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Let $\zeta_{32}$ be the 32th primitive root of unity. I want to calculate the Galois group of the extension $\mathbb Q(\zeta_{32} + \zeta_{32}^{-1})/\mathbb Q$.

First of all, the Galois group of $\zeta_{32}$ is the units of $\mathbb Z_{32}$, so an abelian group of order $16$ consisting of $1, 3, 5, 7, 9, 11, 13, 25, 17, 19, 21, 23, 25, 27, 29, 31$. It can be checked brutally by hand that $\zeta_{32} + \zeta_{32}^{-1}$ is fixed by $\langle 31 \rangle$. Note $\mathbb Q(\zeta_{32} + \zeta_{32}^{-1})/\mathbb Q$ is galois because its fixed field is a subgroup of an abelian group thus normal. Then its degree of extension should be $8$.

How do I calculate its Galois group?

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1 Answer 1

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Hint: Recall from elementary number theory $(\mathbb{Z}/2^n\mathbb{Z})^\times\cong C_{2^{n-2}}\times C_2$ for $n\geq 3$, with the $C_{2^{n-2}}$ generated by $5$ and the $C_2$ generated by $-1$.

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  • $\begingroup$ So the Galois group would be $C_8$ in this case. That is very useful!! Would you mind giving me some reference on the elementary number theory fact? $\endgroup$
    – zach
    Commented Jun 12, 2019 at 15:02
  • $\begingroup$ Most number theory textbook should have them covered. For example, Alan Baker's A concise introduction to the theory of numbers has it in page 25 as a remark (i.e., you should prove it yourself using the $a^{2^{j-2}}\equiv 1\pmod{2^j}$ proved in page 24). If you want a more expanded version, try Ireland and Rosen's A Classical Introduction to Modern Number Theory, page 43-44. $\endgroup$ Commented Jun 12, 2019 at 15:42
  • $\begingroup$ I see, thank you! By the way, is there any faster way to see that the fixed field is $\langle 31 \rangle$ other than checking it element by element? $\endgroup$
    – zach
    Commented Jun 12, 2019 at 15:46
  • $\begingroup$ Note that $\zeta+\zeta^{-1}$ is invariant under $\langle -1\rangle$? $\endgroup$ Commented Jun 12, 2019 at 15:52
  • $\begingroup$ One last question: My argument only shows that $\mathbb Q(\zeta_{32}+\zeta_{32}^{-1})$ is contained in the fixed field of $\langle 31 \rangle$, how to show that it is indeed euqal? $\endgroup$
    – zach
    Commented Jun 17, 2019 at 15:21

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