diff --git a/ws2022/rav/lecture/rav.pdf b/ws2022/rav/lecture/rav.pdf index c213686..7ddf5f9 100644 Binary files a/ws2022/rav/lecture/rav.pdf and b/ws2022/rav/lecture/rav.pdf differ diff --git a/ws2022/rav/lecture/rav.tex b/ws2022/rav/lecture/rav.tex index 5ef474d..e6fbd7c 100644 --- a/ws2022/rav/lecture/rav.tex +++ b/ws2022/rav/lecture/rav.tex @@ -34,6 +34,7 @@ Christian Merten (\href{mailto:cmerten@mathi.uni-heidelberg.de}{cmerten@mathi.un \input{rav16.tex} \input{rav17.tex} \input{rav18.tex} +\input{rav21.tex} \input{rav19.tex} \input{rav20.tex} diff --git a/ws2022/rav/lecture/rav19.tex b/ws2022/rav/lecture/rav19.tex index 53b28b5..bc2de68 100644 --- a/ws2022/rav/lecture/rav19.tex +++ b/ws2022/rav/lecture/rav19.tex @@ -114,22 +114,22 @@ different orderings on $k$, as the next example shows. under isomorphisms of fields. \end{bsp} -The next result will be proved later on. - -\begin{lemma} - Let $(k, \le )$ be an ordered field and $P \in k[t]$ be an irreducible polynomial. - Let $L_1, L_2$ be real-closed extensions of $k$ that are compatible with the ordering of $k$. - Then $P$ has the same number of roots in $L_1$ as in $L_2$. - \label{lemma:number-of-roots-in-real-closed-extension} -\end{lemma} - -\begin{bem} - In particular, if $P \in k[t]$ is an arbitrary polynomial, then if $P$ has a root - in a real-closed extension $L$ of $k$, then it has a root in all real-closed extensions of $k$. - - A polynomial with coefficients in an ordered field $(k, \le)$ might not have roots - in any real-closed extensions of $k$. -\end{bem} +%The next result will be proved later on. +% +%\begin{lemma} +% Let $(k, \le )$ be an ordered field and $P \in k[t]$ be an irreducible polynomial. +% Let $L_1, L_2$ be real-closed extensions of $k$ that are compatible with the ordering of $k$. +% Then $P$ has the same number of roots in $L_1$ as in $L_2$. +% \label{lemma:number-of-roots-in-real-closed-extension} +%\end{lemma} +% +%\begin{bem} +% In particular, if $P \in k[t]$ is an arbitrary polynomial, then if $P$ has a root +% in a real-closed extension $L$ of $k$, then it has a root in all real-closed extensions of $k$. +% +% A polynomial with coefficients in an ordered field $(k, \le)$ might not have roots +% in any real-closed extensions of $k$. +%\end{bem} \begin{lemma} Let $(k, \le)$ be an ordered field, $L / k$ an orderable real-closed extension of $k$ diff --git a/ws2022/rav/lecture/rav21.pdf b/ws2022/rav/lecture/rav21.pdf new file mode 100644 index 0000000..0cd40b7 Binary files /dev/null and b/ws2022/rav/lecture/rav21.pdf differ diff --git a/ws2022/rav/lecture/rav21.tex b/ws2022/rav/lecture/rav21.tex new file mode 100644 index 0000000..c8ed873 --- /dev/null +++ b/ws2022/rav/lecture/rav21.tex @@ -0,0 +1,225 @@ +\documentclass{lecture} + +\begin{document} + +\section{Counting real roots} + +In this section, we will study \emph{Sturm's method} of counting +the number of roots of a separable polynomial with coefficients +in a real-closed field $L$. + +\begin{lemma} + Let $(k, \le)$ be an ordered field and let $P \in k[t]$ be a separable polynomial. + Assume that $P$ has a root $a \in k$. Then there exists $\delta > 0$ such that + \begin{enumerate}[(i)] + \item for $x \in \;]a- \delta, a + \delta[$ and $x \neq a$, $P(x) \neq 0$. + \item for $x \in \;]a, a + \delta[$, $P(x)$ and $P'(x)$ have the same sign. + \item for $x \in \;]a- \delta, a[$, $P(x)$ and $P'(x)$ have opposite signs. + \item for $x \in \;]0, \delta[$, $P(a+h)$ and $P(a-h)$ have opposite signs. + \end{enumerate} + \label{lemma:root-signs-separable} +\end{lemma} + +\begin{proof} + Since $P$ is separable and $P(a) = 0$, it follows that $P'(a) \neq 0$. + By continuity of $P'$, there exists $\delta > 0$ such that + $P'$ has constant sign on $] a- \delta, a + \delta[$. Suppose $P'(x) > 0$. + Since $k$ is real-closed, + this implies that $P$ is strictly increasing on this interval. In particular, + $P(x) < P(a) = 0$ for $x \in \;]a - \delta, a[$ + and $P(x) > P(a) = 0$ for $x \in \;]a, a + \delta[$. The case $P'(x) < 0$ is + similar which concludes the proof. +\end{proof} + +\begin{definition} + Let $(k, \le )$ be an ordered field. A finite sequence $(P_0, \ldots, P_n)$ of polynomials + $P_i \in k[t]$ is called a \emph{Sturm sequence} if it satisfies the following + properties: + \begin{enumerate}[(i)] + \item $P_1 = P_0'$ + \item for all $x \in k$ and $i \in \{0, \ldots, n\}$, if $P_i(x) = 0$, then + $P_{i+1}(x) \neq 0$. + \item for all $x \in k$ and all $i \in \{1, \ldots, n-1\} $, + if $P_i(x) = 0$ then $P_{i-1}(x) P_{i+1}(x) < 0$. + \item $P_n \in k^{\times}$. + \end{enumerate} +\end{definition} + +\noindent If $P \in k[t]$ is separable and $k$ has characteristic $0$, then the greatest +common divisor of $P$ and $P'$ is $1$. To determine a Bézout relation between $P$ and $P'$, +one proceeds by successive Euclidean divisions: + +First set $P_0 = P$ and $P_1 = P'$, next define $P_2$ such that $P_0 = P_1 Q_1 - P_2$ +and $\text{deg}(P_2) < \text{deg}(P_1)$. Inductively, this +defines $P_i = P_{i+1} Q_{i+1} - P_{i+2}$ with $\text{deg}(P_{i+2}) < \text{deg}(P_{i+1})$. +This algorithm stops after at most $\text{deg}(P_0) $ steps +with $P_{n-1} = P_n Q_n$ and $P_n \neq 0$. +Then $P_n$ is a greatest common divisor of $P = P_0$ and $P' = P_1$. Since $P$ and +$P'$ are coprime, $P_n$ is a non-zero constant. + +\begin{korollar} + The sequence of polynomials $(P_0, \ldots, P_n)$ is a Sturm sequence. + This is called the to $P$ associated Sturm sequence. +\end{korollar} + +\begin{proof} + (i) and (iv) are clear. For (ii) observe that if there exists $x \in k$ and + $i \in \{0, \ldots, n\} $ such that $P_i(x) = P_{i+1}(x) = 0$, then + $P_j(x) = 0$ for all $j \ge i$ which contradicts $P_n(x) = P_n \neq 0$. + Finally for (iii), if $P_i(x) = 0$, then $P_{i-1}(x) = - P_{i+1}(x)$, so $P_{i-1}(x)$ + and $P_{i+1}(x)$ have opposite signs. +\end{proof} + +\begin{bem} + Let $(k, \le)$ be an ordered field. + For a finite sequence of elements $(a_0, \ldots, a_n)$ in $k$ with $a_0 \neq 0$, + the number of \emph{sign changes} in this sequence is the number of pairs + $(i, j)$ such that $i < j$, $a_i \neq 0$ and $a_i a_j < 0$ with either $j = i+1$ or + $j > i+1$ and $a_{i+1} = \ldots = a_{j-1} = 0$. +\end{bem} + +\begin{theorem}[Sturm's algorithm] + Let $k$ be a real-closed field equipped with its canonical ordering and let $P \in k[t]$ + be a separable polynomial. Let $(P_0, \ldots, P_n)$ be the associated Sturm sequence. + For all $a \in k$, we denote by $\nu(a)$ the number of sign changes + in the sequence $(P_0(a), \ldots, P_n(a))$. Then, for all pair $a, b \in k$ such that + $a < b$ and $P_i(a)P_i(b) \neq 0$ for all $i$, the number of roots of $P$ in the interval + $[a, b]$ is equal to $\nu(a) - \nu(b)$. + \label{thm:sturm} +\end{theorem} + +\begin{proof} + Let $x_1 < \ldots < x_m$ be the elements of the finite set + \[ + E = \{ x \in \; ]a, b[ \mid \exists i \in \{0, \ldots, n\} , P_i(x) = 0\} + .\] + %For all $x \in E$, we can choose $\delta > 0$ such that + %$] x - \delta, x + \delta[ \cap E = \{x\} $, i.e. + %$]x - \delta , x + \delta [$ contains no other root of one of the $P_i$'s. + There exists a partition of $[a,b]$ in subintervals + $[\alpha_j, \alpha_{j+1}]$ where $\alpha_0 = a$, $\alpha_m = b$, + and for all $j \in \{0, \ldots, m-1\} $, $\alpha_j \not\in E$, + $[\alpha_j, \alpha_{j+1}] \cap E = \{x_j\} $. + Also + \[ + \sum_{j=0}^{m-1} (\nu(\alpha_j) - \nu(\alpha_{j+1})) + = \nu(\alpha_0) - \nu(\alpha_1) + \nu(\alpha_1) - \ldots - \nu(\alpha_m) + = \nu(a) - \nu(b) + .\] Thus it suffices to show that for fixed $j \in \{0, \ldots, m-1\}$, the number of roots + of $P$ in $[\alpha_j, \alpha_{j+1}]$ is equal to $\nu(\alpha_j) - \nu(\alpha_{j+1})$. + By construction, + $P$ has at most one root in $[\alpha_j, \alpha_{j+1}]$, at $x_j$, thus + we want to show + \[ + \nu(\alpha_j) - \nu(\alpha_{j+1}) = \begin{cases} + 0 & P(x_j) \neq 0 \\ + 1 & P(x_j) = 0 + \end{cases} + .\] + If $P(x_j) = 0$, then $P(\alpha_j)$ and $P(\alpha_{j+1})$ must have opposite sign. Indeed, + by \ref{lemma:root-signs-separable} $P(x_j + h) P(x_j - h) < 0$ for all $h > 0$ small + enough, but $P$ cannot change sign on $[\alpha_j, x_j -h]$ nor on + $[x_j + h, \alpha_{j+1}]$, for otherwise the intermediate value theorem would + imply the existence of a root $x \neq x_j$ in $[\alpha_j, \alpha_{j+1}]$. + So $P(\alpha_j) P(\alpha_{j+1}) < 0$. + If $P(\alpha_j) > 0$, then $P(\alpha_{j+1}) < 0$. With $P_1 = P'$ and + \ref{lemma:root-signs-separable}, it follows that $P_1(x) < 0$ for + $x$ close to $x_j$. But $P_1$ cannot change sign in $[\alpha_j, \alpha_{j+1}]$, + otherwise its root in that interval would be $x_j$. Since $P$ is separable + and $P_1 = P'$, this is impossible. Thus $P' < 0$ + and $P$ is strictly decreasing on $[\alpha_j, \alpha_{j+1}]$. So + the sequence of signs in the sequence + $(P_0(\alpha_j), P_1(\alpha_j), \ldots, P_n(\alpha_j))$ starts + with $(+, -, \ldots)$ while the one at $\alpha_{j+1}$ starts + with $(-, -, \ldots)$. Similarly, if $P(\alpha_j) < 0$, then + the sequences are $(-, +, \ldots)$ and $(+, +, \ldots)$. In either case, + there is one more sign change in the sequence corresponding to $\alpha_j$, + so $\nu(\alpha_j) - \nu(\alpha_{j+1}) = 1$. + + Now suppose $P(x_j) \neq 0$. Observe that $P_0(\alpha_j)$ and + $P_0(\alpha_{j+1})$ have the same sign, otherwise by the intermediate value theorem + and the construction, $P_0(x_j) = 0$. Also a difference between + $\nu(\alpha_j)$ and $\nu(\alpha_{j+1})$ only occurs if there exists + $i \in \{0, \ldots, n-1\} $ such that $P_i(\alpha_j) P_i(\alpha_{j+1}) < 0$. In this + case, again by the intermediate value theorem, we have $P_i(x_j) = 0$. By the definition + of a Sturm sequence, we have $P_{i-1}(x_j)P_{i+1}(x_j) < 0$. If $P_{i-1}(x_j) < 0$ + then $P_{i-1} < 0$ on $[\alpha_j, \alpha_{j+1}]$, because $x_j$ is + the only possible root for $P_{i-1}$ in $[\alpha_j, \alpha_{j+1}]$, + so $P_{i-1}$ cannot change sign on that interval. Likewise, + $P_{i+1}$ has the same sign on $[\alpha_j, \alpha_{j+1}]$ as it does at $x_j$. Proceeding + similarly when $P_{i-1}(x_j) > 0$, we arrive at the following possibilities + for the sign sequences of $P_{i-1}(\alpha_j) P_i(\alpha_j)P_{i+1}(\alpha_j)$ + and $P_{i-1}(\alpha_{j+1})P_i(\alpha_{j+1})P_{i+1}(\alpha_{j+1})$: + + \begin{figure}[h!] + \centering + \begin{subfigure}[c]{0.4\textwidth} + \begin{tabular}{c|c|c} + & $P_i(\alpha_j) < 0$ & $P_i(\alpha_j) > 0$ \\ \hline + $P_{i-1}(x_j) < 0$ & $- - +$ & $- + + $ \\ \hline + $P_{i-1}(x_j) > 0$ & $+ - -$ & $+ + -$ + \end{tabular} + \subcaption{Sign sequence at $\alpha_{j}$} + \end{subfigure} + \hspace{1cm} + \begin{subfigure}[c]{0.4\textwidth} + \begin{tabular}{c|c|c} + & $P_i(\alpha_j) < 0$ & $P_i(\alpha_j) > 0$ \\ \hline + $P_{i-1}(x_j) < 0$ & $- + +$ & $- - + $ \\ \hline + $P_{i-1}(x_j) > 0$ & $+ + -$ & $+ - -$ + \end{tabular} + \subcaption{Sign sequence at $\alpha_{j+1}$} + \end{subfigure} + \end{figure} + Since sign sequences located in cells of the two tables corresponding to the same case have + the same number of sign changes, equal to $1$, we see that + $\nu(\alpha_{j}) - \nu(\alpha_{j+1}) = 0$. +\end{proof} + +We deduce from the previous result, this important result: + +\begin{korollar} + Let $(k, \le )$ be an ordered field and let $L_1, L_2$ be real-closed, orderable extensions + of $k$. Let $P \in k[t]$ be an irreducible polynomial over $k$. Then $P$ has the same + number of roots in $L_1$ as it does in $L_2$. + \label{lemma:number-of-roots-in-real-closed-extension} +\end{korollar} + +\begin{proof} + For a polynomial $Q = c_n t^{n} + c_{n-1} t ^{n-1} + \ldots + c_0 \in k[t]$ with + $c_n \neq 0$, the roots + of $Q$ in an ordered real-closed extension $L$ of $k$ are bounded by + \begin{salign*} + M + = 1 + \left| \frac{c_{n-1}}{c_n} \right|_L + \ldots + \left| \frac{c_0}{c_n} \right|_L + = 1 + \left| \frac{c_{n-1}}{c_n} \right|_k + \ldots + \left| \frac{c_0}{c_n} \right|_k + .\end{salign*} + Note that $M$ is independent from $L$. + So given $P \in k[t]$ irreducible and the associated Sturm sequence + $(P_0, P_1, \ldots, P_n)$, there exists $M \in k$ such that all roots + of all $P_i$'s in $L$ are contained in the interval $[-M, M] \subseteq L$. Since + $\text{char } k = 0$, $P$ is separable, by \ref{thm:sturm} + the number of roots of $P$ in $[-M, M] \subseteq L$ is equal + to $\nu(-M) - \nu(M)$. Since $\pm M \in k$, + all $P_i \in k[t]$ and the ordering of $L$ extends the one of $k$, the number + of sign changes $\nu(\pm M)$ in the sequences + $(P_0(-M), P_1(-M), \ldots, P_n(-M))$ + and $(P_0(M), P_1(M), \ldots, P_n(M))$ does not depend on $L$. +\end{proof} + +\begin{bem} + \begin{enumerate}[(i)] + \item + In particular, if $P \in k[t]$ is an arbitrary polynomial, then if $P$ has a root + in a real-closed extension $L$ of $k$, then it has a root in all real-closed extensions of $k$. + + A polynomial with coefficients in an ordered field $(k, \le)$ might not have roots + in any real-closed extensions of $k$. + \item + There is a proof of Sturm's algorithm that does not require $P$ to + be separable. As a consequence \ref{lemma:number-of-roots-in-real-closed-extension} + holds for all $P \in k[t]$, not only the irreducible ones. + \end{enumerate} +\end{bem} + +\end{document}