Dimension of an algebraic closure as a vector space over its base field.

Question

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.

Answer 1

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.

Answer 2

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.