Kaplanksy's definition of normal field extension goes as follows

Let $K\subseteq E$ be fields. $E$ is normal over $K$ if for any $\alpha\in E\setminus K$ there exists $\psi\in \operatorname{Aut}_K(E)$ such that $\psi(\alpha)\neq \alpha$.

By $\operatorname{Aut}_K(E)$ I mean the set of automorphisms of $E$ as a $K$-algebra. A more common definition of normality is

Let $K\subseteq E$ be fields. $E$ is normal over $K$ if any irreducible $f\in K[X]$ that has some roots in $E$ splits in $E$.

The latter definition is equivalent to

  1. $\operatorname{Hom}_K(E,C) = \operatorname{Aut}_K(E)$, where $C$ is an algebraic closure of $E$.
  2. There exists $F\subseteq K[X]$ such that $E=K(R)$, where $R$ is set of all roots in $C$ of polynomials in $F$.

My question is, are Kaplansky normality and common normality equivalent?

  • $\begingroup$ For infinite extensions one might need to invoke Zorn to prove the equivalence. $\endgroup$ – Lord Shark the Unknown Jun 10 '17 at 16:33
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    $\begingroup$ That an irreducible polynomial splits in a field extension $E/K$ doesn't imply there's an automorphism permuting its roots. For example, take $E = \mathbb{R}$ and $K = \mathbb{Q}$. You can't permute the roots of $x^2 - 2$, because the only automorphism of $E$ is the identity. (of course, this example is not a normal extension by either definition) (use $E = \mathbb{R} \cap \overline{\mathbb{Q}}$ if you want to consider only algebraic extensions) $\endgroup$ – Hurkyl Jun 10 '17 at 17:02

These are not equivalent. For instance, let $K=\mathbb{F}_p(t)$ and let $E=K(t^{1/p})$. Then $E$ is common-normal over $K$, but it is not Kaplansky-normal, since there are no automorphisms of $E$ over $K$ besides the identity. Alternatively, let $K=\mathbb{C}$ and let $E$ be the function field of a curve of genus $>1$ over $\mathbb{C}$. Then $E$ has only finitely many automorphisms over $\mathbb{C}$, and so $E$ has finite dimension over the fixed field of all the automorphisms. Since $E$ has infinite dimension over $\mathbb{C}$, the fixed field is larger than $\mathbb{C}$, so $E$ is not Kaplansky-normal. However, it is common-normal, since $\mathbb{C}$ is algebraically closed so trivially every polynomial over $\mathbb{C}$ splits.

What is true is that Kaplansky-normal implies common-normal, and the converse is true assuming $E$ is separable and algebraic over $K$. First, assume $E$ is Kaplansky-normal over $K$, let $f$ be any polynomial over $K$ and let $F\subseteq E$ be the subfield generated by $K$ and all the roots of $f$ in $E$. Every automorphism of $E$ over $K$ maps $F$ to itself, so $F$ is also Kaplansky-normal, and is finite over $K$. If $G=\operatorname{Aut}(F/K)$, we then have $K=F^G$, which implies $F$ is Galois over $K$ by a theorem of Artin. Since $f$ was arbitrary, this implies $E$ is common-normal over $K$.

Conversely suppose $E$ is separable, algebraic, and common-normal over $K$. Then $E$ is Galois over $K$, and Galois theory says the fixed field of the automorphism group of $E$ over $K$ is just $K$. That means exactly that $E$ is Kaplansky-normal over $K$.

  • $\begingroup$ Thanks for your answer. In the second to last sentence in the second paragraph you meant 'finite over $K$', didn't you? Is it true then that algebraic Kaplansky-normal is equivalent to Galois? $\endgroup$ – Human Jun 11 '17 at 7:47
  • $\begingroup$ Oops, yes, I'll correct that. And yes, algebraic and Kaplansky-normal is equivalent to Galois. $\endgroup$ – Eric Wofsey Jun 11 '17 at 16:28

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