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)?

  • 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. $\endgroup$ Apr 10 '18 at 22:02
  • $\begingroup$ Oh, right. That is obvious in retrospect. Thanks. $\endgroup$ Apr 10 '18 at 22:54
  • 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. $\endgroup$ Apr 10 '18 at 23:42

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.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.