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Is it possible to have $\operatorname{Gal}(\overline{K}/K)=\mathbb{Z}$?

My question comes from the link beetween covering and field extensions. For covering the simplest example is $\operatorname{Gal}(\mathbb{R}/\mathbb{S}^1)=\mathbb{Z}$.

Maybe something like $\mathscr{M}(\mathbb{C})/\mathscr{M}(\mathbb{C}^*)$ where $\mathscr{M}(X)$ is the field of meromorphics functions on a Riemann surface?

Gabriel Soranzo
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    To reconcile the similarities between fundamental groups and Galois groups you need to pass to the profinite completion of the fundamental group. The two theories are unified in etale cohomology. – jkramerm47 May 14 '15 at 20:37
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    The whole point of using algebraic fundamental group is that : the galois correspondence between coverings spaces and galois groups isn't good enough for one to have a concrete dictionary between them, as fundamental group of a space might not be profinite, whereas every galois group is. – Balarka Sen May 14 '15 at 20:37
  • ah, @jkramerm47, you beat me by seconds. – Balarka Sen May 14 '15 at 20:38
  • @jkramerm47 So there should be a example in etale cohomology? – Gabriel Soranzo May 15 '15 at 08:26
  • @Soranzo Example of what? – Balarka Sen May 15 '15 at 08:49
  • @jkramerm47 Ok, it seems I don't understand the term "unified" correctly. Thanks. – Gabriel Soranzo May 15 '15 at 11:20
  • Not the same but related: https://math.stackexchange.com/questions/1871485/is-there-a-field-extension-k-bbb-q-such-that-textaut-bbb-qk-cong?noredirect=1&lq=1 – Watson Jun 24 '18 at 10:17

2 Answers2

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No: every Galois group is profinite, and any infinite profinite group has cardinality at least $2^{\aleph_0}$.

Alex Kruckman
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You can get pretty close: there are fields whose absolute Galois groups are the profinite integers $\widehat{\mathbb{Z}}$. In particular every finite field has this property. The finite extensions of finite fields look exactly like the finite covers of $S^1$ (but the infinite extensions are more complicated).

Qiaochu Yuan
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