2
$\begingroup$

I was reading Dummit and Foote and they talked about the study of Class Field Theory, which studies abelian extensions of an arbitrary finite extension of $\mathbb{Q}$ (and related, a theorem that says any abelian $G$ may be realized as the Galois group of some subfield of a cyclotomic extension). So my question is, is there any group that cannot be realized as a Galois group over $\mathbb{Q}$? I know that from above $G$ must be nonabelian, and that if we don't require $\mathbb{Q}$ as our base field specifically, using the Fundamental Theorem of Galois Theory the answer is "no" via realizing $G < S_n$ for some $n$ and then taking the extension $K/K^G$, where $K = \mathbb{Q}[x_1, ... , x_n]$. But if we require the base field to be $\mathbb{Q}$?

$\endgroup$
  • 7
    $\begingroup$ Pretty big open problem. Look up inverse galois theory. Maybe I should add the mention that this only concerns finite groups though (right, finite extensions)... $\endgroup$ – Eoin Feb 2 '17 at 21:09
  • $\begingroup$ But for many groups, including the solvable groups, the groups $A_n$ and the groups $S_n$, it is known that they can be the galois group of a polynomial in $\mathbb Q[x]$ $\endgroup$ – Peter Feb 2 '17 at 21:10
  • 1
    $\begingroup$ Editing the tags, because finite-fields applies only when the fields are finite as sets. Here you only ask about extensions of $\Bbb{Q}$, so they are all infinite, because there are infinitely many rational numbers. $\endgroup$ – Jyrki Lahtonen Feb 3 '17 at 10:02
  • 1
    $\begingroup$ In the same vein as your question, the following approach by J. Minac and his collaborators (*) could be of interest. Let G_F be the absolute Galois group of a number field F containing a primitive q-th root of 1, q being a power of a prime p. The authors study the q-central descending series of G_F using the Bloch-Kato-Milnor conjecture (now a theorem of Voevodsky) on a canonical isomorphism between the cohomology algebra H*(G_F, Z/qZ) and the Milnor K-theory algebra K*(F) mod q. $\endgroup$ – nguyen quang do Feb 4 '17 at 11:58
  • 1
    $\begingroup$ (continued) One surprising result is that the third quotient G[3] in the descending filtration determines the cohomology algebra. This puts constraints on G_F , which allow to construct examples of profinite groups which cannot be absolute Galois groups of number fields. See e.g. Chebolu-Efrat-Minac, Math. Annalen 352 (2012) and Efrat-Minac, Amer. J. of Math. 133 (2011) . $\endgroup$ – nguyen quang do Feb 4 '17 at 12:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.