\documentclass{lecture} \begin{document} \begin{theorem} Let $(k, \le )$ be an ordered field and $k^{r}$ be a real closure of $k$ that extends the ordering of $k$. Let $L$ be a orderable real-closed extension of $k$. Then there exists a unique homomorphism of $k$-algebras $k^{r} \to L$. \label{thm:unique-hom-of-real-closure-in-real-closed} \end{theorem} \begin{proof} Uniqueness: Let $\varphi\colon k^{r} \to L$ be a homomorphism of $k$-algebras and $a \in k^{r}$. Since $a$ is algebraic over $k$, it has a minimal polynomial $P \in k[t]$ over $k$. Denote by $a_1 \le \ldots \le a_n$ the roots of $P$ in $k^{r}$. Since the characteristic of $k$ is $0$, $k$ is perfect, in particular the irreducible polynomial $P$ is separable and thus $a_1 < \ldots < a_n$. Now there exists a unique $1, \ldots, n$ such that $a = a_j$. By \ref{lemma:number-of-roots-in-real-closed-extension}, the polynomial $P$ also has $n$ distinct roots $b_1 < \ldots < b_n$ in the real-closed field $L$. Since $\varphi$ sends roots of $P$ in $k^{r}$ to roots of $P$ in $L$, there is a permutation $\sigma \in S_n$ such that $\varphi(a_i) = b_{\sigma(i)}$. By \ref{lemma:hom-real-closed-fields-respects-orderings}, $\varphi$ respects the ordering of the roots and thus $\sigma = \text{id}$ and $\varphi(a) = \varphi(a_j) = b_j$. Existence: Consider the set $\mathcal{F}$ of all pairs $(E, \psi)$ where $k \subseteq E \subseteq k^{r}$ is a subextension of $k^{r} / k$ and $\psi\colon E \to L$ is a homomorphism of $k$-algebras. Since $(k, k \xhookrightarrow{} L) \in \mathcal{F}$, $\mathcal{F} \neq \emptyset$. Define an inductive ordering on $\mathcal{F}$ by $(E, \psi) \le (E', \psi)$ if there is a commutative diagram \[ \begin{tikzcd} & E' \arrow{d}{\psi'} \\ E \arrow[dashed]{ur} \arrow{r}{\psi} & L \end{tikzcd} \] in the category of $k$-algebras. Then by Zorn, the set $\mathcal{F}$ admits a maximal element $(E, \psi)$. $E$ is real-closed, otherwise it admits a finite real extension $E'$ of $E$. In particular $E' \subseteq k^{r}$. Since $L$ is real-closed, $\psi\colon E \to L$ admits a continuation $\psi'\colon E' \to L$ by \ref{lemma:continuation-in-real-closed}. Thus $(E, \psi) < (E', \psi')$ contradicting the maximality of $(E, \psi)$. Hence $E$ is real-closed and $k^{r} / E$ is real algebraic, thus $E = k^{r}$. So $\psi$ is a homomorphism of $k$-algebras from $k^{r}$ to $L$. \end{proof} \begin{korollar} Let $(k, \le )$ be an ordered field. If $k_1^{r}$ and $k_2^{r}$ are real closures of $k$ whose canonical orderings are compatible with that of $k$, then there exists a unique isomorphism of $k$-algebras $k_1^{r} \xrightarrow{\simeq} k_2^{r}$. \label{kor:unique-iso-of-real-closures} \end{korollar} \begin{proof} By \ref{thm:unique-hom-of-real-closure-in-real-closed} there exist unique homomorphisms of $k$-algebras $\varphi\colon k_1^{r} \to k_2^{r}$ and $\psi\colon k_2^{r} \to k_1^{r}$. Then $\psi \circ \varphi$ and $\text{id}_{k_1^{r}}$ are homomorphisms $k_1^{r} \to k_1^{r}$ of $k$-algebras. By uniqueness in \ref{thm:unique-hom-of-real-closure-in-real-closed}, $\psi \circ \varphi = \text{id}_{k_1^{r}}$. Similarly, $\varphi \circ \psi = \text{id}_{k_2^{r}}$. \end{proof} \begin{bem} Contrary to the situation of algebraic closures of a field $k$, for ordered fields $(k, \le)$ there is a well-defined notion of the real closure of $k$ whose canonical ordering is compatible with that of $k$. As shown by \ref{bsp:different-real-closures-depending-on-ordering}, it is necessary to fix an ordering of the real field $k$ to get the existence of an isomorphism of fields between two orderable real closures of $k$. \end{bem} \begin{korollar} Let $(k, \le )$ be an ordered field and let $k^{r}$ be the real closure of $k$. Then $k^{r}$ has no non-trivial $k$-automorphism. \end{korollar} \begin{proof} Take $k_1^{r} = k_2^{r}$ in \ref{kor:unique-iso-of-real-closures}. \end{proof} \end{document}