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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 $\mathbb{Q}$?

Jyrki Lahtonen
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Tom Gannon
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    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)... – Eoin Feb 02 '17 at 21:09
  • 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 $\mathbb Q[x]$ – Peter Feb 02 '17 at 21:10
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    Editing the tags, because [tag: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. – Jyrki Lahtonen Feb 03 '17 at 10:02
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    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. – nguyen quang do Feb 04 '17 at 11:58
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    (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) . – nguyen quang do Feb 04 '17 at 12:06
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    What about ${\mathbb Z}$? –  Feb 24 '17 at 21:04
  • Related: https://mathoverflow.net/questions/80359/which-small-finite-simple-groups-are-not-yet-known-to-be-galois-groups-over-q – Watson Jul 04 '18 at 17:12
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    There is a remark in Topics in Galois Theory (second edition, p. 16, just before §2.2), by J.-P. Serre, which says that the $p$-adic Lie group $\Bbb Z_p \times \Bbb Z_p$ is not a Galois group over $\Bbb Q$. – Watson Jul 08 '18 at 19:38

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Following the remark by Franz Lemmermeyer, we can easily name an infinite group which is not the Galois group of any (necessarily infinite degree) Galois extension. This is the group of integers, $\Bbb Z$. By infinite Galois theory, the Galois group of an infinite-degree field extension $L\mid K$ becomes an infinite compact Hausdorff topological group, which is uncountable. So it cannot be $\Bbb Z$. This holds in particular for $K=\Bbb Q$.

Dietrich Burde
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