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rav: update lec 17, add lec 19

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christian
2023-01-20 18:40:34 +01:00
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ws2022/rav/lecture/rav.pdf
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@@ -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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ws2022/rav/lecture/rav17.tex
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@@ -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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ws2022/rav/lecture/rav19.tex
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\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}