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Let me start by saying that I have no idea in algebra/number theory/whatever, so, please, forgive my ignorance.

Let $\mathbb{F}$ be a field of characteristic zero and continuum cardinality, which means its algebraic closure is (isomorphic to) $\mathbb{C}$ (complex numbers), right? Now $\mathbb{C}$ is an $\mathbb{F}$-algebra, but what is the $\mathbb{F}$-dimension of $\mathbb{C}$? In other words, if I pick up a Hamel basis in $\mathbb{C}$ over $\mathbb{F}$, what would be the cardinality of that basis? I imagine that it can be infinite, but can it be uncountable?

I discovered about the existence of such fields $\mathbb{F}$ other than $\mathbb{R}$ (real numbers) here:

http://mathforum.org/kb/message.jspa?messageID=7414811

Thank you.

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migrated from mathoverflow.net Mar 23 '16 at 8:36

This question came from our site for professional mathematicians.

  • $\begingroup$ By cardinality I mean not only finite-infinite, but also countable-uncountable. And, please, can anyone tell me how to edit my post? I see no 'Edit' button. $\endgroup$ – Bedovlat Mar 22 '16 at 19:44
  • $\begingroup$ It should be directly below the line with the tags. $\endgroup$ – Sebastian Goette Mar 22 '16 at 19:46
  • $\begingroup$ Oh, yes! Thanks a lot :) $\endgroup$ – Bedovlat Mar 22 '16 at 19:51
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    $\begingroup$ It can be uncountable: just consider a field $K(X)$ of rational functions on 1 indeterminate $X$ with $K$ a field of characteristic zero: then any choice of square roots $\sqrt{X-t}$, $t\in K$, are linearly independent over $K$, so if $K$ has continuum cardinal, then "the" algebraic closure of $K(X)$ has continuum dimension over $K$. It's a bit more subtle to find an example where the dimension is infinite countable, but it is indeed possible (e.g., the field of Laurent series over the complex numbers, by Puiseux's theorem). $\endgroup$ – YCor Mar 22 '16 at 23:40
  • $\begingroup$ Thanks for the comment. If K(X) is the algebraic closure of K, then X is algebraic over K. Let K=R, and X=i, then K(X)=C. Now \sqrt{i-r} are not linearly independent over R. I can't see why they should be. Can you, please, elaborate? $\endgroup$ – Bedovlat Mar 23 '16 at 14:46
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Suppose $F\subsetneq C$ is an arbitrary extension of fields of any characteristic $p\geq 0$, with $C$ algebraically closed and with $[ C:F]=n\lt \infty$.
Then Artin-Schreier's amazing theorem says that the characteristic $p$ is necesarily zero, that $n=2$ and that there exists $j\in C$ such that $j^2=-1$ and $C=F(j)$.
Moreover $F$ must be real closed, which intuitively means that it resembles $\mathbb R$ viewed as a field, stripped of its non-algebraic structures.

Bibliography
This theorem is the very last theorem, numbered 11.14 page 654, of Jacobson's Basic Algebra II.
Be assured however that you don't have to read the preceding 653 pages to understand Artin-Schreier, and even less the 499 pages of Basic Algebra I...

Edit
In a comment the OP asks whether it is possible to find a subfield $F\subset \mathbb C$ with $\mathbb C$ algebraic over $F$ (so that $\mathbb C$ is an algebraic closure of $F$) and satisfying $\dim_F \mathbb C\gt\aleph_0$.
The answer is yes:
Let $(x_i)_{i\in I}$ be a transcendence basis of $\mathbb C$ over $\mathbb Q$, necessarily of cardinality $\vert I\vert=\mathfrak c=2^{\aleph_0}$.
The field $F=\mathbb Q(x_i\vert i\in I)$ is the required field:
Indeed, if one chooses for each $i$ a square root $\sqrt x_i\in \mathbb C$, the family $(\sqrt x_i)_{i\in I}$ is $F$-linearly independent of cardinality $\vert I\vert=\mathfrak c$, which immediately implies that $\dim_F \mathbb C\geq \mathfrak c$ and thus (since the opposite inequality is trivial) $$\dim_F \mathbb C=\mathfrak c=2^{\aleph_0}\gt \aleph_0.$$ (The linear independence of the $\sqrt x_i$'s is proved in the same elementary way that the square roots $\sqrt p$ ($p$ prime) are linear independent over $\mathbb Q $: see for example here)

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  • $\begingroup$ Thank you for your answer. I am aware of Artin-Schreier theorem, so the important part of my question is the case $[C:F]=+\infty$. If $C=\mathbb{C}$, and $C$ is the algebraic closure of $F$, can $[C:F]$ be more than countably infinite? $\endgroup$ – Bedovlat Mar 24 '16 at 17:05
  • $\begingroup$ By the way, Artin-Schreier theorem, as I know it, pertains to the case where C is algebraically closed. This is not 'an arbitrary extension of fields', is it? $\endgroup$ – Bedovlat Mar 24 '16 at 19:06
  • $\begingroup$ Yes, of course you are absolutely right: now corrected. Thanks a lot for your attentive reading. (I had written a first version with $C=\mathbb C$ and then decided to give the general version, but forgot to add that then $C$ must be assumed algebraically closed) $\endgroup$ – Georges Elencwajg Mar 24 '16 at 19:20
  • $\begingroup$ Dear Bedovlat, I have answered "the important part of (your) question" mentioned in your comment above in an Edit. $\endgroup$ – Georges Elencwajg Mar 24 '16 at 22:21
  • $\begingroup$ Dear Prof. Elencwajg, thank you very much for the answer. In order for me to understand (and thus to appreciate) your answer I will need to ask a few questions. First, I am a bit confused with $F=\mathbb{C}(x_i|i\in I)$. I would understand that it is different from $\mathbb{C}$ if it were $F=\mathbb{Q}(x_i|i\in I)$. $\endgroup$ – Bedovlat Mar 24 '16 at 23:19

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