rav: add 17th lecture
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@@ -32,5 +32,6 @@ Christian Merten (\href{mailto:cmerten@mathi.uni-heidelberg.de}{cmerten@mathi.un
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\input{rav10.tex}
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\input{rav15.tex}
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\input{rav16.tex}
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\input{rav17.tex}
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\end{document}
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\documentclass{lecture}
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\begin{document}
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\section{Real-closed fields}
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In this section we study real algebraic extensions of real fields.
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\begin{lemma}
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Let $k$ be a real field and $x \in k \setminus \{0\} $. Then
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$x$ and $-x$ cannot be both sums of squares in $k$.
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\label{lemma:real-field-only-one-is-square}
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\end{lemma}
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\begin{proof}
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If $x \in \Sigma k^{[2]}$ and $-x \in \Sigma k^{[2]}$, then
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\[
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1 = \frac{1}{x^2} (-x) x \in \Sigma k^{[2]}
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\] contradicting that $k$ is real.
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\end{proof}
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\begin{satz}
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Let $k$ be a real field and $a \in k$ such that $a$ is not a square in $k$.
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Then the field
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\[
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k(\sqrt{a}) = k[t] / (t^2 - a)
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\] is real if and only if $-a \not\in \Sigma k^{[2]}$.
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In particular, if $\Sigma k^{[2]} \cup (- \Sigma k^{[2]}) \neq k$,
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then $k$ admits real quadratic extensions.
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\label{satz:quadratic-extensions-of-real-field}
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\end{satz}
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\begin{proof}
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Since $a$ is not a square in $k$, $t^2 - a$ is irreducible in $k[t]$, so
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$k[t] / (t^2 -a)$ is indeed a field. Denote by $\sqrt{a} $ the class of
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$t$ in the quotient.
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($\Rightarrow$): $a$ is a square in $k(\sqrt{a})$, thus by
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\ref{lemma:real-field-only-one-is-square} we have $-a \not\in \Sigma k(\sqrt{a})^2$.
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But $\Sigma k^{[2]} \subseteq \Sigma k(\sqrt{a})^{[2]}$, thus
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$-a \not\in \Sigma k(\sqrt{a})^{[2]}$.
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($\Leftarrow$):
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$-1 \in \Sigma k(\sqrt{a})^{[2]}$
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if and only if there exist $x_i, y_i \in k$, such that
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\[
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-1 = \sum_{i=1}^{n} (x_i + y_i \sqrt{a})^2
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= \sum_{i=1}^{n} (x_i^2 + a y_i^2) + 2 \sqrt{a} \sum_{i=1}^{n} x_i y_i
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.\]
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Since $(1, \sqrt{a})$ is a basis of the $k$-vector space $k(\sqrt{a})$, the previous equality
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implies
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\begin{salign*}
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-1 &= \sum_{i=1}^{n} x_i^2 + a \sum_{i=1}^{n} y_i^2
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.\end{salign*}
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Since $-1 \not\in \Sigma k^{[2]}$, $\sum_{i=1}^{n} y_i^2 \neq 0$, this
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implies
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\[
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-a = \frac{1 + \sum_{i=1}^{n} x_i^2}{\sum_{i=1}^{n} y_i^2}
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= \frac{\left( \sum_{i=1}^{n} y_i^2 \right)\left( 1 + \sum_{i=1}^{n} x_i^2 \right) }
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{\left( \sum_{i=1}^{n} y_i^2 \right)^2}
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\in \Sigma k^{[2]}
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.\]
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\end{proof}
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Simple extensions of odd degree are simpler from the real point of view:
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\begin{satz}
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Let $k$ be a real field and $P \in k[t]$ be an irreducible polynomial of odd degree.
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Then the field $k[t]/(P)$ is real.
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\label{satz:odd-real-extension}
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\end{satz}
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\begin{proof}
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Denote by $n$ the degree of $P$. We proceed by induction on $n \ge 1$.
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If $n = 1$, then $k[t]/(P) \simeq k$ is real. Since $n$ is odd, we
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may now assume $n \ge 3$.
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Let $L \coloneqq k[t]/(P)$. Suppose $L$ is not real. Then there exist
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polynomials $g_i \in k[t]$, of degree at most $n-1$, such that
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$-1 = \sum_{i=1}^{m} g_i^2$
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in $L = k[t]/(P)$. Since $k \subseteq L$ and $k$ is real, at least
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one of the $g_i$ is non-constant.
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By definition of $L$, there exists $Q \in k[t] \setminus \{0\}$
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such that
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\begin{equation}
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-1 = \sum_{i=1}^{m} g_i^2 + P Q
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\label{eq:gi-sq+pq}
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\end{equation}
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in $k[t]$. Since $k$ is real, in $\sum_{i=1}^{m} g_i^2$ no cancellations
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of the terms of highest degree can occur. Thus
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$\sum_{i=1}^{m} g_i^2$ is of positive, even degree at most $2n-2$. By
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\ref{eq:gi-sq+pq}, it follows that $Q$ is of odd degree at most $n-2$.
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In particular, $Q$ has at least one irreducible factor $Q_1$ of odd degree at most
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$n-2$. Since $n \ge 3$, $n-2 \ge 1$. By induction,
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$M \coloneqq k[t]/(Q_1)$ is real. But \ref{eq:gi-sq+pq} implies
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\[
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-1 = \sum_{i=1}^{m} g_i^2
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\] in $M = k[t] / (Q_1)$ contradicting the fact that $M$ is real.
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\end{proof}
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\begin{definition}
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A \emph{real-closed} field is a real field that
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has no proper real algebraic extensions.
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\end{definition}
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\begin{theorem}
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Let $k$ be a field. Then the following conditions are equivalent:
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\begin{enumerate}[(i)]
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\item $k$ is real-closed.
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\item $k$ is real and for all $a \in k$, either $a$ or $-a$
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is a square in $k$ and
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every polynomial of odd degree in $k[t]$ has a
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root in $k$.
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\item the $k$-algebra
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\[
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k[i] \coloneqq k[t] / (t^2+1)
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\] is algebraically closed.
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\end{enumerate}
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\label{thm:charac-real-closed}.
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\end{theorem}
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\begin{proof}
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(i)$\Rightarrow$(ii): Let $a \in k$ such that neither $a$ nor $-a$ is a square in $k$. Then
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by \ref{satz:quadratic-extensions-of-real-field} and (i), $\pm a \in \Sigma k^{[2]}$
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contradicting
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\ref{lemma:real-field-only-one-is-square}. Let $P \in k[t]$ be a polynomial
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of odd degree. $P$ has at least one irreducible factor $P_1$ of odd degree.
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By \ref{satz:odd-real-extension}, $k[t]/(P_1)$ is a real extension of $k$.
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Since $k$ is real-closed, $P_1$ must be of degree $1$ and thus $P$ has a root in $k$.
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(ii)$\Rightarrow$(iii): Since $-1$ is not a square in $k$, the polynomial
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$t^2 + 1$ is irreducible over $k$. Thus $L \coloneqq k[t]/(t^2 + 1)$ is a field. Denote
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by $i$ the image of $t$ in $L$ and for $x = a + ib \in L = k[i]$, denote
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by $\overline{x} = a - ib$. This extends to a ring homomorphism $L[t] \to L[t]$. Let
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$P \in L[t]$ be non-constant. It remains to show, that $P$ has a root in $L$. We
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first reduce to the case $P \in k[t]$.
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Assume every non-constant polynomial in $k[t]$ has a root in $L$. Let $P \in L[t]$. Then
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$P \overline{P} \in k[t]$ has a root $x \in L$, thus either $P(x) = 0$
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or $\overline{P}(x) = 0$. In the first case, we are done.
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In the second case, we have $P(\overline{x}) = \overline{\overline{P}(x)}
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= \overline{0} = 0$, so $\overline{x}$ is a root of $P$ in $L$.
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Thus we may assume $P \in k[t]$. Write $d = \text{deg}(P) = 2^{m} n$ with $2 \nmid n$. We
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proceed by induction on $m$. If $m = 0$, the result is true by (ii). Now assume $m > 0$.
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Fix an algebraic closure $\overline{k}$ of $k$. Since $k$ is real, it is of characteristic
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$0$, thus $k$ is perfect and $\overline{k} / k$ is galois.
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Let $y_1, \ldots, y_d$ be the roots
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of $P$ in $\overline{k}$. Consider for all $r \in \Z$:
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\[
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F_r \coloneqq \prod_{1 \le p < q \le d}^{}
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\left( t - (y_p + y_q) - r y_p y_q) \right) \in \overline{k}[t]
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.\] This polynomial with coefficients in $\overline{k}$ is invariant
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under permutation of $y_1, \ldots, y_d$. Thus its coefficients
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lie in $\overline{k}^{\text{Gal}(\overline{k} / k)} = k$. Moreover
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\[
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\text{deg}(F_r) = \binom{d}{2} = \frac{d(d-1)}{2} = 2^{m-1} n (2^{m} -1)
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.\] with $n (2^{m} -1)$ odd. So the induction hypothesis applies and,
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for all $r \in \Z$, there is a pair $p < q$ in $\{1, \ldots, d\} $
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such that $(y_p + y_q) + r y_p y_q \in L$. Since $\Z$ is infinite,
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we can find a pair $p < q$ in $\{1, \ldots, d\} $ for which
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there exists a pair $r \neq r'$ such that
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\begin{salign*}
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&(y_p + y_q) + r y_p y_q \in L \\
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\text{and } & (y_p + y_q) + r' y_p y_q \in L
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.\end{salign*}
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By solving the system, we get $y_p + y_q \in L$ and $y_p y_q \in L$. But $y_p, y_q$
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are roots of the quadratic polynomial
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\[
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t^2 - (y_p + y_q)t + y_p y_q \in L[t]
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\] and since $i^2 = -1$, the roots of this polynomial lie in $L = k[i]$, by (ii) and the
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usual formulas
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\[
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t_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}
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.\] So $P$ indeed has a root in $k[i]$, which finishes the induction.
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(iii)$\Rightarrow$(i): Denote again by $i$ the image in the algebraically closed field
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$k[t]/(t^2 + 1)$. We first show that $k^{[2]} = \Sigma k^{[2]}$. Let $a, b \in k$. Then
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$a + ib = (c + id)^2$ in $k[i]$ for some $c, d \in k$. Thus
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\[
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a^2 + b^2 = (a+ib)(a-ib) = (c+id)^2(c-id)^2 = (c^2 + d^2)^2
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.\] By induction the claim follows. Since $t^2 + 1$ is irreducible,
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$-1 \not\in k^{[2]} = \Sigma k^{[2]}$ and $k$ is real.
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Let $L$ be a real algebraic extension of $k$. Since $k[i]$ is algebraically closed and contains
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$k$, there exists a $k$-homomorphism $L \xhookrightarrow{} k[i]$. Since
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$[ k[i] : k ] = 2$, either $L = k$ or $L = k[i]$, but $k[i]$ is not real, since $i^2 = -1$
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in $k[i]$. So $L = k$ and $k$ is real-closed.
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\end{proof}
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\begin{korollar}
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A real-closed field $k$ admits a canonical structure of ordered field, in
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which the cone of positive elements is exactly $k^{[2]}$, the set of squares in $k$.
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\end{korollar}
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\begin{proof}
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This was proven in the implication (i)$\Rightarrow$(ii) of \ref{thm:charac-real-closed}.
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\end{proof}
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\begin{bsp}[]
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\begin{itemize}
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\item $\R$ is a real-closed field, because $\R[i] = \mathbb{C}$ is algebraically closed.
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\item The field of real Puiseux series
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\begin{salign*}
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\widehat{\R(t)} \coloneqq \bigcup_{q > 0} \R(t ^{\frac{1}{q}})
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= \left\{
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\sum_{n=m}^{\infty} a_n t ^{\frac{n}{q}} \colon
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m \in \Z, q \in \N \setminus \{0\}, a_n \in \R
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\right\}
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\end{salign*}
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is a real closed field because
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$\widehat{\R(t)}[i] = \widehat{\R[i][t]} = \widehat{\mathbb{C}[t]}$ is the field
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of complex Puiseux series, which is algebraically closed by the
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Newton-Puiseux theorem.
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\end{itemize}
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\end{bsp}
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\end{document}
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