Possible Fields?

Question

Is there an algebraically closed field which is a 1-dimensional vector space (as opposed to complex numbers which are 2-D)?

Also is there a complete $\aleph_0$ field?

Answer 1

If $F$ is an infinite field and $n<|F|$ as cardinal numbers, then the cardinality of any vector space considered over $F$ of dimension $n$ is equal to the size of $F$. In particular, since $\Bbb C\cong\Bbb R^2$ as vector spaces considered over $\Bbb R$ and $2<|\Bbb R|={\frak c}$, we know $\Bbb C$ and $\Bbb R$ are equinumerous.

Given a field $L$ and a subfield $K$, the space $L$ satisfies all of the axioms of being a vector space over the scalar field $K$. Since we can vary $K$, we can vary the field of scalars over which we are considering $L$ as a vector space. Every field is one-dimensional over itself, so in particular any algebraically closed field is one-dimensional over itself.

In order to speak of completeness or Cauchy sequences on a field, there needs to be some kind of metric involved. Often we speak of metrics induced from absolute values on fields. The trivial absolute value induces the discrete metric which is always complete. Otherwise, any algebraic extension of $\Bbb Q$ or $\Bbb F_p(T)$ will have nontrivial absolute values classified by Ostrowski's theorem and under none of them is the field complete.

More generally, if the topology on $F$ induced from some metric makes $F$ a topological field (so that open sets remain open upon translation), and $F$ is countable, then the metric cannot be complete but not discrete, as Kevin remarks in the comments. The Baire category theory implies that every complete metric space with no isolated points is uncountable, the contrapositive of which is that every countable complete metric space has isolated points. If $x\in F$ is isolated, then by translation-invariance every point is isolated, so the topology is discrete and hence the metric is too.