TITLE: Does every algebraically closed field contain the field of complex numbers?
QUESTION [15 upvotes]: Does every algebraically closed field contain the field of complex numbers? Thank you very much.

REPLY [11 votes]: The following is close to a positive answer to your question.
Let $\kappa$ be any uncountable cardinal. Then any two algebraically closed fields of characteristic $0$ and cardinality $\kappa$ are isomorphic.  
In particular, any algebraically closed fields of characteristic $0$ and cardinality $c$ (the continuum) is isomorphic to the complex numbers.  
And any algebraically closed field of characteristic $0$ that is big enough (cardinality $\ge c$) contains a copy of the complex numbers.
Note that we need $\kappa$ to be uncountable. There are non-isomorphic countable algebraically closed fields of characteristic $0$.
But any algebraically closed field of characteristic $0$ contains an isomorphic copy of the field of algebraic numbers.
Naturally, if the field $K$ has characteristic $p\ne 0$, there cannot be an embedding of $\mathbb{C}$ in $K$. And there are algebraically closed fields of characteristic $p$ and cardinality $\kappa$ for every prime $p$ and every infinite cardinal $\kappa$.