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Let $k$ be an infinite field and $\bar{k}$ its algebraic closure. The Artin-Schreier Theorem tells us (among other things) that $[\bar{k}:k]=1,2,\infty$. There's a natural interpretation of $[\bar{k}:k]$ as an infinite cardinal when $[\bar{k}:k]=\infty$. We of course have by elementary considerations that $[\bar{k}:k] \leq |k|$ since $|\bar{k}|=|k|$. My question is the following:

Assuming that $[\bar{k}:k]$ is not finite and given an infinite cardinal $\kappa \leq [\bar{k}:k]$ can we find an algebraic extension $k \subset F$ such that $[\bar{k}:F]=\kappa$.

I imagine it suffices to answer the question when $K$ is the prime subfield of $\bar{k}$ and $k=K(\{t_\lambda\}_{\lambda \in \Lambda})$ where $\{t_\lambda\}$ is a transcendence basis for $\bar{k}$. But I haven't been able to come up with anything clever enough to do the job.

Edit: I think this solves one case. Let $k$ be an uncountable algebraically closed field of cardinality $\kappa$ and consider $k(t)$. Note that the algebraic closure of $k(t)$ is $k$, up to isomorphism. We have that $\mathrm{Gal}(k/k(t))$ is the free profinite group on $\kappa$ generators. We can take the subgroup $H \subset \mathrm{Gal}(k/k(t)$ generated by $\mu \leq \kappa$ generators and take the fixed field of $H$ to get our desired extension.

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Here are two similar non-trivial examples of what may happen .

$\bullet$ For an algebraic closure $\overline {\mathbb Q_p}$ of the $p$-adic field $\mathbb Q_p$, we have $\text {card} \:{\mathbb Q_p}=\text {card} \: \overline {\mathbb Q_p}=2^{\aleph_0}$ but $[\overline {\mathbb Q_p}:\mathbb Q_p]=\aleph_0$.

$\bullet$$\bullet$ For the field of Puiseux series, which is an algebraic closure $\overline {\mathbb C((t))}=\bigcup_{n=1,2,3,\cdots} \mathbb C((t^{1/n}))$ of the field of Laurent series $\mathbb C((t))$, we have $\text {card}\:\overline {\mathbb C((t))}=\text {card} \:{\mathbb C((t))}=2^{\aleph_0}$ but $[ \overline {\mathbb C((t))}:\mathbb C((t))]=\aleph_0$.

Edit
Let me show that for any infinite cardinal $\aleph$ there exists a field $k$ with cardinality $\aleph$, so that $$\text {card}(k)=\text {card}(\bar k)=\aleph$$ We just take for $k$ the field of rational functions $k=\mathbb Q(x_a\mid a\in A)$ in a family of indeterminates $x_a$ indexed by a set $A$ of cardinality $\aleph$.
To prove that $k$ has cardinality $\aleph$ it is enough to prove that the corresponding polynomial ring $P=\mathbb Q[x_a\mid a\in A]$ has cardinality $\aleph$.
To prove that $P$ has cardinality $\aleph$, it is enough to prove that the set of monomials $q\cdot x_{a_1}^{n_{a_1}}\cdots x_{a_r}^{n_{a_r}}\quad (q\in \mathbb Q, r\in \mathbb N) $ has cardinality $\aleph$ and this is true because that set of monomials has cardinality $\text {card} (\mathbb Q)\cdot \aleph=\aleph_0\cdot \aleph=\aleph $
[A little cardinal arithmetic has to be used to show that the set of pure monomials $ x_{a_1}^{n_{a_1}}\cdots x_{a_r}^{n_{a_r}}$ indeed has cardinality $\aleph$, the key fact for this result being that the set $\mathcal P_{fin} (A)$ of finite subsets of $A$ has the same cardinality as $A$, namely $\aleph$]

The above field has characteristic zero but to obtain a field $k$ of cardinality $\aleph$ and characteristic $p$ just replace $\mathbb Q$ by $\mathbb F_p$ in the above construction.

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    $\begingroup$ It should also be easy to show that in your construction that $[\bar{k}:k]=\aleph$. Indeed I think that the set $\{\sqrt{t_a}\}$ should be a $k$-linearly independent set of cardinality $\aleph$. $\endgroup$ – JSchlather Mar 14 '13 at 13:23
  • $\begingroup$ Dear JSchlather, I think you are right and this remark of yours is extremely elegant and interesting . Since I can't upvote comments, I have upvoted your question (which deserves it, anyway) ! $\endgroup$ – Georges Elencwajg Mar 14 '13 at 16:13
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    $\begingroup$ Thanks for the kind remarks, although you can upvote comments it just doesn't count for anything. Also the question is answered in in the affirmative for purely transcendental extensions such as your construction. It follows that if you fix $c \in A$ set $L=\overline{\mathbb Q(\{t_a : a \in A \setminus \{c\})}$ then $ k \subset L(t_c)$ and my edit answers the question for extensions of the form $L(t)$ with $L$ algebraically closed. This also provides another proof that $[\bar{k}:k]=\aleph$. $\endgroup$ – JSchlather Mar 14 '13 at 17:28
  • $\begingroup$ I am upvoting your comments :-) $\endgroup$ – Georges Elencwajg Mar 14 '13 at 20:02
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Transfinite iteration to the rescue!

Let $F_0 = k$. To compute $F_\alpha$ for an ordinal $\alpha$:

  • If $\alpha$ is a limit ordinal, define $F_\alpha = \bigcup_{\beta < \alpha} F_\beta$
  • If $\alpha$ is not a limit ordinal and $F_{\alpha-1} = \bar{k}$, then do not define $F_\beta$ for any $\beta \geq \alpha$.
  • If $\alpha$ is not a limit ordinal and $F_{\alpha-1} \neq \bar{k}$, let $F_\alpha$ be any finite extension of $F_{\alpha-1}$.

I claim that, for infinite $\alpha$, $[F_\alpha : k] = |\alpha|$, where $|\alpha|$ is the cardinality of $\alpha$.

The hard part of the inductive proof of my claim is the first bullet point. It feels obvious, but sketchy enough that I'm wary. I think things become much more clear if we construct a new sequence of $k$-vector spaces $V_\alpha$:

  • If $\alpha$ is a limit ordinal, then $V_\alpha = \bigcup_{\beta < \alpha} V_\beta$
  • If $\alpha$ is not a limit ordinal and $V_{\alpha-1} = \bar{k}$, then do not define $V_\beta$ for any $\beta \geq \alpha$.
  • If $\alpha$ is not a limit ordinal and $V_{\alpha-1}$ is not a field, let $V_\alpha = V_{\alpha-1} \oplus k \xi $ where $\xi$ is any element of the field generated by $V_{\alpha-1}$ but not actually in $V_{\alpha-1}$.
  • If $\alpha$ is not a limit ordinal, $V_{\alpha-1} \neq \bar{k}$, and $V_{\alpha-1}$ is a field, let $V_\alpha = V_{\alpha-1} \oplus k \xi $ where $\xi$ is any element of $\bar{k} \setminus V_{\alpha-1}$.

Now, it's clear that $V_\alpha$ has an $\alpha$-indexed basis.

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    $\begingroup$ It seems to me that you've answered the dual of what I'm asking, but maybe there's an easy reduction I'm not seeing. Since I want to find $F$ such that $[\bar{k}:F]=\alpha$. $\endgroup$ – JSchlather Mar 13 '13 at 23:52
  • $\begingroup$ Whoops. :( That's a harder looking question! $\endgroup$ – user14972 Mar 14 '13 at 8:42
  • $\begingroup$ I think it may be the case that a modification of this argument will answer this related question I just asked. $\endgroup$ – JSchlather Mar 14 '13 at 18:51

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