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Let $\xi\in\overline{\mathbb{Q}}\backslash\mathbb{Q}$ be an algebraic number of degree $d\geq2$ and let $\mathbb{K}=\mathbb{Q}\left(\xi\right)$.

What are some conditions on $\xi$ and $d$ which will guarantee that the galois group $\textrm{Gal}\left(\mathbb{K}/\mathbb{Q}\right)$ of $\mathbb{K}$ over $\mathbb{Q}$ is cyclic? A list of multiple such conditions would be fine, assuming that there isn't already a comprehensive answer to this question.

MCS
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  • possible duplicate of https://math.stackexchange.com/questions/2204162/real-cyclic-extensions-of-mathbb-q-of-given-degree – jijijojo Jan 09 '19 at 01:36
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    What you want to use is the fact that every abelian extension of $\Bbb Q$ is contained in a cyclotomic extension $\Bbb Q(\zeta_n)$, where $\zeta_n$ is a primitive $n$-th root of unity. There are wrinkles, but the task is not hard. – Lubin Jan 09 '19 at 02:31
  • I am an analyst. I have no idea how to do any of the things of which you speak. I am looking for an answer so I can press forward. I don't like being stuck like this, unable to do anything until someone finally deigns to answer my question, but I have no other option. – MCS Jan 09 '19 at 20:31
  • @mouthetics: No, this is not a duplicate. I am not asking to construct a field extension with a given galois group, I am asking, "given a field extension of this sort, when will it be a cyclic extension". This isn't a homework problem, it's a question I need answered in order to press on with what I'm doing. – MCS Jan 09 '19 at 20:33
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    As it is presented, your question cannot be answered, because the given data are too "fuzzy". For instance, you define your number field as $K=Q(x)$. But such a primitive element $x$ is not unique. To make this point clearer, just start from the opposite point of view, take a cyclic $K$ and try to find an adequate $x$. You are immediately brought back to such trials as the one suggested by @Lubin, namely use the Kronecker-Weber theorem. But even this powerful tool couldn't always give you a canonical answer. Unless you particularize the situation by adding further hypotheses... – nguyen quang do Jan 10 '19 at 07:59
  • A down-to-earth approach would be to compute the Galois group $G$ of $K/Q$ and check that it is cyclic (after all, this is the definition of a cyclic extension). In the general case (i.e. without extra information) this would require to give not only $x$ but also its minimal polynomial $f$. Then $G$ could be determined by sending $x$ to all the roots of $f$. But again this is unnatural, because it is hard to compute the roots of a polynomial. After all, the fundamental idea of Galois theory is to replace the manipulation of roots by that of groups acting on them. – nguyen quang do Jan 10 '19 at 08:35
  • @nguyenquangdo: Yes, of course, starting with the galois group would be the more sensible way of doing things, but the things I am doing (elliptic curves) take the field extension as a given, rather than the galois group. That being said, I did find a partial answer on wikipedia. Cyclotomic extensions by a primitive $n$th root of unity have (Z/nZ)* as their galois groups, and Gauss established exactly for which $n$ for which those groups are cyclic. That actually proved to be enough for my purposes. Thanks for your time, though. :) – MCS Jan 11 '19 at 21:15
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    This approach was suggested by @Lubin. Precisely, the "wrinkles" he alludes to can be smoothed by the theory of Gauss periods (see Wiki.) – nguyen quang do Jan 11 '19 at 21:30
  • MCS You don't get ALL the cyclic extensions by constraining yourself to cyclotomic fields. What Lubin and Nguyen Quang Do suggested is to take those subfields of cyclotomic fields that also are cyclic. If $p$ is a prime you know that $\Bbb{Q}(\zeta_{p^n})$ is cyclic. Hence, so are all its subfields. For example $\Bbb{Q}(2\cos(2\pi/p))$ for any prime $p$. Also, you don't need $\Bbb{Z}_n^*$ to be cyclic in order to find new cyclic subfields. – Jyrki Lahtonen Jan 14 '19 at 20:45
  • (cont'd) For example $\Bbb{Z}{35}^*$ is the direct product of the cyclic subgroups by the residue classes of $3$ and $-1$ respectively. Implying that the fixed field of the automorphism corresponding to $-1$ is a cyclic extension of degree $ord(3)=12$. The automorphism corresponding to $-1$ is the usual complex conjugation, so the fixed field $$L=\Bbb{Q}(\zeta{35})\cap\Bbb{R}=\Bbb{Q}(2\cos(2\pi/35))$$ is a cyclic extension of degree twelve. – Jyrki Lahtonen Jan 14 '19 at 20:48
  • What they were also saying is that if you pick a random element $\alpha$ of the above field $L$, it is a bit inconvenient (but in principle relatively simple, depending) to decide whether $L=\Bbb{Q}(\alpha)$. No matter, a subfield of a cyclic extension is itself cyclic, so you still get a cyclic field if you don't get all of $L$. – Jyrki Lahtonen Jan 14 '19 at 20:51

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