\documentclass{lecture}

\begin{document}

\section{Real closures}

\begin{satz}
    Let $k$ be a real field. Then there exists a real-closed
    algebraic orderable extension $k^{r}$ of $k$.
    \label{satz:existence-alg-closure}
\end{satz}

\begin{proof}
    Let $\overline{k}$ be an algebraic closure of $k$ and $E$ be the set of intermediate
    extensions $k \subseteq L \subseteq \overline{k}$ such that $L$ is real and algebraic over $k$.
    $E \neq \emptyset$ since $k \in E$. Define $L_1 < L_2$ on $E$ if and only if
    $L_1 \subseteq L_2$ and $L_2 / L_1$ is ordered, i.e.
    the order relation on $L_1$ coincides with the on induced by $L_2$.
    Then
    every totally ordered familiy $(E_i)_{i \in I}$ has an upper bound, namely
    $\bigcup_{i \in  I} E_i$. By Zorn, $E$ has a maximal element, which we
    denote by $k^{r}$ and which is an algebraic extension of $k$. Such
    a $k^{r}$ is real-closed, because otherwise it would admit a proper real
    algebraic extension contradicting the maximality of $k^{r}$ as a real algebraic extension of $k$.
\end{proof}

\begin{definition}[]
    A real-closed real algebraic extension of a real field $k$ is called
    a \emph{real closure} of $k$.
\end{definition}

\begin{bem}
    By the construction in the proof of \ref{satz:existence-alg-closure},
    a real closure of a real field $k$ can be chosen as a subfield
    $k^{r}$ of an algebraic closure of $\overline{k}$.
    Since $k^{r}[i]$ is algebraically closed and algebraic over $k^{r}$, so also over $k$,
    it follows $k^{r}[i] = \overline{k}$.
\end{bem}

\begin{satz}
    Let $k$ be a real field and $L$ be a real-closed extension of $k$. Let
    $\overline{k}^{L}$ be the relative algebraic closure of $k$ in $L$, i.e.
    \[
    \overline{k}^{L} = \{ x \in L \mid x \text{ algebraic over } k\} 
    .\] Then $\overline{k}^{L}$ is a real closure of $k$.
\end{satz}

\begin{proof}
    It is immediate that $\overline{k}^{L}$ is a real algebraic extension of $k$. Let
    $x \in \overline{k}^{L}$. Then $x$ or $-x$ is a square in $L$, since
    $L$ is real-closed. Without loss of generality, assume that
    $x \in L^{[2]}$. Then $t^2 - x \in \overline{k}^{L}[t]$
    has a root in $L$. Since this root is algebraic over $\overline{k}^{L}$, hence over $k$,
    it belongs to $\overline{k}^{L}$. Thus $x$ is in fact a square in $\overline{k}^{L}$. By
    the same argument every polynomial of odd degree has a root in $\overline{k}^{L}$.
\end{proof}

\begin{bsp}[]
    \begin{enumerate}[(i)]
        \item $\overline{\Q}^{\R} = \overline{\Q}^{\mathbb{C}} \cap \R$
            is a real closure of $\Q$. In particular, $\overline{\Q}^{\mathbb{C}}
            = \overline{\Q}^{\R}[i]$ as subfields of $\mathbb{C}$.
        \item Consider the real field $k = \R(t)$ and the real-closed extension
            \begin{salign*}
            \widehat{\R(t)} =
            \bigcup_{q > 0} \R((t ^{t/q}))
            .\end{salign*} Then the subfield
            $\overline{\R(t)}^{\widehat{\R(t)}}$, consisting of all those real
            Puiseux series that are algebraic over $\R(t)$, is a real closure of $\R(t)$.

            The field of real Puiseux series itself is a real closure of the field $\R((t))$
            of real formal Laurent series.
    \end{enumerate}
\end{bsp}

\begin{lemma}
    Let $L_1, L_2$ be real-closed fields and let $\varphi\colon L_1 \to L_2$ be a homomorphism
    of fields. Then $\varphi$ is compatible with the canonical orderings of $L_1$ and $L_2$.
    \label{lemma:hom-real-closed-fields-respects-orderings}
\end{lemma}

\begin{proof}
    It suffices to prove that $x \ge_{L_1} 0$ implies $\varphi(x) \ge_{L_2} 0$
    for all $x \in L_1$. This follows from the fact that in a real-closed field $L$,
    for all $x \in L$, $x \ge 0$ if and only if $x$ is a square.
\end{proof}

If $k$ is a real field and $k^{r}$ is a real closure of $k$, then
$k$ inherits an ordering from $k^{r}$. However, different real closures may induce
different orderings on $k$, as the next example shows.

\begin{bsp}[]
    \label{bsp:different-real-closures-depending-on-ordering}
    Let $k = \Q(t)$. This is a real field, since $\Q$ is real. Since $\pi$
    is transcendental over $\Q$, we can embed $\Q(t)$ in $\R$ by sending $t$ to $\pi$.
    \[
    i_1\colon \Q(t) \xhookrightarrow{\simeq} \Q(\pi) \subseteq \R
    .\] Since $\R$ is real-closed, the relative algebraic closure
    $i_1(\Q(t))^{\R}$ is a real closure of $i_1(\Q(t))$.

    We can also embed $\Q(t)$ in the field $\widehat{\R(t)}$ of real Puiseux series via
    a homomorphism $i_2$ and then
    $\overline{i_2(\Q(t))}^{\widehat{\R(t)}}$ is a real closure of $i_2(\Q(t))$.
    However, the ordering on $\overline{i_1(\Q(t))}^{\R}$
    is Archimedean, because it is a subfield of $\R$,
    while the ordering on $\overline{i_2(\Q(t))}^{\widehat{\R(t)}}$
    is not Archimedean (it contains infinitesimal elements, such as $t$ for instance).

    The fields $\overline{i_1(\Q(t))}^{\R}$
    and $\overline{i_2(\Q(t))}^{\widehat{\R(t)}}$ cannot be isomorphic as fields.
    Indeed, when two real-closed fields $L_1, L_2$ are isomorphic as fields,
    then they are isomorphic as ordered fields, since positivity on a real
    closed field is defined by the condition of being a square, which is preserved
    under isomorphisms of fields.
\end{bsp}

%The next result will be proved later on.
%
%\begin{lemma}
%    Let $(k, \le )$ be an ordered field and $P \in k[t]$ be an irreducible polynomial.
%    Let $L_1, L_2$ be real-closed extensions of $k$ that are compatible with the ordering of $k$.
%    Then $P$ has the same number of roots in $L_1$ as in $L_2$.
%    \label{lemma:number-of-roots-in-real-closed-extension}
%\end{lemma}
%
%\begin{bem}
%    In particular, if $P \in k[t]$ is an arbitrary polynomial, then if $P$ has a root
%    in a real-closed extension $L$ of $k$, then it has a root in all real-closed extensions of $k$.
%
%    A polynomial with coefficients in an ordered field $(k, \le)$ might not have roots
%    in any real-closed extensions of $k$.
%\end{bem}

\begin{lemma}
    Let $(k, \le)$ be an ordered field, $L / k$ an orderable real-closed extension of $k$
    and $\varphi\colon k \to L$ a morphism of $k$-algebras.
    If $E / k$ is a finite, real extension of $k$, then $\varphi$ admits a continuation, i.e.
    a morphism of $k$-algebras $\varphi'$ such that the following diagram commutes:
    \[
    \begin{tikzcd}
        k \arrow[hook]{d} \arrow{r}{\varphi} & L \\
        E \arrow[dashed, swap]{ur}{\varphi'}
    \end{tikzcd}
    .\] 
    \label{lemma:continuation-in-real-closed}
\end{lemma}

\begin{proof}
    Since $k$ is perfect, $E / k$ is separable. Moreover $E / k$ is finite, thus by the
    primitive element theorem, $E = k[a]$ for $a \in E$. Let
    $P \in k[t]$ be the minimal polynomial of $a$ over $k$. Let $E^{r}$ be
    an orderable real-closure of $E$. Thus $E^{r}$ is
    a real-closed extension of $k$ that contains a root of $P$. By
    \ref{lemma:number-of-roots-in-real-closed-extension},
    $P$ has a root $b \in L$. Now
    define $\psi\colon k[t] \to L$ by $t \mapsto b$ and
    $\psi|_k = \varphi$. Since $b$ is a root of $P$,
    $\psi$ factors through $E = k[a] = k / (P)$ and gives $\varphi'\colon E \to L$.
\end{proof}

\end{document}