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rav: add new lecture, reorder some lemmas

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christian
2023-01-26 16:41:20 +01:00
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ws2022/rav/lecture/rav.pdf
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@@ -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}
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ws2022/rav/lecture/rav19.tex
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@@ -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}
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\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}