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Let $K$ be a real closed field of transcendence degree 1 over $\mathbb{R}$. It is not difficult to see that $K$ has the following "minimality property": Whenever $L$ is a real closed field that realizes all cuts of the rational numbers, then $K$ embeds into $L$.

Question: Is $K$ up to isomorphism uniquely determined by this minimality property?

Terminology: By a cut of $\mathbb{Q}$ I mean a pair $(L,R)$ of subsets of $\mathbb{Q}$ with the property that $l < r$ for all $l\in L$, $r\in R$ and such that $R\cup L=\mathbb{Q}$

The question asks whether $K$ can be identified parallel to the way one can define $\mathbb{R}$: The real field $\mathbb{R}$ is up to isomorphism the unique real closed field that embeds into any real closed field which realizes every non-principal cut of $\mathbb{Q}$; a cut $(L,R)$ is non-principal if $L$ does not have a supremum in $\mathbb{Q}\cup \{\pm\infty\}$

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Yes. If $F$ is another such field, then $F$ would embed over ${\mathbb R}$ into $K$ so $F$ would also be a transcendence degree 1 extension of $\mathbb R$. We can find $s\in K$ and $t\in F$ with ${\mathbb R} < r,s$. The ordered fields ${\mathbb R}(s)$ and ${\mathbb R} (t)$ are isomorphic. Thus their real closures $K$ and $F$ are isomorphic.

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@Dave The minimality assumption on $F$ only ensures that $\mathbb{R}$ can be embedded into $F$. Why can you embed it so that $F$ has transcendence degree 1 over the image of $\mathbb{R}$ under this embedding? I should also mention that there are embeddings of $K$ (being a real closed transcendence degree 1 extension of $\mathbb{R}$) into $K$ that are not surjective. How does your argument work if $F$ is a real closed field between $K$ and the image of such an embedding? – Marcus Nov 4 '12 at 20:01

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