rav: add new lecture, reorder some lemmas

ws2022/rav/lecture/rav.tex

@@ -35,5 +35,6 @@ Christian Merten (\href{mailto:cmerten@mathi.uni-heidelberg.de}{cmerten@mathi.un
 \input{rav17.tex}
 \input{rav18.tex}
 \input{rav19.tex}
+\input{rav20.tex}
 
 \end{document}

ws2022/rav/lecture/rav19.tex

@@ -73,13 +73,24 @@
     \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
+\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 $\pi$.
     \[
@@ -105,10 +116,46 @@ different orderings on $k$, as the next example shows.
 
 The next result will be proved later on.
 
-\begin{lemma}[]
+\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}

ws2022/rav/lecture/rav20.tex

@@ -0,0 +1,87 @@
+\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}