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I tried to find all intermediate fields of $\mathbb{Q}(\zeta_{20})$, where $\zeta=\zeta_{20}$ is the primitive 20th root of unity. I found the Galois group of $\mathbb{Q}(\zeta)$ over $\mathbb{Q}$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$ and it has 6 subgroups. If the generater of $\mathbb{Z}/2\mathbb{Z}$ is $\sigma$ and that of $\mathbb{Z}/4\mathbb{Z}$ is $\tau$, correspondence between subgroups and intermediate fields is following.

$$ <\sigma^2>\leftrightarrow \mathbb{Q}(\zeta+\zeta^9) $$ $$ <\tau\sigma^2>\leftrightarrow \mathbb{Q}(\zeta+\zeta^{19}) $$ $$ <\sigma>\leftrightarrow \mathbb{Q}(\zeta+\zeta^3+\zeta^7+\zeta^9) $$ $$ <\tau\sigma>\leftrightarrow \mathbb{Q}(\zeta+\zeta^{9}+\zeta^{13}+\zeta^{17}) $$

The question I have is what a subgroup $<\sigma^2,\tau>,<\tau>$ corresponds. Please give me some advice.

user26857
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masutarou
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    There are many ways to write this Galois group as a direct product. It would be more useful to tell us what $\tau$ and $\sigma$ are as automorphisms. What are $\tau(\zeta)$ and $\sigma(\zeta)$? – Jyrki Lahtonen Aug 10 '17 at 13:10
  • I first thought that $\tau(\zeta)=\zeta^{11}$ and $\sigma(\zeta)=\zeta^3$. Then $\tau$ would have order two and $\sigma$ order four contradicting your description. Furthermore, $\langle \tau\sigma\rangle$ appears twice on your list. Please check all this. – Jyrki Lahtonen Aug 10 '17 at 13:16
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    Yet another problem is that $\zeta^{10}=-1$, so $\zeta+\zeta^{11}=0$. Therefore the first item on your list must be wrong. – Jyrki Lahtonen Aug 10 '17 at 13:19
  • I'm sorry. Your are right. I made some mistakes. Please wait some minutes. I edit soon. – masutarou Aug 10 '17 at 13:32

1 Answers1

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There are apparently some confusions about the identification of the elements. In an attempt to clarify those let's denote by $\sigma_j$ the automorphism determined by $\sigma_j(\zeta)=\zeta^j$. Here $j\in S:=\{1,3,7,9,11,13,17,19\}$, and the Galois group consists of all the elements $\sigma_j, j\in S$.

The subgroups of order four are then

  • $H_1=\langle\sigma_3\rangle=\{\sigma_j\mid j\in\{1,3,9,7\}\}$ (cyclic),
  • $H_2=\langle\sigma_{13}\rangle=\{\sigma_j\mid j\in\{1,13,9,17\}\}$ (cyclic),
  • $H_3=\langle\sigma_9,\sigma_{11}\rangle=\{\sigma_j\mid j\in\{1,9,11,19\}\}$ (isomorphic to Klein Viergruppe).

The fixed fields of these must be quadratic extensions of $\Bbb{Q}$. What could those be? Clearly $i\in L:=\Bbb{Q}(\zeta_{20})$, so $\Bbb{Q}(i)$ is one of them. It is a standard exercise to show that $\sqrt5\in\Bbb{Q}(\zeta_5)\subset\Bbb{Q}(\zeta_{20})$, so $\Bbb{Q}(\sqrt5)$ is another quadratic subfield. The third quadratic subfield is thus $\Bbb{Q}(\sqrt{-5})$.

Which subfield correspondens to which subgroup of order four? A most obvious thing is that $\sigma_{19}$ is the usual complex conjugation. Therefore its fixed points are all real numbers. This allows us to deduce that the fixed field of $H_3$ must be $\Bbb{Q}(\sqrt5)$. $i=\zeta_{20}^5$, so $\sigma_j(i)=i^j$. The exponents $1,13,9,17$ are all congruent to $1\pmod4$, so we see that $i$ is a fixed point for all of the automorphisms in $H_2$. Hence $\Bbb{Q}(i)$ is the fixed field of $H_2$. By a process of elimination we then see that the fixed field of $H_1$ must be $\Bbb{Q}(\sqrt{-5})$.

The group $G$ has three elements of order two, so there are three subgroups of order two, yielding three quartic subfields. The Galois correspondence goes as follows:

  • The subgroup $K_1=\langle\sigma_9\rangle=H_1\cap H_2\cap H_3$, so its fixed field must be $\Bbb{Q}(i,\sqrt5)$.
  • The subgroup $K_2=\langle\sigma_{19}\rangle$ has complex conjugation as its only non-trivial automorphism, so its fixed field is the intersection $L\cap\Bbb{R}$. That field is generated by the number $z=2\cos(\pi/10)=\zeta+\zeta^{-1}$. The minimal polynomial of $z$ is $x^4-5x^2+5$ (irreducible by Eisenstein).
  • The automorphism $\sigma_{11}$ maps the fifth root of unity $\zeta_5=\zeta^4$ to itself, so the fixed field of $K_3=\langle\sigma_{11}\rangle$ is the quartic field $\Bbb{Q}(\zeta_5)$.
Jyrki Lahtonen
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  • You can recall that $$\Bbb{Q}(\sqrt5)=\Bbb{Q}(2\cos(2\pi/5))=\Bbb{Q}(\zeta_5+\zeta_5^{-1}).$$ As an extra exercise, using $\zeta_5=\zeta^4$, you can then verify that $$\frac{1+\sqrt5}2=2\cos(2\pi/5)=\zeta^4+\zeta^{-4}$$ is invariant under all the automorphisms of $H_3$. – Jyrki Lahtonen Aug 10 '17 at 13:49
  • Thank you for your help. One of three subgroups of oder two is $<\sigma_{9}>=H_1\cap H_2$. So, by the fundamental theorem of Galois theory, corresponding subfield is $\mathbb{Q}(i,\sqrt{5})$. – masutarou Aug 10 '17 at 14:08
  • Correct, @masutarou. Well done! Can you figure out which automorphisms fix $\zeta_5=\zeta^4$? – Jyrki Lahtonen Aug 10 '17 at 14:09
  • It is $\tau$ above, that is ,$\zeta\rightarrow \zeta^{11}$. – masutarou Aug 10 '17 at 14:19
  • Correct again! Updating my answer... – Jyrki Lahtonen Aug 10 '17 at 14:20
  • I have one more question. I can understand $\zeta^4$ is fixed by $\sigma_{11}$ thanks to your hint. But how do you find $\zeta^4$. Is thorough search necessary? – masutarou Aug 10 '17 at 14:32
  • A good question @masutarou. The idea was, again, to start listing subfields of $L=\Bbb{Q}(\zeta_{20})$ Because $4\mid 20$ and $5\mid20$ we have both $\Bbb{Q}(\zeta_5)$ and $\Bbb{Q}(\zeta_4)=\Bbb{Q}(i)$ among the subfields. Using that extra piece of information tells that some subgroup (of a size we can predict) has them as fixed fields. The question then becomes to identify that subgroup. This kind of reverse engineering is often a part of the game. It becomes trickier when factors (here of $20$) and the real subfield are not enough. – Jyrki Lahtonen Aug 10 '17 at 14:38
  • I see. Thank you many times! – masutarou Aug 10 '17 at 14:45