rav: add lecture 18

ws2022/rav/lecture/rav18.tex

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+\documentclass{lecture}
+
+\begin{document}
+
+\section{Extensions of ordered fields}
+
+If a field $k$ admits a structure of ordered field, we will say that $k$ is \emph{orderable}.
+For an \emph{ordered} field $k$, an extension $L / k$ is called \emph{oderable} if the
+field $L$ is orderable such that the induced order on $k$ coincides with the fixed order
+on $k$.
+
+\begin{definition}[]
+    Let $k$ be a field. A quadratic form $q\colon k^{n} \to k$
+    is called \emph{isotropic} if there exists
+    $x \in k \setminus \{0\} $ such that $q(x) = 0$. Otherwise, the quadratic
+    form is called \emph{anisotropic}.
+\end{definition}
+
+\begin{bem}[]
+    Recall that, given a quadratic form $q$ on a finite-dimensional
+    $k$-vector space $E$, there always exists a basis of $E$ in which
+    $q(x_1, \ldots, x_n) = a_1x_1^2 + \ldots + a_r x_r^2$, where
+    $r = \text{rg}(q) \le n = \text{dim } E$ and $a_1, \ldots, a_r \in k$.
+    The form $q$ is non-degenerate on $E$ if and only if $r = \text{dim } E$.
+\end{bem}
+
+\begin{bsp}[]
+    \begin{itemize}
+        \item A field $k$ is real if and only if for all $n \in \N$, the form
+            $x_1^2 + \ldots + x_n^2$ is anisotropic.
+        \item A degenerate quadratic form is isotropic.
+        \item If $k$ is algebraically closed and $n \ge 2$,
+            all quadratic forms on $k^{n}$ are isotropic.
+        \item If $(k, \le )$ is an ordered field and
+            $q(x_1, \ldots, x_n) = a_1 x_1^2 + \ldots + a_n x_n^2$ with
+            $a_i > 0$ for all $i$, then $q$ and $-q$ are anisotropic on $k^{n}$.
+    \end{itemize}
+\end{bsp}
+
+\begin{definition}
+    Let $k$ be a field and $L$ an extension of $k$. A quadratic form
+    $q\colon k^{n} \to k$ induces a quadratic form $q_L \colon L^{n} \to L$. The form
+    $q$ is called \emph{anisotropic over $L$} if $q_L$ is anisotropic.
+\end{definition}
+
+It can be checked that, on an ordered field $(k, \le )$, a quadratic form
+$q$ is anisotropic if and only if it is non-degenerate and of constant sign. The interest
+of this notion for us is given by the following result.
+
+\begin{theorem}
+    \label{thm:charac-orderable-extension}
+    Let $(k, \le )$ be an ordered field and $L$ be an extension of $k$. Then the following
+    conditions are equivalent:
+    \begin{enumerate}[(i)]
+        \item The extension $L / k$ is orderable.
+        \item For all $n \ge 1$ and all $a = (a_1, \ldots, a_n) \in k^{n}$ such that
+            $a_i > 0$ for all $i$, the quadratic form
+            $q(x_1, \ldots, x_n) = a_1x_1^2 + \ldots + a_n x_n^2$ is anisotropic over $L$
+            (i.e. all positive definite quadratic forms on $k$ are anisotropic over $L$).
+    \end{enumerate}
+\end{theorem}
+
+\begin{proof}
+    (i)$\Rightarrow$(ii): Assume that there is an ordering of $L$ that extends
+    the ordering of $k$ and let $n \ge 1$. Let $a = (a_1, \ldots, a_n) \in k^{n}$
+    with $a_i > 0$ for all $i$. Then $a_i > 0$ still holds in $L$. Since
+    squares are non-negative for all orderings, the sum
+    $a_1 x_1^2 + \ldots + a_n x_n^2$ is a sum of positive terms in $L$. Therefore
+    it can only be $0$, if all of its terms are $0$. Since $a_i \neq 0$, it follows
+    $x_i = 0$ for all $i$.
+
+    (ii)$\Rightarrow$(i): Define
+    \[
+    P = \bigcup_{n \ge 1} \left\{ \sum_{i=1}^{n} a_i x_i^2 \colon a_i \in k, a_i > 0, x_i \in L \right\} 
+    .\] The set $P$ is stable by sum and product and contains all squares of $L$,
+    so it is a cone in $L$. Suppose $-1 \in P$. Then there exists
+    $n \ge 1$ and $a = (a_1, \ldots, a_n) \in k^{n}$ with $a_i > 0$
+    and $x = (x_1, \ldots, x_n) \in L^{n}$ such that
+    $-1 = \sum_{i=1}^{n} a_i x_i^2$. So
+    \[
+    a_1 x_1^2 + \ldots + a_n x_n^2 + 1 = 0
+    ,\] meaning that the quadratic form $a_{1} x_1^2 + \ldots + a_n x_n^2 + x_{n+1}^2$
+    is isotropic on $L^{n+1}$, contradicting (ii).
+    Thus $P$ is a positive cone containing all positive elements of $k$. By
+    embedding $P$ in a maximal positive cone, the claim follows.
+\end{proof}
+
+\begin{satz}[]
+    Let $(k, \le )$ be an ordered field and let $c > 0$ be a positive element in $k$.
+    Then $k[\sqrt{c}]$ is an orderable extension of $k$.
+\end{satz}
+
+\begin{proof}
+    If $c$ is a square in $k$, there is nothing to prove. Otherwise, $k[\sqrt{c}]$ is
+    indeed a field. Let $n \ge 1$ and let $a = (a_1, \ldots, a_n) \in k^{n}$
+    with $a_i > 0$ for all $i$. Assume that $x = (x_1, \ldots, x_n) \in k[\sqrt{c}]^{n}$
+    satisfies
+    \[
+    a_1 x_1^2 + \ldots + a_n x_n^2 = 0
+    .\] Since $x_i = u_i + v_i \sqrt{c} $ for some $u_i, v_i \in k$, we can rewrite
+    this equation as
+    \[
+    \sum_{i=1}^{n} a_i (u_i^2 + c v_i^2) + 2 \sum_{i=1}^{n} u_i v_i \sqrt{c} = 0
+    .\] 
+    Since $1$ and $\sqrt{c}$ are linearly independent over $k$, we get
+    $\sum_{i=1}^{n} a_i (u_i^2 + c v_i^2) = 0$, hence
+    $u_i = v_i = 0$ for all $i$, since all terms in the previous sum are non-negative.
+    So $x_i = 0$ for all $i$ and (ii) of \ref{thm:charac-orderable-extension}
+    is satisfied.
+\end{proof}
+
+\begin{satz}
+    Let $(k, \le )$ be an ordered field and let $P \in k[t]$ be an irreducible
+    polynomial of odd degree. Then the field $L \coloneqq k[t]/ (P)$ is an orderable
+    extension of $k$.
+\end{satz}
+
+\begin{proof}
+    Denote by $d$ the degree of $P$ and proceed by induction on $d \ge 1$. If $d = 1$, then
+    $L = k$. Now assume $d \ge 2$. Let $n \ge 1$ and $a_1, \ldots, a_n \in k$ with $a_i > 0$.
+    Denote by $q_L$ the quadratic form
+    \[
+    q_L(x_1, \ldots, x_n) = a_1 x_1^2 + \ldots + a_n x_n^2
+    \] on $L^{n}$. If $q_L$ is isotropic over $L$, then there exist
+    polynomials $g_1, \ldots, g_n \in k[t]$ with $\text{deg}(g_i) < d$
+    and $h \in k[t]$ such that
+    \begin{equation}
+        q_L(g_1, \ldots, g_n) = h P
+        \label{eq:quad-form}
+    \end{equation}
+    Let $g$ be the greatest common divisor of $g_1, \ldots, g_n$. Since $q_L$ is
+    homogeneous of degree $2$,
+    $g^2$ divides $q_L(g_1, \ldots, g_n)$. Since $P$ is irreducible, $g$ divides $h$.
+    We may thus assume that $g = 1$. The leading coefficients of the terms on
+    the left hand side of (\ref{eq:quad-form}) are non-negative, thus
+    the sum has even degree $< 2d$. Since the degree of $P$ is odd,
+    $h$ must be of odd degree $< d$. Therefore, $h$ has an irreducible factor
+    $h_1 \in k[t]$ of odd degree. Let $\alpha$ be a root of $h_1$. By evaluating
+    (\ref{eq:quad-form}) at $\alpha$, we get
+    \[
+    q_{k[\alpha]}(g_1(\alpha), \ldots, g_n(\alpha)) = 0
+    \] in $k[\alpha]$. Since the $gcd(g_1, \ldots, g_n) = 1$ and $k[t]$ is a principal ideal
+    domain, there exist $h_1, \ldots, h_n \in k[t]$ such that
+    \[
+    h_1 g_1 + \ldots + h_n g_n = 1
+    .\] In particular
+    \[
+    h_1(\alpha) g_1(\alpha) + \ldots + h_n(\alpha) g_n(\alpha) = 1
+    ,\] so not all $g_i(\alpha)$ are $0$ in $k[\alpha]$. Thus $q_{k[\alpha]}$
+    is isotropic over $k[\alpha] = k[t] / (h_1)$ contradicting the induction hypothesis.
+\end{proof}
+
+\end{document}