I already tried to ask this question, but it was considered a duplicate. I didn't find the associated post very explanatory, and probably I expressed myself poorly in the previous question. I try to give a better context now; my issue is the part written in bold, and I'm interested in understanding the argument in that part, not in another way to prove the same thing. The proof is in fact reconstructed from the sketchy notes of my teacher, so I would like to have it clear: the only part that I really can't fill is the one in bold. Thanks for any hint.
Let $\alpha\in \mathbb C$ be the real cubic root of $2$. Its minimal polynomial over $\mathbb Q$ is: $$f:=X^3-2\in \mathbb Q[X].$$
Let $\zeta\in \mathbb C$ be a cubic root of $1$ (not $1$). Its minimal polynomial over $\mathbb Q$ is: $$g:=X^2+X+1\in \mathbb Q[X].$$ Set $K:=\mathbb Q(\alpha,\zeta)$. We show in seven steps that the only subfield $E\subset K$ with $[E:\mathbb Q]=2$ is $E=\mathbb Q(\zeta)$.
Lemma. Let $F\subset \mathbb C$ be a field, $F\subset L$ be a field extension and $\omega\in \mathbb C$ an element with minimal polynomial $P\in F[X]$ over $F$. Call $Z$ the set of roots of $P\in L[X]$ over $L$; then there is a bijection $$\mathrm{Hom}_F(F(\omega),L)\to Z$$ $$\varphi\mapsto \varphi(\omega).$$
From the Lemma we initially note that:
- There is a bijection $$\operatorname{Aut}_{\mathbb Q(\zeta)}K\to \{\alpha, \alpha\zeta,\alpha\zeta^2\}.$$ In fact $[K:\mathbb Q(\zeta)]=3$, so the minimal polynomial of $\alpha$ over $\mathbb Q(\zeta)$ has degree $3$ and it divides $f$, implying it is $f$ itself.
- Analogously there is a bijection $$\operatorname{Aut}_{\mathbb Q(\alpha)}K\to \{\zeta,\zeta^2\},$$ as the minimal polynomial of $\zeta$ over $\mathbb Q(\alpha)$ is $g$.
Call then $\sigma\in \operatorname{Aut}_{\mathbb Q(\alpha)}K$ the automorphism fixing $\alpha$ such that $\sigma(\zeta)=\zeta^2$, and call $\tau\in \operatorname{Aut}_{\mathbb Q(\zeta)}K$ the automorphism fixing $\zeta$ such that $\tau(\alpha)=\alpha\zeta$.
Before starting the proof, recall that any automorphism of $K$ is equal to $\tau^i\circ\sigma^j$, for some $0\le i\le 2$ and $0\le j\le 1$.
Step 1. $\alpha \notin E$: in fact $\alpha\in E$ implies $\mathbb Q\subset \mathbb Q(\alpha)\subset E$ but $[\mathbb Q(\alpha):\mathbb Q]=3$ does not divide $[E:\mathbb Q]=2$, a contradiction.
Step 2. $E(\alpha)=K$: in the equality $$[E(\alpha):\mathbb Q]=[E(\alpha):E][E:\mathbb Q]$$ we know $[E(\alpha):E]\gt 1$, as $\alpha\notin E$, and $[E:\mathbb Q]=2$; so $[E(\alpha):\mathbb Q]\ge 4$, but $[E(\alpha):\mathbb Q]$ divides $[K:\mathbb Q]=6$, so $[E(\alpha):\mathbb Q]=6$ and $E(\alpha)=K$.
Step 3. The minimal polynomial of $\alpha$ over $E$ is $f$. In fact $$[E(\alpha):E]=[E(\alpha):\mathbb Q][E:\mathbb Q]^{-1}=3;$$ so the minimal polynomial of $\alpha$ over $E$ has degree $3$, and it divides $f$, proving that it is $f$ itself.
Step 4. $K=E(\alpha)$, so by the Lemma there is a bijection $$\operatorname{Aut}_EK\to \{\alpha,\alpha\zeta,\alpha\zeta^2\}.$$ Let $\nu\in \operatorname{Aut}_EK$ be the automorphism fixing $E$ such that $\nu(\alpha)=\alpha\zeta$.
Step 5. The fixed field $F$ of $\nu$ is precisely $E$: it's obvious as $[K:E]=3$ is prime, and $[K:F]\gt 1$ because $\nu$ does not fix $\alpha$ for example.
Step 6. This is the problematic step, in which we prove that $\nu$ fixes $\zeta$. First note that, writing $\nu$ as $\tau^i\circ \sigma^j$, we have $$\zeta=\nu(\alpha)\alpha^{-1}=\tau^i(\alpha)\alpha^{-1}=\zeta^i,$$ hence $i=1$. Then either $\nu=\tau\circ\sigma$ or $\nu=\tau$. The text observes: $$\tau\circ\sigma(\alpha\zeta^2)=\tau(\alpha\zeta)=\alpha\zeta^2$$ implies that $\nu=\tau$.
Likely I should get a contradiction combining the observation above with the assumption that $\nu=\tau\circ\sigma $, knowing that $\nu(\alpha)=\alpha\zeta$ (but the same holds for $\tau\circ\sigma$) and it fixes $E$.
Step 7. As $\nu$ fixes $\zeta$, by Step 5 we have $\mathbb Q\subset\mathbb Q(\zeta)\subset E$; then $E=\mathbb Q(\zeta)$ because $[\mathbb Q(\zeta):\mathbb Q]=[E:\mathbb Q]=2$.
I would argue that the automorphisms of $K/E$ form a group, which is then necessarily the cyclic group of order three. As $\nu^2$ is not the identity, it must send $\alpha$ to $\alpha\zeta^2$. Now $$ \alpha\zeta^2 = \nu^2(\alpha) = \nu(\alpha\zeta) = \nu(\alpha)\nu(\zeta) = \alpha\zeta\nu(\zeta). $$ Thus $\nu(\zeta)=\zeta$.