Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

To be more specific, given a field $k$, must we have the algebraic closure $\bar{k}$ contained in $\mathbb{C}$ up to isomorphism?

share|improve this question
3  
$\Bbb Z/2\Bbb Z$? –  Brian M. Scott Feb 4 '13 at 3:09
    
Perhaps it would be good to mention a theorem of Steinitz: an (uncountable) algebraically closed field is characterized solely by its characteristic and its cardinality. –  user641 Feb 4 '13 at 3:25
1  
@SteveD: Or in particular its transcendence degree and characteristics. Then you can include the countable case as well. –  Asaf Karagila Feb 4 '13 at 3:27
1  
Your title should be «Is every field...». Otherwise the answer is trivially yes, because there do exist fields contained in $\mathbb C$, like $\mathbb C$ itself. –  Mariano Suárez-Alvarez Feb 4 '13 at 3:36
    
An apt question for Asaf: I suppose that the truth of the OP’s statement depends crucially on the Axiom of Choice? If I remember correctly, the standard proof of Steinitz’s Theorem relies upon Zorn’s Lemma. –  Haskell Curry Feb 4 '13 at 3:45
add comment

3 Answers

There are fields of arbitrarily large cardinality, in particular there are fields whose cardinality is much larger than that if the complex numbers, and therefore cannot be embedded there.

It is true, however, that every field of characteristics zero whose cardinality is at most $2^{\aleph_0}$ can be embedded into the complex numbers.

share|improve this answer
add comment

We have fields with arbitrarily high cardinality, so pick a field $ \mathbb{F} $ whose cardinality is greater than that of $ \mathbb{C} $ and take its algebraic closure $ \overline{\mathbb{F}} $. Clearly, we cannot embed $ \overline{\mathbb{F}} $ into $ \mathbb{C} $.

To construct an $ \mathbb{F} $ that satisfies the aforementioned requirement, pick a set $ I $ of indeterminates whose cardinality is greater than that of $ \mathbb{C} $. Then $ \mathbb{C}[I] $ is an integral domain. Letting $ \mathbb{F} $ be the fraction field of $ \mathbb{C}[I] $, we are done. :)

share|improve this answer
    
I am amused that we posted the same answer almost simultaneously and then gave complementing edits almost simultaneously as well. :-) and then you had to ruin it by editing again! ;-p –  Asaf Karagila Feb 4 '13 at 3:31
    
@Asaf: This is really amusing! –  Haskell Curry Feb 4 '13 at 3:32
add comment

Certainly not. For instance any field of non-zero characteristic is not contained in $\mathbb C$, such as $\mathbb F_q$ the finite field with $q=p^t$ elements. But there are also fields of characteristic $0$ not contained in $\mathbb C$ such as $\mathbb C(\{t_\alpha\}_{\alpha \in 2^{\mathfrak c}})$ the field of rational functions in $2^{\mathfrak c}$ variables.

share|improve this answer
3  
$\Bbb{C}(t)$ is in fact isomorphic to a subfield of $\Bbb{C}$. –  Chris Eagle Feb 4 '13 at 3:15
    
@ChrisEagle, I forgot about that. Okay I adjoined more variables than their are elements in $\mathbb C$, that should take care of it. –  JSchlather Feb 4 '13 at 3:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.