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Let $K = \mathbb{R}(X)$ be the field of rational functions with real coefficients, and let $$F = \mathbb{R}\left(X^4 - \frac{1}{X^4}\right)$$ be a subfield of $K.$ Let $L$ be the Galois closure of the field extension $K/F.$ I am asked to show that the Galois group $\Gamma(L/F)$ has an element of order $8$, and to find all intermediate extensions of degree $4$.

First, I'd need to know what $L$ is. Note that $K = F(X),$ so $L$ is the splitting field of the minimal polynomial of $X$ over $F.$ A polynomial that does indeed kill $X$ is $$f(t) = t^8 - \left(X^4 - \frac{1}{X^4}\right)t^4 - 1.$$ We need to show that this is irreducible. Note that $$[K:F(X)] = [\mathbb{R}(X):\mathbb{R}(X^4)][\mathbb{R}(X^4):\mathbb{R}(X^4 - X^{-4})] = 4\times2=8.$$ So by degree considerations, $f(t) \in F[t]$ is indeed the minimal polynomial of $X$.

After some calculation, we can show that the roots of $f(t)$ are $$\pm X, \pm iX, \pm \frac {\zeta_8} X, \pm \frac {i\zeta_8} X,$$ where $\zeta_8 = e^{2\pi i/8}.$ Therefore, $L = \mathbb{C}(X),$ which means $$[L:F] = [\mathbb{C}(X):\mathbb{R}(X)][\mathbb{R}(X):F] = 16.$$ I believe/hope I am right so far.

However, I can't seem to show that $\Gamma(L/F)$ has an element of order $8$. Any field automorphism of $L$ fixing $F$ must permute the roots of $f(t)$ as the coefficients are fixed. However, I can't seem to find a permutation that has order $8:$

  • $X \mapsto -X$ has order $2$.
  • $X \mapsto \pm iX$ has order $4$.
  • $X \mapsto \zeta_8/X$ has order $2$.
  • All the rest also seems to have order $2$ because $X \mapsto aX^{-1} \mapsto X$ for any constant $a$?

What exactly am I doing wrong? It's been a while since I've done Galois theory, so I feel like I might be making a simple mistake. Any help will be much appreciated.

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    $\begingroup$ Hmm. Don't know for sure whether it plays a role, but $L=F(X,i)$. So an automorphism is not uniquely determined by where it maps $X$, you also need to specify the image of $i$. $\endgroup$ – Jyrki Lahtonen Aug 16 '18 at 12:16
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    $\begingroup$ Yes! Let $\sigma$ be the automorphism that maps $x\mapsto \zeta/X$ and $i\mapsto -i$. Then $\zeta\mapsto \overline{\zeta}=1/\zeta$, and hence $$\sigma: X\mapsto \frac{\zeta}X\mapsto \frac{1/\zeta}{\zeta/X}=\frac{X}{\zeta^2}=-iX.$$ Therefore $\sigma^2(X)=-iX$. Clearly $\sigma^2(i)=i$. It looks like $\sigma$ has order eight. $\endgroup$ – Jyrki Lahtonen Aug 16 '18 at 12:22
  • $\begingroup$ Is it clear to you why an automorphism like $\sigma$ above exists? $\endgroup$ – Jyrki Lahtonen Aug 16 '18 at 12:23
  • $\begingroup$ @JyrkiLahtonen right, of course I need to check where $i$ goes to! $\endgroup$ – dodo628 Aug 16 '18 at 12:59
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You already observed that $[L:F]=16$. As $L=F(X,i)$ any element $\tau$ of the Galois group $G=Gal(L/F)$ is determined uniquely, if we know the images $\tau(X)$ and $\tau(i)$.

Clearly $\tau(X)$ has to be a zero of $f(t)$, so there are eight choices. Equally clearly $\tau(i)=\pm i$, two choices. That's sixteen combinations, so they all must occur, each with a uniquely determined automorphism.

You asked for an automorphism of order eight. The automorphism $\sigma$ determined by $\sigma(i)=-i$, $\sigma(X)=\zeta/X$ turns out to be one such. The restriction of $\sigma$ to $\Bbb{C}$ is the usual complex conjugation. Therefore $\sigma(\zeta)=\overline{\zeta}=1/\zeta$. Therefore $$ \sigma^2(X)=\sigma(\zeta/X)=\frac{\sigma(\zeta)}{\sigma(X)}=\frac{1/\zeta}{\zeta/X}=\frac{X}{\zeta^2}=-iX. $$ It follows that $\sigma$ permutes the roots of $f(t)$ as follows $$ \sigma:X\mapsto \frac{\zeta}{X}\mapsto -iX\mapsto\frac{i\zeta}{X}\mapsto-X\mapsto-\frac{\zeta}X\mapsto iX\mapsto-\frac{i\zeta}X\mapsto X, $$ that is, as an 8-cycle.

As you observed, the automorphism $\alpha$ defined by $X\mapsto \zeta/X$, $i\mapsto i$, has order two. Unless I made a mistake, we have $\alpha\sigma\alpha=\sigma^{-1}$ implying that $G\simeq D_8$, the group of symmetries of a regular octagon. Here $\sigma$ corresponds to a rotation by $\pi/4$, and $\alpha$ is one of the reflections.

The degree four extension fields are the fixed fields of subgroups of order four. By the well known structure of the dihedral group we know that the only cyclic subgroup of order four is generated by $\sigma^2$. Also, reflections w.r.t. any pair of orthogonal symmetry axes of the octagon generate a copy of the Klein four group. There are a total of four such subgroups.

Hope this gets you started with the hunt for the degree four intermediate fields!

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  • $\begingroup$ Thank you so much for your answer! Cleared up a lot of stuff for me. Just one thing, why is it clear that $\tau(i) = \pm i?$ Those certainly are the most obvious choices, but why can't $\tau(i) = X + i,$ for example? $\endgroup$ – dodo628 Aug 16 '18 at 13:05
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    $\begingroup$ $\Bbb{R}$ is contained in $F$. Therefore the coefficients of $g(t)=t^2+1=(t+i)(t-i)$ are in $F$. Consequently $\tau$ must permute the zeros of $g(t)$. $\endgroup$ – Jyrki Lahtonen Aug 16 '18 at 13:10

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