The field of real numbers has a nice property - for a fixed $n\in \mathbb{N}$, every $r\in \mathbb{R}$ is either an $n$-th power (if $n$ is odd or if $r>0$) or -$r$ is an $n$-th power. I want to generalize this - let $K$ be a field extension of $F$, then $a\in K$ is almost an $n$-th power if there is $b\in F$ and $c\in K$ such that $ab=c^n$. In other words, if $A=\\{ x^n \mid x\in K^\times \\}$ then $A F^\times = K^\times$ as groups (In the example above the smaller field is $\mathbb{Q}$).

My first question is how to find such field extensions $F\leq K$ where every element in $K$ is almost an $n$-th power for a fixed $n$ (or even better - for all $n$).

Two such examples are when $K$ is algebraically closed field (so every element has an $n$-th root), or if it is a formally real closed field. Another such example is when $K$ is finite and $n$ is prime to $|K|-1$ (since $K^\times$ is cyclic of order $|K|-1$), though I'm more interested with fields of char $0$.

Since my first two examples are fields that are algebraically closed or "almost" closed, I thought that I should start with such fields $F$ where $F(\alpha)$, for $\alpha$ algebraic over $F$, is algebraically closed (or is formally real closed in the case where $F$ can be ordered). The problem is that besides the real numbers I never encounter such fields. I mean, it is "easy" to start with a field $F$ and ask what are the properties of $F(\alpha)$ but I never saw someone do it in the opposite direction.

So my second question is what are the properties of fields that have finite algebraic extensions which are algebraically closed or formally real closed, and specially properties that can help with my first question.


Your first question is interesting but I don't know any answers.

The answer to your second question is well-known. If a field $F$ has a finite extension which is real-closed then it has a finite extension which is algebraically closed (by adjoining $\sqrt{-1}$ to the real-closed field). But the only fields which have finite extensions $K$ which are algebraically closed are either real-closed or algebraically closed already, and the degree of $K/F$ is $2$ or $1$. This is a theorem of Artin and Schreier and can be found in the more comprehensive standard texts, for instance Lang's Algebra.

  • $\begingroup$ thanks. at least I have a good reason for not knowing any other fields that are "almost" algebraically closed. $\endgroup$ – Ofir Oct 15 '10 at 15:07

There exist lots of fields $K$ that are closed under taking $n$-th roots for fixed $n$. They can be contructed using valuation theory: take a field $K$ which is henselian with respect to some valuation $v$ and has the following additional properties:

  1. $n$ is not divisible by the characteristic of the residue field $\overline{K}$.
  2. The residue field is algebraically closed.
  3. The value group $v(K)$ is divisible by $n$.

Then $K$ is closed under taking $n$-th roots by Hensel's lemma.



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