A field is said to be perfect if it has no non-separable algebraic extensions. Aside from being very useful as ground fields, they have a nice classification: they are exactly the fields of characteristic $0$ and the fields of characteristic $p>0$ that are closed under $p^{th}$ roots.

Along these lines I am wondering if there is a nice characterization of fields that are 'normally perfect' - fields with no non-normal algebraic extensions.

Just as a field being perfect is equivalent to it having no non-separable finite extensions, a field being 'normally perfect' is equivalent to it having no non-normal finite extensions, so it is easy to see that all finite fields are 'normally perfect'. It is also a trivial fact that all algebraically closed fields and all real closed fields are 'normally perfect'. But beyond this I'm having a difficult time seeing what fields could be 'normally perfect' or finding nice necessary conditions for these fields.

  • $\begingroup$ If $char(K)=p$ then replace $K$ by $K^{1/p^\infty}$ to get that $K$ is perfect. Next, the two theorems en.wikipedia.org/wiki/Dedekind_group and kconrad.math.uconn.edu/blurbs/galoistheory/artinschreier.pdf imply that $\overline{K}/K$ must be abelian. $\endgroup$
    – reuns
    Jan 17, 2021 at 11:29
  • $\begingroup$ @reuns Are you saying that all 'normally perfect' fields of positive characteristic are also perfect? If so, I don't see how this follows. Isn't $K^{\frac{1}{p^{\infty}}}$ a normal extension of $K$ regardless of whether or not $K$ is 'normally perfect'? $\endgroup$ Jan 17, 2021 at 13:10

1 Answer 1


Not a complete answer, but it might interest you (and anyway, it's too long for a commment)

If $F$ is normally perfect, Let $G$ be its absolute Galois group, and let $H$ be a closed subgroup of $G$. Then $F_{sep}^H/F$ is spearable, and normal by assumption. In other words, this extension is Galois, hence $H$ is normal.

Now let us assume that the following is true:

Claim (?) If $H$ is a non normal subgroup of $G$, then its closure is non norml.

In this case, every subgroup of $G$ is normal. But groups whose all subgroups are normal are classified (https://en.wikipedia.org/wiki/Dedekind_group); they are either abelian, or of the form $E_2\times A\times Q_8$, where $A$ is a an abelian torsion group where are elements have odd order, and $E_2$ is a $2$-elementary abelian group.

An consequence of Artin Schrier theorem ( https://kconrad.math.uconn.edu/blurbs/galoistheory/artinschreier.pdf) implies that the torsion part of $G$ is $2$-elementary abelian, so $Q_8$ does not appear and $A$ does not appear either.

Hence, in any case $G$ is abelian. Conversely, if $G$ is abelian, any separable extension of $F$ is Galois.

Hence, if the claim above is true, a field $F$ is perfect and normally perfect if and only if its absolute Galois group is abelian.

This would solve your question in zero characteristic.

Anyway, indepedently of the veracity of the claim, fields $F$ of characteristic zero with abelian absolute Galois group give you another family of examples. You can take for example $F=F_0((t_1))((t_2))\cdots((t_n)),$$ where $F_0$ has charactertstic zero.


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