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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?

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You ask about, but never define, $F_1$ and $F_2$. Please rewrite the question so it makes sense. – Gerry Myerson Jun 24 '12 at 12:14
up vote 2 down vote accepted

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.

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Thank you Marc. I felt like I'm already assuming every field generated by $D$ is ismorphic to $F$. Because of that, I was uncertain that the monomorphism between $F$ and $F_1$ is necessarily an isomorphism, even if they're already known to be isomorphic. Do you know why that is? – Dedede Jun 24 '12 at 9:50
@Dedede: I think you already answered your own question. The monomorphism between $F$ and $F_1$ is necessarily an isomorphism because it is a monomorphism (injective) by the way it is obtained, and surjective because by assumption $D$ generates $F_1$; that's all. Because the monomorphism is unique, the isomorphism is so (given its restriction to $D$) as well: any bijection $F\to F_1$ extends its restriction to $D$, which is a monomorphism. – Marc van Leeuwen Jun 24 '12 at 13:29

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.

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Thanks Gerry, I tried to use what you said and added my thoughts to the OP. Do you mind giving it a look over? – Dedede Jun 24 '12 at 9:27

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