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