\mathbb{Q}(i,\sqrt{3})$. Here's one way to show this (there's probably a better way, but this is all I can think of at the moment). Note that $|\alpha|=\sqrt[4]{3}$ so $\alpha\bar{\alpha}=\sqrt{3}$ and so if $\alpha=(a+b\sqrt{3})+i(c+d\sqrt{3})$ were in $\mathbb{Q}(i,\sqrt{3})$ we would have $$(a+b\sqrt{3})^2+(c+d\sqrt{3})^2=\sqrt{3}.$$ But expanding the left-hand side and comparing coefficients in the basis $\{1,\sqrt{3}\}$ of $\mathbb{Q}(\sqrt{3})$ over $\mathbb{Q}$, we get $$a^2+3b^2+c^2+3d^2=0.$$ This implies $a=b=c=d=0$ so $\alpha=0$, which is obviously not correct.

Answer 2

For the 1st question, the answer is No!

For $n=12$, we have

$\Phi_{12}(x)=x^4-x^2+1$, and

Over $k:=\mathbb{Q}(\sqrt{3})$, we have

$x^4-x^2+1=(x^2 - \sqrt{3}x + 1)(x^2 + \sqrt{3}x + 1)$.

Let $\xi \in \overline{k}$ be a root of $p(x)=x^2 - \sqrt{3}x + 1$, and $\zeta \in \overline{k}$ be a root $g(x)=x^2 + \sqrt{3}x + 1$.

Note that $L=\mathbb{Q}(\sqrt{3},i)$ is the splitting field of $p(x)=x^2 - \sqrt{3}x + 1$, and of $g(x)=x^2 + \sqrt{3}x + 1$

We have $[K:k]=2$, so $K/k$ is Galois with $Gal(K/k)=\{1,\sigma\}$, and clearly $\sigma (\xi)\neq \zeta$.