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\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}

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