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Let $K = \mathbb{Q}(\sqrt{5})$. I'm trying to find an algebraic number $z \in \mathbb{C}$ such that $K(z)/K \cong \mathbb{Z}/7\mathbb{Z}$. How to go about this? The only thing I could think of as useful is the fact that $\mathbb{Q} \subset \mathbb{Q}(\sqrt{5}) \subset \mathbb{Q}(\zeta_5)$ is a tower of field extensions.

Sigurd
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How about this? Take $\mathbb{Q}(\zeta_{29})$. Its Galois group over $\mathbb{Q}$ is cyclic of order 28, so it has a unique subgroup of order $4$; let $L$ be the fixed field of this subgroup.

Now $L$ is linearly disjoint from $K$, so $LK$ (as an extension of $K$) has Galois group $\mathbb{Z}/7\mathbb{Z}$ as desired.

hunter
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  • Nice idea, what does the last line mean? In particular, what does it mean to be linearly disjoint? – Sigurd Jun 25 '20 at 13:36
  • If say $L = \mathbb{Q}(\sqrt[3]{3})$ and $K = \mathbb{Q}(\sqrt[3]{3}\zeta_3)$ then the degree of the compositum over $L$ would not be the same as the degree of $K$ over $\mathbb{Q}$. This can't happen in our case since both fields are Galois over $\mathbb{Q}$. – hunter Jun 25 '20 at 13:38
  • To check: we have two extension. One is $\mathbb{Q} \subset \mathbb{Q}(\sqrt{5})$ of degree 2 and one is $\mathbb{Q} \subset \mathbb{Q}(\eta_4)$, of degree 7, where $\eta_4$ is the 4th Gaussion perdiod of $\zeta_{28}$. This extension has Galois group $\mathbb{Z}/7\mathbb{Z}$. Now we are claiming that then the extension $\mathbb{Q}(\sqrt{5}) \subset \mathbb{Q}(\sqrt{5},\eta_4)$ also has Galois group $\mathbb{Z}/7\mathbb{Z}$? I don't see why that's the case. – Sigurd Jun 25 '20 at 13:43
  • @Sigurd one has a "diamond" of fields with linearly disjoint ones on the side; the Galois groups of "opposite" sides are isomorphic. See e.g. here: https://kconrad.math.uconn.edu/blurbs/galoistheory/galoiscorrthms.pdf – hunter Jun 25 '20 at 13:46
  • Nice trick, thanks! I actually meant $\eta_7$, the 7th Gaussion period, with $H_7 \subset (\mathbb{Z}/29\mathbb{Z})^\ast$ as subgroup of index 7, and order 4? – Sigurd Jun 25 '20 at 13:52
  • @Sigurd I haven't heard this phrase Gaussian period before, but what you want is the fixed field of the subgroup of order $4$. Concretely, the field is generated by $\zeta + \zeta^{12} + \zeta^{-1} + \zeta^{17}$ if I am not mistaken. – hunter Jun 25 '20 at 13:54
  • Yes, this is what is meant I think. If you're interested, I just posted a question about Gaussian periods, including its definition: https://math.stackexchange.com/questions/3734153/finding-a-counter-example-for-gaussion-perdiods-for-non-primes – Sigurd Jun 25 '20 at 14:01