# Galois group of the sum of the 32th primitive of unity and its inverse

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?

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 $$-1$$.

• 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?
– zach
Commented Jun 12, 2019 at 15:02
• 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. Commented Jun 12, 2019 at 15:42
• 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?
– zach
Commented Jun 12, 2019 at 15:46
• Note that $\zeta+\zeta^{-1}$ is invariant under $\langle -1\rangle$? Commented Jun 12, 2019 at 15:52
• 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?
– zach
Commented Jun 17, 2019 at 15:21