rav: add 17th lecture

ws2022/rav/lecture/rav.tex

@@ -32,5 +32,6 @@ Christian Merten (\href{mailto:cmerten@mathi.uni-heidelberg.de}{cmerten@mathi.un
 \input{rav10.tex}
 \input{rav15.tex}
 \input{rav16.tex}
+\input{rav17.tex}
 
 \end{document}

ws2022/rav/lecture/rav17.tex

@@ -0,0 +1,217 @@
+\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}
+
+\end{document}