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index 2504e42..333f886 100644
--- a/ws2022/rav/lecture/rav.tex
+++ b/ws2022/rav/lecture/rav.tex
@@ -33,5 +33,7 @@ Christian Merten (\href{mailto:cmerten@mathi.uni-heidelberg.de}{cmerten@mathi.un
 \input{rav15.tex}
 \input{rav16.tex}
 \input{rav17.tex}
+\input{rav18.tex}
+\input{rav19.tex}
 
 \end{document}
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index 6106523..19751c6 100644
--- a/ws2022/rav/lecture/rav17.tex
+++ b/ws2022/rav/lecture/rav17.tex
@@ -201,7 +201,7 @@ Simple extensions of odd degree are simpler from the real point of view:
         \item $\R$ is a real-closed field, because $\R[i] = \mathbb{C}$ is algebraically closed.
         \item The field of real Puiseux series
             \begin{salign*}
-            \widehat{\R(t)} \coloneqq \bigcup_{q > 0} \R(t ^{\frac{1}{q}})
+            \widehat{\R(t)} \coloneqq \bigcup_{q > 0} \R((t ^{\frac{1}{q}}))
             = \left\{ 
                 \sum_{n=m}^{\infty} a_n t ^{\frac{n}{q}} \colon
                 m \in \Z, q \in \N \setminus \{0\}, a_n \in \R
@@ -214,4 +214,14 @@ Simple extensions of odd degree are simpler from the real point of view:
     \end{itemize}
 \end{bsp}
 
+\begin{bem}[]
+    By \ref{thm:charac-real-closed}, if $k$ is a real-closed field, then the absolute galois
+    group of $k$ is
+    \[
+    \text{Gal}(\overline{k} / k) = \text{Gal}(k[i] / k) \simeq \Z / 2 \Z
+    .\] The Artin-Schreier theorem shows that if $\overline{k} / k$
+    is a non-trivial extension of \emph{finite} degree,
+    then $k$ is real-closed.
+\end{bem}
+
 \end{document}
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--- /dev/null
+++ b/ws2022/rav/lecture/rav19.tex
@@ -0,0 +1,114 @@
+\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}
+
+Real-closed fields $L$ admit a canonical structure of ordered field, where $x \ge 0$
+in $L$, if and only if $x$ is a square. In particular,
+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}[]
+    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 $L_2$.
+\end{lemma}
+
+\end{document}