The field of rationals inside an algebraically closed field Let $F$ be an algebraically closed field of characteristic zero.
 Is there a unique copy of the field of rational numbers $\mathbb{Q}$ inside $F$ or there are many?
PS: By  a copy of $\mathbb{Q}$ I mean a subfield of $F$ which looks like (i.e. isomorphic as a field to) the rational numbers field.
A related question is: is $\mathbb{Q}$ necessarily a subfield of $F$?
Edit:
 A third question is: is the real numbers field $R \supset \mathbb{Q}$ unique in $F$?
 A: Let $F$ be a field(not necessarily algebraically closed) and $K\subset F$ a subfield. Then the multiplicative neutral elements $1_F$ and $1_K$ must coincide, otherwise $1_F-1_K\neq0$ would be a zero-divisor in $F$.
Now suppose that $K_1$ and $K_2$ are subfields of $F$ both isomorphic to $\Bbb Q$. By what we just said, must have $L=K_1\cap K_2\neq\emptyset$. If $K_1\neq K_2$, the field $L$ would be identified to a proper subfield of $\Bbb Q$, of which there is none. Therefore $K_1=K_2$.

Moreover, given any field $F$ of characteristic $0$ the natural map
$$
\phi:\Bbb Z\longrightarrow F,\qquad
\phi(n)=n\cdot1_F
$$
is injective (this is actually the definition of characteristic $0$), giving an embedding $\Bbb Z\subset F$. 
If $n\neq0$ the element $\phi(n)\in F$ has an inverse in $F$, and thus $\phi$ extends to a map
$$
\Phi:\Bbb Q\longrightarrow F,\qquad
\Phi(\frac mn)=\Phi(m)\Phi(n)^{-1}
$$
which is easily seen to be well-defined and a field embedding.
A: Any field $F$ of characteristic 0 necessarily includes exactly one isomorphic copy of $\mathbb{Q}$ as a subfield; it is the prime field of $F$ (the smallest subfield containing $1_F$).
