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Let $\zeta = e^{\frac{2i\pi}{7}}$. I know that the automorphisms of $\mathbb{Q}(\zeta)$ are isomorphic to the cyclic group with $6$ elements so that the subfields of $\mathbb{Q}(\zeta)$ correspond to the subgroups of $C_6$. I found one of the proper subfields is $\mathbb{Q}(\zeta + \zeta^2 + \zeta^4)$. I was able to determine this by observing the form of the elements that are fixed by the mapping that takes $\zeta$ to $\zeta^2$. How can I determine the other subfield that corresponds to the group of $2$ automorphisms consisting of the trivial mapping and the mapping that sends $\zeta$ to $\zeta^6$?

Robin
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Geoffrey Critzer
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    Note that $\zeta^6 = \zeta^{-1}$, so $\zeta + \zeta^6$ is invariant under your map. – Magdiragdag Mar 20 '16 at 20:23
  • @Magdiragdag Thanks I agree with the above statement but I do not understand how this will help me determine the subfield. – Geoffrey Critzer Mar 20 '16 at 20:28
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    In exactly the same way as you determined the other subfield: there you found the element $\zeta + \zeta^2 + \zeta^4$ that is invariant under the map $\zeta \mapsto \zeta^2$ and hence the field ${\mathbb Q}(\zeta + \zeta^2 + \zeta^4$) is the field corresponding to the subgroup $C_3$ of $C_6$. – Magdiragdag Mar 20 '16 at 20:30
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    What Magdiragdag says! Also, in this answer I give a more familiar description for the quadratic subfield that you found. – Jyrki Lahtonen Mar 20 '16 at 20:46
  • I think that the elements of Q(zeta) that are fixed by this mapping that takes zeta to zeta^6 must have the form a + b(zeta^2 + zeta^5) + c(zeta^3 + zeta^4) for some a,b,c in Q. So is the other field: Q( zeta^2 + zeta^5, zeta^3 + zeta^4) ) ? – Geoffrey Critzer Mar 20 '16 at 20:47
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    Geoffrey, you can also just use the element $u=\zeta+\zeta^{-1}=2\cos(2\pi/7)$ that Magdiragdag suggested. Observe that $u^2=\zeta^2+2+\zeta^5$ and $$u^3=\zeta^3+3\zeta+3\zeta^{-1}+\zeta^{-3},$$ so $u^3-3u=\zeta^3+\zeta^4$. – Jyrki Lahtonen Mar 20 '16 at 21:59
  • Let z = e^(2PiI/5). I have that the only proper subfield (other than Q) of Q(z) is Q(z^2 + z^3) = Q(sqrt(5)). Does anyone agree? – Geoffrey Critzer Mar 21 '16 at 17:34

1 Answers1

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  1. Basis Information:

    • $\mathbb{Q}(\zeta_7)$ is obtained by adjoining a primitive 7th root of unity, denoted as $\zeta_7$, to the rational numbers $\mathbb{Q}$.

  2. Galois Group:

    • The Galois group of the extension is isomorphic to $\mathbb{Z}_7^*$, the multiplicative group of integers modulo 7.

  3. Identifying Subgroups:

    • The subgroups of $\mathbb{Z}_7^*$ are $\{1\}, \{1, 2, 4\}, \{1, 3, 5, 2, 6, 4\}, \{1, 2, 3, 4, 5, 6\}$

  4. Fixed Fields:

    • Corresponding to each subgroup, we find the fixed fields under the action of the Galois group.

  5. Subfield 1: $\mathbb{Q}(\zeta_7)$:

    • The full field extension $\mathbb{Q}(\zeta_7)$.

  6. Subfield 2: $\mathbb{Q}(\zeta_7 + \zeta_7^{-1})$:

    • The fixed field under complex conjugation, which is represented by the subgroup $\{1\}$.

  7. Subfield 3: $\mathbb{Q}(\zeta_7 + \zeta_7^2 + \zeta_7^4)$:

    • The fixed field under a cyclic subgroup of order $3$, represented by $\{1, 2, 4\}$.

  8. Subfield 4: $\mathbb{Q}(\zeta_7^3)$:

    • The fixed field under a cyclic subgroup of order 2, represented by $\{1, 3, 5, 2, 6, 4\}$.

These subfields correspond to distinct intermediate field extensions between $\mathbb{Q}$ and $\mathbb{Q}(\zeta_7)$ and are determined by the fixed elements under specific subgroups of the Galois group.

Sapphire_Gaming21
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