Let $F$ be an ordered field, $R/F$ a real closure and $R'$ an arbitrary real-closed field extension of $F$. Need there be an (ordered) $F$-homomorphism $R\to R'$? Does this follow from the essential uniqueness of the real closure?
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2026-03-27 06:08:18.1774591698
Real closure is the smallest real-closed field extension
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Yes, this is almost immediate from the uniqueness of the real closure. Let $R'_0$ be the subfield of $R'$ consisting of the elements that are algebraic over $F$. Then $R'_0$ is also real-closed (how you prove this depends on how you define "real-closed"; for many definitions it is trivial), and so it is a real closure of $F$. Thus there is an isomorphism $R\to R'_0$ over $F$.
(Alternatively, this should follow immediately from the proof that real closures are unique. I don't know what proof of that fact you have seen, but every proof that I know of can be trivially modified to prove this statement instead.)