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Show that $\sigma(\sqrt[7]{2}), \sigma(e^{\frac{2\pi i}{7}})$ determine an automorphism $\sigma$ on $\mathbb{Q}(\sqrt[7]{2}, e^{\frac{2\pi i}{7}})$.

My Attempt: Let $\alpha=\sqrt[7]{2}$ and $\zeta=e^{\frac{2\pi i}{7}}$. Indeed, for every automorphism $\sigma$ on $\mathbb{Q}(\alpha, \zeta)$, we must have $\sigma(\alpha)\in A:=\{\alpha, \zeta\alpha, \zeta^2\alpha, \zeta^3\alpha, \dots, \zeta^6\alpha\}$, and $\sigma(\zeta)\in B:=\{\zeta, \zeta^2, \dots, \zeta^6\}$.

Let $\sigma(\alpha)=a\in A$ and $\sigma(\zeta)=b \in B$, then define $\sigma(\frac{p(\alpha, \zeta)}{q(\alpha, \zeta)})=\frac{p(a, b)}{q(a, b)}$, where $p, q \in \mathbb{Q}[t_1, t_2]$. We claim that $\sigma$ is an automorphism.

  1. First we must show that $\sigma$ is well-defined. Suppose $\frac{f(\alpha, \zeta)}{g(\alpha, \zeta)}=\frac{p(\alpha, \zeta)}{q(\alpha, \zeta)}$, where $g(\alpha, \zeta)\neq0\neq q(\alpha, \zeta)$. We claim that $\sigma(\frac{f(\alpha, \zeta)}{g(\alpha, \zeta)})=\sigma(\frac{p(\alpha, \zeta)}{q(\alpha, \zeta)})$, i.e., $\frac{f(a, b)}{g(a, b)}=\frac{p(a, b)}{q(a, b)}$.

Proof:

Lemma: Suppose $p, f\in\mathbb{Q}[t_1, t_2]$, then $f(\alpha. \zeta)=p(\alpha, \zeta)\implies f(a, b)=p(a ,b)$.

Proof of Lemma: Note $a=\alpha\zeta^i, b=\zeta^j$ for some $0\leq i\leq6, 1\leq j\leq6$. Note each term in $f, p$ has the form $ct_1^mt_2^n$, where $c\in\mathbb{Q}$; hence for $f(\alpha, \zeta)$, each term has the form $c\alpha^m\zeta^n=c2^k\alpha^{m'}\zeta^{n'}$, where $0\leq m', n'\leq6$, $m'\equiv_7m, n'\equiv_7n$, and $m-m'=7k$. The same term in $\mathbf{f(a, b)}$ becomes $\mathbf{ca^mb^n=c(\alpha\zeta^i)^m(\zeta^j)^n=c\alpha^m\zeta^{im}\zeta^{jn}=c\alpha^m\zeta^{im+jn}}=c2^k\alpha^{m'}\zeta^t$, where $\mathbf{0\leq m', t\leq6}$, $\mathbf{m'\equiv_7m, t\equiv_7im+jn}$. Thus $f(\alpha, \zeta)=c_0+c_1\alpha+c_2\zeta+c_3\alpha\zeta+c_4\alpha^2+\dots+c_s\alpha^6\zeta^6$, where $c_i\in\mathbb{Q}$. Similarly, $p(\alpha, \zeta)=d_0+d_1\alpha+d_2\zeta+d_3\alpha\zeta+d_4\alpha^2+\dots+d_s\alpha^6\zeta^6$, where $d_i\in\mathbb{Q}$. If $f(\alpha, \zeta)=p(\alpha, \zeta)$, then we must have $c_0=d_0, c_1=d_1, \dots, c_s=d_s$.

Now, by the boldfaced text, the only difference between the terms in $f(\alpha, \zeta)$ and $f(a, b)$ corresponding to $ct_1^mt_2^n$ is that $\zeta^{n'}$ becomes $\zeta^t$. Screeching halt.

My Question: Should I define $\sigma$ explicitly in a different way? Or is there a trick that I could use? Any help would be greatly appreciated.

Dick Grayson
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    I think this is more or less a duplicate of this older, more general version. By the way, I did not see any finite fields in the question so I removed the tag. – Jyrki Lahtonen Apr 08 '22 at 05:28
  • @JyrkiLahtonen I appreciate the help and the reference! But like this answer here, the solution involves machinery that I am not supposed to use, so I wonder if there is any elementary approach to the problem. – Dick Grayson Apr 08 '22 at 05:34
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    Yeah, I think we have an example thread somewhere that may help you more. Gimme a minute... or five... – Jyrki Lahtonen Apr 08 '22 at 05:36
  • @JyrkiLahtonen Hey no rush at all! – Dick Grayson Apr 08 '22 at 05:37
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    May be start from here or here? Five in place of seven changes a few things but not that much actually. – Jyrki Lahtonen Apr 08 '22 at 05:40
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    Each $\Bbb Q$-automorphism $\sigma$ of the given field $K=\Bbb Q(a, u)$ - with $a=\sqrt[7]2$ and $u=\exp\frac 172\pi i$ - is determined by the two images $\sigma(a)$ and $\sigma(u)$ of the two field generators of $K$. – dan_fulea Apr 08 '22 at 10:23
  • @dan_fulea Thank you so much for the tip! I see you point; but I still struggle to prove that such $\sigma$'s are well-defined on $\mathbb{Q}(a, u)$-see the added details in my post. – Dick Grayson Apr 08 '22 at 20:34
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    You can realize $K=\Bbb Q(u)$ as the polynomial ring $\Bbb Q[U]$ modulo the (ideal $J$ generated by the) irreducible cyclotomic polynomial $\Phi_7=U^6+\dots +U+1$. To show that $u\to u^k$ is indeed a morphism, it is enough to check the map $f$ given by $U\to U^k$ in the polynomial ring. (That it lands again in the ideal $J$. Show that $f(\Phi_7)$ has a factor $\Phi_t$, check the roots.) For the rest work in $L=K(a)$, seen as $K[A]$ modulo the polynomial $A^7-2$. – dan_fulea Apr 08 '22 at 20:59
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    Well every member of $K=\mathbb{Q} (\alpha, \zeta) $ is a rational function of $\alpha, \zeta$ with rational coefficients. Thus if $f(\alpha, \zeta) \in K$ where $f(x, y) \in\mathbb {Q} (x, y) $ and $\sigma $ is a $\mathbb {Q} $-automorphism of $K$ then by definition (of an automorphism) we have $\sigma(f(\alpha, \zeta)) =f(\sigma (\alpha), \sigma(\zeta)) $. I don't think we need to show anything more than that. – Paramanand Singh Apr 10 '22 at 12:29
  • @ParamanandSingh Thank you very much for the tips! However, my primary objective here is too find an explicit formula/description for the $\mathbb{Q}$-automorphism, and that's why I have to be rigorous. – Dick Grayson Apr 10 '22 at 13:11
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    @DickGrayson: my previous comment is fully rigorous and I don't think some examiner would require more detail in a typical exam related to abstract algebra course. The way to explicitly give automorphisms in this case is to give the values of $\sigma(\alpha), \sigma(\zeta) $ explicitly. – Paramanand Singh Apr 10 '22 at 15:55
  • @ParamanandSingh I believe the word I should have used is "complete", not "rigorous". And you are absolutely correct-this is more of a personal project. – Dick Grayson Apr 10 '22 at 18:11

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