Let $K=\mathbb{Q}(\omega)$ be given, where $\omega^3=1$. I want to know:

(1) Whether there is a Galois extension $L/\mathbb{Q}$ containing $K$ such that $\mathrm{Gal}(L/\mathbb{Q})\cong\mathbb{Z}_4$?

(2) Whether there is a Galois extension $L/\mathbb{Q}$ containing $K$ such that $\mathrm{Gal}(L/\mathbb{Q})\cong Q$, where $Q$ is the quaternion group with 8 elements.

I have no idea how to solve such problem.

  • 1
    $\begingroup$ Can you share with us what you know about Galois extensions, Galois groups, the Galois correspondence etc.? $\endgroup$ – Alon Amit Jun 23 '13 at 6:31
  • $\begingroup$ @AlonAmit, I'm new to Galois theory, but I can say I know the definitions of the concepts you mentioned, and have basic understandings. $\endgroup$ – hxhxhx88 Jun 23 '13 at 6:44
  • $\begingroup$ Why are both of the questions labeled (1)? For the first (1), have you looked at the extension by 12th roots of 1? $\endgroup$ – Gerry Myerson Jun 23 '13 at 7:22
  • $\begingroup$ @GerryMyerson, sorry, another typo, should be (1) and (2). I thought $\mathrm{Gal}(\mathbb{Q}(\zeta_{12})/\mathbb{Q})\cong \mathbb{Z}_2\times\mathbb{Z}_2$. Because it is $\mathbb{Z}_{12}^*$ $\endgroup$ – hxhxhx88 Jun 23 '13 at 8:40
  • $\begingroup$ Sorry, you're right about 12th roots. $\endgroup$ – Gerry Myerson Jun 23 '13 at 11:18

The answer to part 1 is that such a field $L$ cannot exist. This is seen as follows. Assume that $L=\mathbb{Q}(\alpha)$ were such a field (the extension is known to be simple). Then the complex conjugate $\overline{\alpha}$ would also be an element of $L$ as is shares the minimal polynomial (over the rationals!) with $\alpha$. The argument can be repeated for all the elements of $L$, so we have shown that $L$ is stable under complex conjugation.

But then the fixed field of complex conjugation, $\mathbb{R}\cap L$, is necessarily a quadratic extension of the rationals. Therefore $L$ contains at least two intermediate quadratic fields contradicting the fact that a cyclic group contains a unique subgroup of each possible order.

The non-existence of the field in part 2 is a bit different. As in part 1 we see that $L$ has to be stable under complex conjugation. The restriction of complex conjgation to $L$ is clearly of order two (can't be identity, when $K\subset L$). But the quaternion group $Q$ has a unique element of order two ($-1=i^2=j^2=k^2$), so the field $M=\mathbb{R}\cap L$ is the only quartic subfield of $L$. The other subgroups of $Q$ are all cyclic of order four, these are $\langle i\rangle$, $\langle j\rangle$ and $\langle k\rangle$. Therefore $L$ has exactly three subfields that are quadratic extensions of the rationals. The catch is that Galois correspondence forces all these to be subfields of $M$, and thus real. Hence $K$ cannot be one of them.

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    $\begingroup$ I have to confess that I spent close to an hour trying to construct such a beast inside a suitable cyclotomic field, following Gerry Myerson's suggestion (motivated by Kronecker-Weber theorem). I couldn't :-) $\endgroup$ – Jyrki Lahtonen Jun 23 '13 at 12:47
  • $\begingroup$ I really should have come up with hints here, but I feel that this was a useful exercise for me, so I'm putting the teacher's hat on the table for a while. Apparently I never learned Galois theory well enough, as this took way too long for me :-( $\endgroup$ – Jyrki Lahtonen Jun 23 '13 at 13:41
  • $\begingroup$ I am making a wild guess that shortly before these exercises the observation about all Galois extensions of $\mathbb{Q}$ being stable under complex conjugation was given :-) $\endgroup$ – Jyrki Lahtonen Jun 23 '13 at 13:51
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    $\begingroup$ Dear Jyrki, One can also disprove (1) via Kronecker--Weber (the Galois group of the (total) cyclotomic extension of $\mathbb Q$ is $\widehat{\mathbb Z}^{\times} = \prod_{\ell} \mathbb Z_{\ell}^{\times}$, and its not hard to see that this has no cyclic quotient of order 4 which in turn surjects onto the order two quotient $(\mathbb Z/3)^{\times}$ --- because any such quotient would have to a quotient of the $\mathbb Z_3^{\times}$ factor, but it doesn't have a quotient of order $4$.) But your argument is obviously much more direct! Cheers, $\endgroup$ – Matt E Jun 23 '13 at 13:52
  • $\begingroup$ @MattE: I first realized the difficulty of the construction on the Galois group side. But only one group at the time. I tried $\mathbb{Z}_m^*$ with $m=24$, $15$, $48$, $96$, and only then truth dawned on me. $\endgroup$ – Jyrki Lahtonen Jun 23 '13 at 13:59

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