If $K$ has only finitely many Galois extensions, then $K$ must be either separably closed or real closed. Are there any other fields whose abelianizations are finite extensions (i.e. whose absolute Galois groups have finite abelianizations)?
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4$\begingroup$ The maximal solvable extension of, say, $\mathbb Q$ has only one abelian extension (viz. itself), but is not separably closed or real closed. $\endgroup$– Emil JeřábekCommented Apr 10, 2018 at 22:02
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$\begingroup$ Oh, right. That is obvious in retrospect. Thanks. $\endgroup$– Alex MennenCommented Apr 10, 2018 at 22:54
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2$\begingroup$ This suggests the followup question of whether there are any fields $K$ not real closed that have at least 1 but only finitely many proper abelian extensions. $\endgroup$– Alex MennenCommented Apr 10, 2018 at 23:42
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1 Answer
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Here is another example, which also answers the "followup question": Let $K$ be the field of Laurent series over $\mathbb{R}$. Its absolute Galois group is the infinite profinite dihedral group $\hat{\mathbb{Z}}\rtimes(\mathbb{Z}/2\mathbb{Z})$, where the action is by inversion. This group is the free profinite product of two groups of order $2$. Its abelianization is therefore $(\mathbb{Z}/2\mathbb{Z})^2$. In particular, $K$ has the desired property.
