A field of characteristic zero can be embedded into $\mathbb C$? I am working through Silverman's Arithmetic of Elliptic Curves. Exercise 3.8(b) says that $\deg[m]=m^2$ where $[m]$ is the multiplication map of en elliptic curve. Actually 3.8(a) gives this result for the field $\mathbb C$ and the hint of (b) says that if $K$ can be embedded into $\mathbb C$ then the results follows from (a).

So how can one embed $K$, a field with characteristic $0$ into $\mathbb C$?

 A: Not every field of characteristic $0$ can embed in $\mathbb{C}$.  For instance, if $S$ is a set of variables of cardinality greater than that of $\mathbb{C}$, then the field $\mathbb{Q}(S)$ of rational functions in the variables in $S$ with coefficients in $\mathbb{Q}$ clearly cannot embed in $\mathbb{C}$.  However, it turns out that this cardinality constraint is the only constraint: 

Theorem: Let $K$ be a field of characteristic $0$.  Then the following are equivalent:
  
  
*
  
*$K$ embeds in $\mathbb{C}$.
  
*The cardinality of $K$ is $\leq2^{\aleph_0}$.
  
*The transcendence degree of $K$ over $\mathbb{Q}$ is $\leq2^{\aleph_0}$.

Proof: It is trivial that (1) implies (2) and (2) implies (3).  So suppose the transcendence degree of $K$ over $\mathbb{Q}$ is at most $2^{\aleph_0}$; let $S\subset K$ and $T\subset \mathbb{C}$ be transcendence bases over $\mathbb{Q}$.  By hypothesis, there exists an injection $S\to T$, which extends uniquely to an embedding of fields $\mathbb{Q}(S)\to \mathbb{Q}(T)\subset\mathbb{C}$.  Since $K$ is algebraic over $\mathbb{Q}(S)$ and $\mathbb{C}$ is algebraically closed, this embedding extends to an embedding $K\to\mathbb{C}$.
