uni/ws2022/rav/lecture/rav17.tex

\documentclass{lecture}

\begin{document}

\section{Real-closed fields}

In this section we study real algebraic extensions of real fields.

\begin{lemma}
    Let $k$ be a real field and $x \in k \setminus \{0\} $. Then
    $x$ and $-x$ cannot be both sums of squares in $k$.
    \label{lemma:real-field-only-one-is-square}
\end{lemma}

\begin{proof}
    If $x \in \Sigma k^{[2]}$ and $-x \in \Sigma k^{[2]}$, then
    \[
    1 = \frac{1}{x^2} (-x) x \in \Sigma k^{[2]}
    \] contradicting that $k$ is real.
\end{proof}

\begin{satz}
    Let $k$ be a real field and $a \in k$ such that $a$ is not a square in $k$.
    Then the field
    \[
    k(\sqrt{a}) = k[t] / (t^2 - a)
    \] is real if and only if $-a \not\in \Sigma k^{[2]}$.
    In particular, if $\Sigma k^{[2]} \cup (- \Sigma k^{[2]}) \neq k$,
    then $k$ admits real quadratic extensions.
    \label{satz:quadratic-extensions-of-real-field}
\end{satz}

\begin{proof}
    Since $a$ is not a square in $k$, $t^2 - a$ is irreducible in $k[t]$, so
    $k[t] / (t^2 -a)$ is indeed a field. Denote by $\sqrt{a} $ the class of
    $t$ in the quotient.

    ($\Rightarrow$): $a$ is a square in $k(\sqrt{a})$, thus by
    \ref{lemma:real-field-only-one-is-square} we have $-a \not\in \Sigma k(\sqrt{a})^2$.
    But $\Sigma k^{[2]} \subseteq \Sigma k(\sqrt{a})^{[2]}$, thus
    $-a \not\in \Sigma k(\sqrt{a})^{[2]}$.

    ($\Leftarrow$):
    $-1 \in \Sigma k(\sqrt{a})^{[2]}$
    if and only if there exist $x_i, y_i \in k$, such that
    \[
    -1 = \sum_{i=1}^{n} (x_i + y_i \sqrt{a})^2
    = \sum_{i=1}^{n} (x_i^2 + a y_i^2) + 2 \sqrt{a} \sum_{i=1}^{n} x_i y_i
    .\]
    Since $(1, \sqrt{a})$ is a basis of the $k$-vector space $k(\sqrt{a})$, the previous equality
    implies
    \begin{salign*}
        -1 &= \sum_{i=1}^{n} x_i^2 + a \sum_{i=1}^{n} y_i^2
    .\end{salign*}
    Since $-1 \not\in \Sigma k^{[2]}$, $\sum_{i=1}^{n} y_i^2 \neq 0$, this
    implies
    \[
    -a = \frac{1 + \sum_{i=1}^{n} x_i^2}{\sum_{i=1}^{n} y_i^2}
    = \frac{\left( \sum_{i=1}^{n} y_i^2 \right)\left( 1 + \sum_{i=1}^{n} x_i^2 \right) }
    {\left( \sum_{i=1}^{n} y_i^2 \right)^2}
    \in \Sigma k^{[2]}
    .\]
\end{proof}

Simple extensions of odd degree are simpler from the real point of view:

\begin{satz}
    Let $k$ be a real field and $P \in k[t]$ be an irreducible polynomial of odd degree.
    Then the field $k[t]/(P)$ is real.
    \label{satz:odd-real-extension}
\end{satz}

\begin{proof}
    Denote by $n$ the degree of $P$. We proceed by induction on $n \ge 1$.
    If $n = 1$, then $k[t]/(P) \simeq k$ is real. Since $n$ is odd, we
    may now assume $n \ge 3$.
    Let $L \coloneqq k[t]/(P)$. Suppose $L$ is not real. Then there exist
    polynomials $g_i \in k[t]$, of degree at most $n-1$, such that
    $-1 = \sum_{i=1}^{m} g_i^2$
    in $L = k[t]/(P)$. Since $k \subseteq L$ and $k$ is real, at least
    one of the $g_i$ is non-constant.
    By definition of $L$, there exists $Q \in k[t] \setminus \{0\}$
    such that
    \begin{equation}
    -1 = \sum_{i=1}^{m} g_i^2 + P Q
    \label{eq:gi-sq+pq}
    \end{equation}
    in $k[t]$. Since $k$ is real, in $\sum_{i=1}^{m} g_i^2$ no cancellations
    of the terms of highest degree can occur. Thus
    $\sum_{i=1}^{m} g_i^2$ is of positive, even degree at most $2n-2$. By
    \ref{eq:gi-sq+pq}, it follows that $Q$ is of odd degree at most $n-2$.
    In particular, $Q$ has at least one irreducible factor $Q_1$ of odd degree at most
    $n-2$. Since $n \ge 3$, $n-2 \ge 1$. By induction,
    $M \coloneqq k[t]/(Q_1)$ is real. But \ref{eq:gi-sq+pq} implies
    \[
    -1 = \sum_{i=1}^{m} g_i^2
    \] in $M = k[t] / (Q_1)$ contradicting the fact that $M$ is real.
\end{proof}

\begin{definition}
    A \emph{real-closed} field is a real field that
    has no proper real algebraic extensions.
\end{definition}

\begin{theorem}
    Let $k$ be a field. Then the following conditions are equivalent:
    \begin{enumerate}[(i)]
        \item $k$ is real-closed.
        \item $k$ is real and for all $a \in k$, either $a$ or $-a$
            is a square in $k$ and
            every polynomial of odd degree in $k[t]$ has a
            root in $k$.
        \item the $k$-algebra
            \[
            k[i] \coloneqq k[t] / (t^2+1)
            \] is algebraically closed.
    \end{enumerate}
    \label{thm:charac-real-closed}.
\end{theorem}

\begin{proof}
    (i)$\Rightarrow$(ii): Let $a \in k$ such that neither $a$ nor $-a$ is a square in $k$. Then
    by \ref{satz:quadratic-extensions-of-real-field} and (i), $\pm a \in \Sigma k^{[2]}$
    contradicting
    \ref{lemma:real-field-only-one-is-square}. Let $P \in k[t]$ be a polynomial
    of odd degree. $P$ has at least one irreducible factor $P_1$ of odd degree.
    By \ref{satz:odd-real-extension}, $k[t]/(P_1)$ is a real extension of $k$.
    Since $k$ is real-closed, $P_1$ must be of degree $1$ and thus $P$ has a root in $k$.

    (ii)$\Rightarrow$(iii): Since $-1$ is not a square in $k$, the polynomial
    $t^2 + 1$ is irreducible over $k$. Thus $L \coloneqq k[t]/(t^2 + 1)$ is a field. Denote
    by $i$ the image of $t$ in $L$ and for $x = a + ib \in L = k[i]$, denote
    by $\overline{x} = a - ib$. This extends to a ring homomorphism $L[t] \to L[t]$. Let
    $P \in L[t]$ be non-constant. It remains to show, that $P$ has a root in $L$. We
    first reduce to the case $P \in k[t]$.

    Assume every non-constant polynomial in $k[t]$ has a root in $L$. Let $P \in L[t]$. Then
    $P \overline{P} \in k[t]$ has a root $x \in L$, thus either $P(x) = 0$
    or $\overline{P}(x) = 0$. In the first case, we are done.
    In the second case, we have $P(\overline{x}) = \overline{\overline{P}(x)}
    = \overline{0} = 0$, so $\overline{x}$ is a root of $P$ in $L$.

    Thus we may assume $P \in k[t]$. Write $d = \text{deg}(P) = 2^{m} n$ with $2 \nmid n$. We
    proceed by induction on $m$. If $m = 0$, the result is true by (ii). Now assume $m > 0$.
    Fix an algebraic closure $\overline{k}$ of $k$. Since $k$ is real, it is of characteristic
    $0$, thus $k$ is perfect and $\overline{k} / k$ is galois.
    Let $y_1, \ldots, y_d$ be the roots
    of $P$ in $\overline{k}$. Consider for all $r \in \Z$:
    \[
    F_r \coloneqq \prod_{1 \le p < q \le d}^{}
    \left( t - (y_p + y_q) - r y_p y_q) \right) \in \overline{k}[t]
    .\] This polynomial with coefficients in $\overline{k}$ is invariant
    under permutation of $y_1, \ldots, y_d$. Thus its coefficients
    lie in $\overline{k}^{\text{Gal}(\overline{k} / k)} = k$. Moreover
    \[
    \text{deg}(F_r) = \binom{d}{2} = \frac{d(d-1)}{2} = 2^{m-1} n (2^{m} -1)
    .\] with $n (2^{m} -1)$ odd. So the induction hypothesis applies and,
    for all $r \in \Z$, there is a pair $p < q$ in $\{1, \ldots, d\} $
    such that $(y_p + y_q) + r y_p y_q \in L$. Since $\Z$ is infinite,
    we can find a pair $p < q$ in $\{1, \ldots, d\} $ for which
    there exists a pair $r \neq r'$ such that
    \begin{salign*}
        &(y_p + y_q) + r y_p y_q \in L \\
        \text{and } & (y_p + y_q) + r' y_p y_q \in L
    .\end{salign*}
    By solving the system, we get $y_p + y_q \in L$ and $y_p y_q \in L$. But $y_p, y_q$
    are roots of the quadratic polynomial
    \[
    t^2 - (y_p + y_q)t + y_p y_q \in L[t]
    \] and since $i^2 = -1$, the roots of this polynomial lie in $L = k[i]$, by (ii) and the
    usual formulas
    \[
        t_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}
    .\] So $P$ indeed has a root in $k[i]$, which finishes the induction.

    (iii)$\Rightarrow$(i): Denote again by $i$ the image in the algebraically closed field
    $k[t]/(t^2 + 1)$. We first show that $k^{[2]} = \Sigma k^{[2]}$. Let $a, b \in k$. Then
    $a + ib = (c + id)^2$ in $k[i]$ for some $c, d \in k$. Thus
    \[
    a^2 + b^2 = (a+ib)(a-ib) = (c+id)^2(c-id)^2 = (c^2 + d^2)^2
    .\] By induction the claim follows. Since $t^2 + 1$ is irreducible,
    $-1 \not\in k^{[2]} = \Sigma k^{[2]}$ and $k$ is real.

    Let $L$ be a real algebraic extension of $k$. Since $k[i]$ is algebraically closed and contains
    $k$, there exists a $k$-homomorphism $L \xhookrightarrow{} k[i]$. Since
    $[ k[i] : k ] = 2$, either $L = k$ or $L = k[i]$, but $k[i]$ is not real, since $i^2 = -1$
    in $k[i]$. So $L = k$ and $k$ is real-closed.
\end{proof}

\begin{korollar}
    A real-closed field $k$ admits a canonical structure of ordered field, in
    which the cone of positive elements is exactly $k^{[2]}$, the set of squares in $k$.
\end{korollar}

\begin{proof}
    This was proven in the implication (i)$\Rightarrow$(ii) of \ref{thm:charac-real-closed}.
\end{proof}

\begin{bsp}[]
    \begin{itemize}
        \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}}))
            = \left\{ 
                \sum_{n=m}^{\infty} a_n t ^{\frac{n}{q}} \colon
                m \in \Z, q \in \N \setminus \{0\}, a_n \in \R
            \right\} 
            \end{salign*}
            is a real closed field because
            $\widehat{\R(t)}[i] = \widehat{\R[i][t]} = \widehat{\mathbb{C}[t]}$ is the field
            of complex Puiseux series, which is algebraically closed by the
            Newton-Puiseux theorem.
    \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}