Unique isomorphism between fields generated by a domain. Suppose $F$ and $K$ are fields both generated by a common subring $D$, which is a domain. My question is, why is there a unique isomorphism between $F$ and $K$ which is the identity on $D$?
Wouldn't the field generated by $D$ be unique? So $F=K$, and then any isomorphism is determined by its images on a generating set, that is, $D$, so such an isomorphism is unique? If this is so, how can we be sure that there exists an isomorphism which restricts to the identity on $D$?

Second thoughts: Let $F$ be the field of fractions of $D$. Then I let $\iota\colon D\to F_1$ be the identity embedding, which is known to have a unique extension to a monomorphism of $F$ into $F_1$. Since $F_1$ is generated by $D$, it is isomorphic to $F$, so this unique monomorphism is an isomorphism? Likewise, there is a unique isomorphism between $F$ and $F_2$ which is the identity on $D$, so there is a unique isomorphism between $F_1$ and $F_2$ which is the isomorphism on $D$. Is this right?
 A: Yes  you are right: every field generated by a given domain $D$ is canonically isomorphic to the field of fractions of $D$, by the argument you given under "Second thoughts". And this isomorphism is the unique one that extends the embedding of $D$. So as long as your two fields have the same domain $D$ in common, there is a unique way to exend this identification on $D$ to one on all of the fields.
Should your fields instead just have isomorphic generating subdomains, then you must deal with the possibility that there are multiple isomorphisms between those subdomains, each of which will extend differently to isomorphisms between the fields. But that does not seem to be the situation given in the question.
A: You don't start with the isomorphism, then see what happens when you restrict it to $D$; you start with the identity on $D$, and see that you can extend it, uniquely, to an isomorphism from $F$ to $K$. 
On another issue, I wouldn't say the field generated by $D$ is unique; I'd say it's unique up to isomorphism. So we don't have $F=K$; we have $F$ isomorphic to $K$. This may seem like hair-splitting, but it's actually a distinction worth keeping in mind. 
