TITLE: Smallest real closed field realizing all cuts of the rational numbers
QUESTION [5 upvotes]: 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 $\mathbb{Q}\cup \{\pm\infty\}$

REPLY [4 votes]: 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.