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This is part of an exercise from Hungerford's Algebra:

Find all intermediate fields in the field extension $F_5/\mathbb{Q}$, where $F_5$ is the cyclotomic extension of $\mathbb{Q}$ of order $5$.

I figured out the automorphism group, it's $\mathbb{Z}_4$. So, since there is only one non-trivial subgroup of order $2$, by the Fundamental Theorem, I should find precisely one intermediate field of degree $2$, which (in my understanding) in this case means I should find an element with degree $2$ minimal polynomial. But, apparently I couldn't. I'll appreciate any help.

Also, in general, is my approach the only way to find the intermediate fields? In other words, do we always need to do a trick, and find an element with a minimal polynomial of desired degree, or is there another approach that doesn't require a separate trick in each different case? Thanks!

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Remember what the fundamental theorem of Galois theory says: if $L/K$ is a finite Galois extension, then given a subgroup $H\subseteq\mathrm{Gal}(L/K)$, the intermediate field corresponding to it is $$L^H=\{\alpha\in L:\sigma(\alpha)=\alpha\text{ for all }\sigma\in\mathrm{Gal}(L/K)\}$$ Observe that it's not enough to know what $\mathrm{Gal}(L/K)$ is isomorphic to, you have to actually know what the automorphisms are (which is good advice regardless).

In your situation, it's not enough to know that $\mathrm{Gal}(F_5/\mathbb{Q})\cong\mathbb{Z}/4\mathbb{Z}$, you should know that $$\mathrm{Gal}(F_5/\mathbb{Q})=\{\sigma_r:F_5\to F_5, \;\sigma_r(\zeta_5)=\zeta_5^r:r=1,2,3,4\}$$ You should figure out an explicit isomorphism $\mathrm{Gal}(F_5/\mathbb{Q})\cong\mathbb{Z}/4\mathbb{Z}$, i.e., write down which $\sigma_r$ goes to which element of $\mathbb{Z}/4\mathbb{Z}$. Then you'll know which $\sigma_r$ comprise the subgroup of $\mathrm{Gal}(F_5/\mathbb{Q})$ of order $2$. (Then you can get to work finding the subfield fixed by them.)

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    $\begingroup$ I see, thanks a lot! So, in this case it turns to be $\sigma_4$ which definitely has the fixed field $\mathbb{Q}(\zeta_5+\zeta_5^{-1})$ mainly because $\zeta_5^{-1}=\zeta_5^4$. It really reminded me how to use fixed field concept, again bunch of thanks! $\endgroup$ – vgmath Jun 20 '15 at 17:52
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    $\begingroup$ Yup, that's exactly right! Glad I could help :) And the minimal polynomial of $\zeta_5+\zeta_5^{-1}$ over $\mathbb{Q}$ is $x^2+x-1$, of degree $2$ as we expected. $\endgroup$ – Zev Chonoles Jun 20 '15 at 17:56
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    $\begingroup$ Actually, if you don't mind, I had another question that I remembered to ask after your this comment. How did you quickly come up with the minimal polynomial? I just wondered what kind of a trick you're generally using? Sorry for a second question. $\endgroup$ – vgmath Jun 20 '15 at 17:59
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    $\begingroup$ This one I happened to know of the top of my head, but I explain a general approach in my post here. $\endgroup$ – Zev Chonoles Jun 20 '15 at 18:21
  • $\begingroup$ I see, thanks a lot! $\endgroup$ – vgmath Jun 20 '15 at 18:26

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