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Actually I am interested in degree $ 8 $ cyclic extension over $ \mathbb{Q} $. Let $ L $ be such extension. At first I was thinking to take basis as normal basis, as we can determine the galois group and how the group acted on $ L $ but we don't know how the multiplication can be defined. Next to make easy calculation in multiplication I take primitive element as this is simple extension but we dont know how the Galois group will act on $ L $. Then I consider a particular cyclic extension, $ \mathbb Q(\eta) $ where $ \eta $ be $ 17^{\text{th}} $ primitive roots of unity. Then by galois fundamental theorem we have an extension(cyclic Galois) $ L $ over $ \mathbb{Q} $ of degree $ 8 $. Now in this extension we can determine the multiplication rule and the action of galois group also, but multiplication between two elements like $b $ and $\sigma(b)$ are so much calculative. So does there any cyclic extension $ L $ over $ \mathbb{Q} $ where we know the multiplication rule and also the galois group action and also the calculation is easier. I.e we can calculate the term $ b\sigma(b) $ easily where $ b \in L $.

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    $\begingroup$ I am confused as to why the real subfield of $\mathbb{Q}(\zeta_{17})$ doesn't satisfy. Set $c_a = \zeta^a + \zeta^{-a}$ for $a \in (\mathbb{Z}/17 \mathbb{Z})^{\times}/(\pm 1)$. Then the $c_a$ are a basis, the Galois action on that basis is obvious, and multiplication is $c_a c_b = c_{a+b} + c_{a-b}$ (for $a \neq b$) and $c_a^2 = c_{2a}+2= c_{2a} - 2 \sum_{a=1}^8 c_a$. What would make this nicer? $\endgroup$ Commented Jan 27, 2022 at 11:34
  • $\begingroup$ You can make this look more familiar by using the basis $\tfrac{1}{2} c_a = \cos \tfrac{2 \pi a}{17}$ instead, so that the mulitplication rule is the familiar trig formula sosmath.com/trig/prodform/prodform.html . But this is probably not a good idea, since $c_a$ is an algebraic integer and $\tfrac{1}{2} c_a$ isn't. $\endgroup$ Commented Jan 27, 2022 at 11:37
  • $\begingroup$ Yes @david E Speyer , your 1st idea is good, I have also get similar . Okay now I will definitely use ur idea. $\endgroup$
    – Sky
    Commented Jan 27, 2022 at 11:49

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