diff --git a/galois-theory-notes.pdf b/galois-theory-notes.pdf index 60b6f90..69bf452 100644 Binary files a/galois-theory-notes.pdf and b/galois-theory-notes.pdf differ diff --git a/galois-theory-notes.tex b/galois-theory-notes.tex index 740342d..40155ae 100644 --- a/galois-theory-notes.tex +++ b/galois-theory-notes.tex @@ -73,7 +73,12 @@ {\renewcommand\theinnerlemma{#1}\innerlemma} {\endinnerlemma} -\newtheorem{innercor}{Lemma} +\newtheorem{innerprop}{Proposition} +\newenvironment{prop}[1] +{\renewcommand\theinnerprop{#1}\innerprop} +{\endinnerprop} + +\newtheorem{innercor}{Corollary} \newenvironment{cor}[1] {\renewcommand\theinnercor{#1}\innercor} {\endinnercor} @@ -83,6 +88,11 @@ {\renewcommand\theinnereg{#1}\innereg} {\endinnereg} +\newtheorem{innerex}{Exercise} +\newenvironment{ex}[1] +{\renewcommand\theinnerex{#1}\innerex} +{\endinnerex} + \title{Galois Theory notes} \author{arnaucube} @@ -103,7 +113,7 @@ \tableofcontents \section{Galois Theory notes} -\subsection{Chapters 4-6} +\subsection{Chapters 4-12} (Definitions, theorems, lemmas, corollaries and examples enumeration follows from Ian Stewart's book \cite{ianstewart}). \begin{defn}{4.10} @@ -134,6 +144,24 @@ Therefore, $r-s=0$, so $r=s$, proving uniqueness. \end{proof} +\begin{thm}{5.10} + $\forall 0 \neq f \in \frac{K[t]}{},~~ \exists f^{-1}$ iff $m$ is irreducible in $K[t]$.\\ + Then $\frac{K[t]}{}$ is a field. +\end{thm} +\begin{thm}{5.12} \label{5.12} + Let $K(\alpha):K$ simple algebraic extension, let $m$ minimal polynomial of $\alpha$ over $K$.\\ + $K(\alpha):K$ is isomorphic to $\frac{K[t]}{}$.\\ + The isomorphism $\frac{K[t]}{} \longrightarrow K(\alpha)$ can be chosen to map $t$ to $\alpha$. + +\end{thm} +\begin{cor}{5.13} \label{5.13} + Let $K(\alpha):K$ and $K(\beta):K$ be simple algebraic extensions.\\ + If $\alpha,~ \beta$ have same minimal polynomial $m$ over $K$, then the two extensions are isomorphic, and the isomorphism of the larger fields map $\alpha$ to $\beta$. +\end{cor} +\begin{proof} + By \ref{5.12}, both extensions are isomorphic to $\frac{K[t]}{}$. +\end{proof} + \begin{lemma}{5.14} Let $K(\alpha):K$ be a simple algebraic extension, let $m$ be the minimal polynomial of $\alpha$ over $K$, let $\delta m =n$. @@ -176,9 +204,15 @@ So the elements $x_i y_j$ are linearly independent over $K$. \item prove that $x_i y_j$ span $M$ over $K$:\\ - Any $x \in M$ can be written $x=\sum_j \lambda_j y_j$ for $\lambda_j \in L$, because $y_j$ spans $M$ over $L$. - Similarly, $\forall j\in J,~ \lambda_j = \sum_i \lambda_{ij} x_i y_j$ for $\lambda_{ij} \in K$.\\ - Putting the pieces together, $x=\sum_{ij} \lambda_{ij} x_i y_j$ as required. + Any $x \in M$ can be written + $$x=\sum_j \lambda_j y_j$$ + for $\lambda_j \in L$, because $y_j$ spans $M$ over $L$. + Similarly, + $$\forall j\in J,~ \lambda_j = \sum_i \lambda_{ij} x_i y_j$$ + for $\lambda_{ij} \in K$.\\ + Putting the pieces together, + $$x=\sum_{ij} \lambda_{ij} x_i y_j$$ + as required. \end{enumerate} \end{proof} @@ -190,10 +224,210 @@ From \ref{shorttowerlaw}. \end{proof} -[...] TODO: pending to add key parts up to Chapter 15. +\begin{thm}{6.7} + if $K(\alpha):K$ + \begin{itemize} + \item transcendental $\Longrightarrow~~[K(\alpha):K] = \inf$ + \item algebraic $\Longrightarrow~~[K(\alpha):K] = \delta m$ + \end{itemize} + (where $m$ is the minimal polynomial of $\alpha$ over $K$). +\end{thm} + +\begin{defn}{8.1} + $L:K$, a \emph{$K$-automorphism} of $L$ is an automorphism $\alpha$ of $L$ such that $\alpha(k)=k ~~ \forall k \in K$.\\ + ie. $\alpha$ \emph{fixes} $k$. +\end{defn} + +\begin{thm}{8.2, 8.3} + The set of all $K$-automorphisms of $L$ forms a group, $\Gamma(L:K)$, the Galois group of $L:K$. +\end{thm} + +\begin{defn}{8.12}(Radical Extension) \label{8.12} + $L:K$ is radical if $L=K(\alpha_1, \ldots, \alpha_m)$ where for each $j=1, \ldots, m$, $\exists~ n_j$ such that $\alpha_j^{n_j} \in K(\alpha_1, \ldots, \alpha_{j-1})~~(j\geq 1)$ +\end{defn} + +\begin{lemma}{8.18} + Let $q \in L$. The minimal polynomial of $q$ over $K$ \emph{splits} into linear factors over L. +\end{lemma} + +\begin{ex}{E.8.7} + TODO +\end{ex} + + +\begin{defn}{9.1} + For $K \subseteq \mathbb{C}$, and $f \in K[t]$, $f$ \emph{splits} over $K$ if it can be expressed as a product of linear factors + $$f(t) = k \cdot (t- \alpha_1) \cdot \ldots \cdot (t - \alpha_n)$$ + where $k, \alpha_i \in K$. + + $\Longrightarrow$ (Thm 9.3) if $f$ splits over $\Sigma$, $\Sigma$ is the \emph{splitting field}.\\ + If $K \subseteq \Sigma' \subseteq \Sigma$ and $f$ splits over $\Sigma'$, then $\Sigma' = \Sigma$. +\end{defn} + +\begin{thm}{9.6} \label{9.6} + TODO +\end{thm} + +\begin{defn}{9.8} + $L:K$ is \emph{normal} if every irreducible polynomial $f \in K[t]$ that has at least one zero in $L$, splits in $L$. +\end{defn} + +\begin{thm}{9.9} \label{9.9} + TODO +\end{thm} + +\begin{thm}{9.10} + An irreducible polynomial $f \in K[t]$ ($K \subseteq \mathbb{C}$) is \emph{separable over} $K$ if it has simple zeros in $\mathbb{C}$, or equivelently, simple zeros in its splitting field. +\end{thm} + +\begin{lemma}{9.13} + $f \in K[t]$ with splitting field $\Sigma$. $f$ has multiple zeros (in $\Sigma$ or $\mathbb{C}$) iff $f$ and $Df$ have a common factor of degree $\geq 1$ in $\Sigma[t]$.\\ + More details at Rolle's theorem (\ref{rolle}) section. +\end{lemma} + +\begin{thm}{10.5} \label{10.5} + $|\Gamma(K:K_0)| = [K:K_0]$, where $K_0$ is the fixed field of $\Gamma(K:K_0)$. +\end{thm} + + +\begin{defn}{11.1} + $K \subseteq L$, $K \subseteq L$. A $K$-monomorphism of $M$ into $L$ is a field monomorphism + $$\phi: M \longrightarrow L$$ + such that $\phi(k)=k ~~ \forall k \in K$. +\end{defn} + +\begin{thm}{11.3} \label{11.3} + $L:K$ normal, $K \subseteq M \subseteq L$. Let $\tau$ any $K$-monomorphism $\tau: M \longrightarrow L$.\\ + Then, $\exists$ a $K$-automorphism $\sigma$ of $L$ such that $\sigma\biggr\vert_M=\tau$. +\end{thm} +\begin{proof} + $L:K$ normal $\Longrightarrow$ by Thm \ref{9.9}, $L$ splitting field for some poly $f \in K[t]$. + + Hence, $L$ is splitting field over $M$ for $f$ and over $\tau(M)$ for $\tau(f)$. + + Since $\tau \biggr\vert_K$ is the identity, $\tau(f)=f$. + + \vspace{0.5cm} + We have + + \begin{tikzpicture}[node distance=1.5cm, auto] + \node (M) {$M$}; + \node (L) [right of=M] {$L$}; + \node (t) [below of=M] {$\tau(M)$}; + \node (L2) [below of=L] {$L$}; + + \draw[->] (M) to node {$ $} (L); + \draw[->] (M) to node {$\tau$} (t); + \draw[->] (t) to node {$ $} (L2); + \draw[->] (L) to node {$ $} (L2); + \end{tikzpicture} + + with $\sigma$ yet to be formed. + + By Theorem \ref{9.6}, $\exists$ isomorphism $\sigma: L \longrightarrow L$ such that $\sigma \biggr\vert_M = \tau$.\\ + Therefore, $\sigma$ is an automorphism of $L$, and since $\sigma\biggr\vert_K = \tau\biggr\vert_K=id$, $\sigma$ is a $K$-automorphism of $L$. +\end{proof} + +\begin{prop}{11.4} \label{11.4} + $L:K$ finite normal, $\alpha,~\beta$ are zeros in $L$ of the irreducible polynomial $p \in K[t]$. + + Then, $\exists$ a $K$-automorphism $\sigma$ of $L$ such that $\sigma(\alpha)=\beta$. +\end{prop} +\begin{proof} + By Corollary \ref{5.13}, $\exists$ isomorphism $\tau: K(\alpha) \longrightarrow K(\beta)$ such that $\tau \biggr\vert_K$ is the identity, and $\tau(\alpha)=\beta$. + + By Theorem \ref{11.3}, $\tau$ extends to a $K$-automorphism $\sigma$ of $L$. +\end{proof} + +\begin{lemma}{11.8} \label{11.8} + $K \subseteq L \subseteq N \subseteq M$, $L:K$ finite, $N$ normal closure of $L:K$.\\ + Let $\tau$ any $K$-monomorphism $\tau: L \longrightarrow M$.\\ + Then $\tau(L) \subseteq N$. +\end{lemma} +\begin{proof} + $\alpha \in L$, $m$ minimal polynomial of $\alpha$ over $K$.\\ + $\Longrightarrow ~~ m(\alpha)=0$, so $\tau(m(\alpha))=0$ + + (since $\tau$ is a $K$-automorphism, ie. maps the zeros of $m(t)$).\\ + + Since $\tau$ is a $K$-monomorphism, $\tau(m(\alpha))=m(\tau(\alpha))=0$ + + $\Longrightarrow~~ \tau(\alpha)$ is a zero of $m$.\\ + + Therefore, $\tau(\alpha)$ lies in $N$, since $N:K$ is normal.\\ + Henceforth, $\tau(L) \subseteq N$. +\end{proof} + +\begin{thm}{11.9} + The following are equivalent: + \begin{enumerate} + \item $L:K$ normal + \item $\exists$ finite normal extension $N$ of $K$ containing $L$,\\ + such that every $K$-monomorphism $\tau: L \longrightarrow N$ is a $K$-automorphism of $L$. + \item for every finite extension $M$ of $K$ containing $L$,\\ + every $K$-monomorphism $\tau: L \longrightarrow M$ is a $K$-automorphism of $L$. + \end{enumerate} +\end{thm} + +\begin{thm}{11.10} + $[L:N]=1,~ N$ normal closure of $L:K$. Then, + + $\exists~ n~ K$-monomorphisms $L \longrightarrow N$.\\ + (the ones proven by Lemma \ref{11.8}). +\end{thm} + +\begin{cor}{11.11} \label{11.11} + $|\Gamma(L:K)| = [L:K]$ (if $L:K$ is normal). + + ie. there are precisely $[L:K]$ distinct $K$-automorphisms of $L$. +\end{cor} + +\begin{thm}{11.12} + $\Gamma(L:K) = G$. If $L:K$ normal, then $K$ is the fixed field of $G$. +\end{thm} +\begin{proof} + let $K_0$ be the fixed field of $G$. Let $[L:K]=n$.\\ + By \ref{11.11}, $|G| = [L:K] = n$.\\ + By \ref{10.5}, $[L:K_0]=n$ ($K_0$ fixed field).\\ + Since $K \subseteq K_0$, we must have $K=K_0$. + + $\Longrightarrow$ thus $K$ is the fixed field of $G$. +\end{proof} + +\begin{thm}{11.14} + if $L$ any field, $G$ any finite group of automorphisms of $L$, and $K$ its fixed field, + + then $L:K$ is \emph{finite} and \emph{normal}, with Galois group $G$. +\end{thm} + + +\begin{thm}{12.2}(Fundamental Theorem of Galois Theory) + if $L:K$ finite and normal inside $\mathbb{C}$, with $\Gamma(L:K)=G$, then: + \begin{enumerate} + \item $|\Gamma(L:K)| = [L:K]$ + (by Corollary \ref{11.11}) + \item the maps * and $\dagger$ are mutual inverses, and setup an order-reversing one-to-one correspondence between $\mathcal{F}$ and $\mathcal{G}$. + \item if $M$ an intermediate field, then + $$[L:M] = |M^*|~~~~~~~ [M:K]=\frac{|G|}{|M^*|}$$ + \item for $M$ an intermediate field, $M:K$ normal iff + $$\underbrace{\Gamma(M:K)}_{=M^*} \lhd \underbrace{\Gamma(L:K)}_{=G}$$ + \item for $M$ intermediate, if $M:K$ normal, then + $$\Gamma(M:K) \cong \frac{G}{M^*}$$ + ie. + $$\Gamma(M:K) \cong \frac{\Gamma(L:K)}{\Gamma(L:M)}$$ + \end{enumerate} +\end{thm} +\begin{proof} + TODO +\end{proof} + +\subsection{Chapter 13 - Full example} + +(Chapter 13 is basically a full example. More examples can be found at section \ref{ex:galoisgroups}) + \subsection{Detour: Isomorphism Theorems} -\begin{thm}{}(\emph{First Isomorphism Theorem}) \label{1stisothm} +\begin{thm}{i.1}(\emph{First Isomorphism Theorem}) \label{1stisothm} \begin{minipage}{0.75\textwidth} If $\psi: G \longrightarrow H$ a group homomorphism, then $ker(\psi) \triangleleft G$.\\ @@ -263,7 +497,7 @@ \end{proof} -\begin{thm}{}(\emph{Second Isomorphism Theorem}) \label{2ndisothm} +\begin{thm}{i.2}(\emph{Second Isomorphism Theorem}) \label{2ndisothm} Let $H \subseteq G$, $N \triangleleft G$. Then \begin{enumerate}[i.] \item $HN \subseteq G$ @@ -329,7 +563,7 @@ \end{proof} -\begin{thm}{}(\emph{Third Isomorphism Theorem}) \label{2ndisothm}\\ +\begin{thm}{i.3}(\emph{Third Isomorphism Theorem}) \label{2ndisothm}\\ Let $H \subseteq K$ and $K \triangleleft G,~ H \triangleleft G$.\\ Then $\frac{K}{H} \triangleleft \frac{G}{H}$ @@ -387,6 +621,218 @@ \subsection{Chapter 14} +\begin{defn}{14.1} \label{14.1} + a group $G$ is soluble if it has a finite series of subgroups + $$1=G_0 \subseteq G_1 \subseteq \ldots \subseteq G_n = G$$ + such that + \begin{enumerate}[i.] + \item $G_i \lhd G_{i+1}$ for $i=0,\ldots,n-1$ + \item $\frac{G_{i+1}}{G_{i+1}}$ is Abelian for for $i=0,\ldots,n-1$ + \end{enumerate} + + (Note: $G_i \lhd G_{i+1} \lhd G_{i+2}$ does not imply $G_i \lhd G_{i+2}$) +\end{defn} + +\begin{thm}{14.4} + $H \subseteq G,~~ N \triangleleft G$, then + \begin{enumerate} + \item if $G$ soluble $\Longrightarrow H$ soluble + \item if $G$ soluble $\Longrightarrow G/N$ soluble + \item if $N$ and $G/N$ soluble $\Longrightarrow G$ soluble + \end{enumerate} +\end{thm} +\begin{proof} + \begin{enumerate} + \item Since $G$ soluble, by definition: $\exists~~ 1=G_0 \triangleleft G_1 \triangleleft \ldots \triangleleft G_r = G$ with Abelian quotients $\frac{G_{i+1}}{G_i}$. + + Let $H_i = G_i \cap H$, then $H$ has a series $1=H_0 \triangleleft H_1 + \triangleleft \ldots \triangleleft H_r = H$, next we show that the + quotients $\frac{H_{i+1}}{H_i}$ are Abelian (so that H is soluble): + + $$\frac{H_{i+1}}{H_i} = \frac{G_{i+1} \cap H}{G_i \cap H} \stackrel{(*)}{=} \frac{G_{i+1} \cap H}{G_i \cap (G_{i+1}\cap H)} + \stackrel{(**)}{\cong} \frac{G_i(G_{i+1} \cap H)}{G_i} \subseteq \frac{G_{i+1}}{G_i} + $$ + + (*): to see why, $H_i = G_i \cap H = G_i \cap H_i = G_i \cap H_{i+1} = G_i \cap (G_{i+1} \cap H)$. + (**): by the 2nd Isomorphism Theorem (\ref{2ndisothm}). + + [TODO: diagram of subgroups] % TODO % + + Notice that $\frac{G_{i+1}}{G_i}$ is Abelian, thus the left-hand-side of the congruence is also Abelian. + Therefore, $\frac{H_{i+1}}{H_i}$ is Abelian, thus $H$ is soluble. + + \item For $G/N$ to be soluble, (by definition) it would have the series $\frac{N}{N} = G_0 \frac{N}{N} \triangleleft G_1 \frac{N}{N} \triangleleft \ldots \triangleleft G_r \frac{N}{N} = \frac{G}{N}$, + and any quotient being $\frac{G_{i+1}\frac{N}{N}}{G_i \frac{N}{N}}$. + + The series clearly exists, so now we show that the quotients are Abelian, so that $G/N$ is soluble: + + $$ + \frac{G_{i+1} N}{G_i N} = \frac{G_{i+1}(G_i N)}{G_i N} \stackrel{(*)}{\cong} + \frac{G_{i+1}}{G_{i+1} \cap (G_i N)} \cong \frac{G_{i+1}/G_i}{(G_{i+1} + \cap (G_i N))/G_i} + $$ + + (*): by the 2nd Isomorphism Theorem (\ref{2ndisothm}). + + The last quotient is a quotient of the Abelian group $G_{i+1}/G_i$, so it is Abelian. + + Hence, $\frac{G_{i+1}N}{G_i N}$ is also Abelian; so $\frac{G}{N}$ is soluble. + + \item + By the definition of $N$ and $G/N$ being soluble, + \begin{align*} + N \text{soluble} \Longrightarrow~~ 1 &= N_0 \triangleleft N_1 \triangleleft \ldots \triangleleft N_r = N\\ + G/N \text{soluble} \Longrightarrow~~ 1= \frac{N}{N} &= \frac{G_0}{N} \triangleleft \frac{G_1}{N} \triangleleft \ldots \triangleleft \frac{G_r}{N} = \frac{G}{N} + \end{align*} + both with Abelian quotients. + + Consider the series of $G$ given by combining the two previous series: + + $$ + 1 = N_0 \triangleleft N_1 \triangleleft \ldots \triangleleft N_r = N = G_0 \triangleleft G_1 \triangleleft \ldots \triangleleft G_r = G + $$ + the quotients are either + \begin{itemize} + \item $\frac{N_{i+1}}{N_i}$, Abelian + \item $\frac{G_{i+1}}{G_i}$, isomorphic to $\frac{G_{i+1}/N}{G_i/N}$, + which is Abelian. + \end{itemize} + \end{enumerate} + Therefore, the quotients are always Abelian; hence $G$ is soluble. +\end{proof} + + +\begin{defn}{14.5} + $G$ is \emph{simple} if it's nontrivial and it's only normal subgroups are $1$ and $G$. +\end{defn} + +\begin{thm}{14.6} \label{14.6} + A \emph{soluble} group is \emph{simple} iff it is cyclic of prime order. +\end{thm} +\begin{thm}{14.7} \label{14.7} + if $n \geq 5$, then $\mathbb{A}_n$ is simple. +\end{thm} +\begin{cor}{14.8} + $\mathbb{S}_n$ is not soluble if $n \geq 5$. +\end{cor} +\begin{proof} + if $\mathbb{S}_n$ were soluble, then $\mathbb{A}_n$ would be soluble by Theorem \ref{14.1}(i), and simple by Theorem \ref{14.7}, hence of prime order by Theorem \ref{14.6}. + + But observe: $|\mathbb{A}_n|=\frac{1}{2}(n!)$ is not prime if $n \geq 5$.\\ + Thus $\mathbb{S}_n$ is not soluble if $n \geq 5$. +\end{proof} + +\begin{lemma}{14.14} \label{14.14} + if $A$ finite and abelian group with $p \biggr\vert |A|$ ($p$ prime), then $A$ has an element of order $p$. +\end{lemma} +\begin{proof} + \begin{enumerate}[i.] + \item if $|A|$ prime and Abelian $\Longrightarrow$ then $A$ is cyclic. + + Since $p \vert |A| ~~ \Longrightarrow ~~ \exists! ~ B \subseteq A$ such that $|B|=p$, where $B = $ with $ord(b)=p$. So the lemma is proven. + + \item if $|A|$ non-prime: + + take $M \subseteq A$ with $|M|=m$, $m$ maximal. Then + + \begin{enumerate}[a.] + \item if $p|m ~~ \Longrightarrow ~~ \exists! ~ B'=,~ b' \in A$ with $|B'|=p$ and $ord(b')=p$. + \item if $p \nmid m$: + Let $b \in A \not M$ and $B=$.\\ + Then $MB \supseteq M$, and by maximility must be $MB=A$. + + By the 1st Isomorphism Theorem (\ref{1stisothm}), + $$|A| = |MB| = \frac{|M| \cdot |B|}{|M \cap B|}$$ + both $|A|$ and $|B|$ are divisible by $p$ (but recall that $p \nmid m=|M|$), since $B$ is cyclic and $p \vert |B|$ + + $\Longrightarrow$ thus, $B$ has an element of order $p$. + + So, if $|B|=r$, and $p|r ~~ \Longrightarrow~~ ord(b^{r/p}) = p$.\\ + + Hence, in all cases i, ii.a, ii.b, $A$ contains an element of order $p$. + \end{enumerate} + \end{enumerate} +\end{proof} + +\begin{thm}{14.15}(Cauchy's Theorem) + if $p \biggr\vert |G|$ ($p$ prime), then $\exists ~ x \in G$ such that $ord(x)=p$. +\end{thm} +\begin{proof} + (induction on $|G|$) + + For $|G|=1,2,3$, trivial. + + Induction step: class equation + $$|G|=1+|C_2|+ \ldots + |C_r|$$ + since $p \biggr\vert |G|$, must have $p \nmid |C_j|$ for some $j \geq 2$. + + If $x \in C_j ~~\Longrightarrow p \biggr\vert |C_G(x)|$ (since $|C_j|=|G|/|C_G(x)|$, recall $p\biggr\vert |G|$). + + \begin{enumerate}[i.] + \item if $C_G(x) \neq G$:\\ + (by induction) since $p \biggr\vert |C_G(x)|$, + + $\exists a \in C_G(x)$ with $ord(a)=p$, and $a \in G$ (since $C_G(x) \subset G$). + \item otherwise, $C_G(x)=G$:\\ + implies $x \in Z(G)$, by choice $x\neq 1$, so $Z(G)\neq 1$. + + Then either + \begin{enumerate}[I.] + \item $p \biggr\vert |Z(G)| ~~ \longrightarrow$ Abelian case, Lemma \ref{14.14}. + \item $p \not\biggr\vert |Z(G)|$: + by induction, $\exists x \in G$ such that $\hat{x} \in G/Z(G)$, with $ord(\hat{x})=p$. (where $\hat{x}$ is the image of $x$). + + $\Longrightarrow~~ x^p \in Z(G)$, but $x \not\in Z(G)$. + + Let $X=$, cyclic.\\ + $XZ(G)$ is Abelian, and $p \biggr\vert |XZ(G))|$\\ + $\Longrightarrow$ by Lemma \ref{14.14}, it has an element of order $p$, and this element belongs to $G$. + \end{enumerate} + + \end{enumerate} +\end{proof} + +\begin{defn}{15.1}(Soluble by radicals) + let $f \in K[t],~ K \subseteq \mathbb{C}$, and $\Sigma$ a splitting field of $f$ over $K$. + + $f$ is \emph{soluble by radicals} if\\ + $\exists$ a field $M$ with $\Sigma \subseteq M$ such that $M:K$ is a radical extension (\ref{8.12}). + + Note: not required $\Sigma:K$ to be radical. +\end{defn} + +\begin{lemma}{15.3} + $L:K$ radical extension $\mathbb{C}$, and $M$ normal enclosure of $L:K$, then $M:K$ is radical. +\end{lemma} +\begin{proof} + let $L=K(\alpha_1, \ldots, \alpha_r)$ with $\alpha_i^{n_i} \in K(\alpha_1, \ldots, \alpha_{j-1})$ (by definition of $L:K$ being a radical extension). + + Let $f_i$ be the minimal polynimal of $\alpha_i$ over $K$. + + Then, $M \supseteq L$ is spliting field of $\Prod_{i=1}^r f_i$, since $M$ is normal enclosure of $L:K$. + + For every zero $\beta_{ij}$ of $f_i$ in $M$,\\ + $\exists$ an isomorphism $\sigma: K(\alpha_i) \longrightarrow K(\beta_{ij})$ by Corollary \ref{5.13}. + + By Proposition \ref{11.4}, since $K(\alpha_i),~K(\beta_{ij}) \subset M$, + $\sigma$ extends to a $K$-automorphism + $$\tau: M \longrightarrow M$$ + since $M$ is splitting field (ie. contains the zeros of $\Prod f_i$). + + Since $\alpha$ is a member of radical sequence for a subfield of $M$, so it is $\beta_{ij}$. + + By combining the sequences for $M$, $M:K$ is a radical extension. +\end{proof} + + +\vspace{0.5cm} + +The next two lemmas show that certain Galois groups are Abelian. + +\begin{lemma}{15.4} +\end{lemma} +\begin{proof} +\end{proof} \newpage @@ -607,7 +1053,54 @@ $s(6) = 1+2+3+6 = 12$; henceforth, the total amount of subgroups is $d(n)+s(n) = \vspace{0.3cm} For $n \geq 3, ~~\mathbb{D}_n \subseteq \mathbb{S}_n$ (subgroup of the Symmetry group). +\subsection{Rolle's theorem} \label{rolle} + +\begin{thm}{}(Rolle's Theorem) + if a real-valued function $f$ is + \begin{itemize} + \item \emph{continuous} on a proper closed interval $[a,b]$ + \item \emph{differentiable} on the open interval $(a,b)$ + \item $f(a)=f(b)$ + \end{itemize} + then, $\exists$ at least one $c$ in $(a,b)$ such that $Df(c)=0$. +\end{thm} +\begin{proof} (proof source: cue math website) + Notice that when $Df(x_i)=0$ occours, is a maximum or minimum (extrema) value of $f$. + $\Longrightarrow$ if a function is continuous, it is guaranteed to have both a maximum and a minimum point in the interval.\\ + + Two possibilities: + \begin{enumerate}[i.] + \item $f$ is constant\\ + $\Longrightarrow$ ie. a horizontal line ($f(a)=f(b)$), ie. no slope $\Longrightarrow$ $Df=0$ everywhere in $[a,b]$. + \item $f$ is not constant:\\ + since $f$ not constant, must change directions in ordder to start and end at the same $y$-value ($f(a)=f(b)$).\\ + Thus at some point between $a$ and $b$ it will either have a minimum, maximum or both. + \begin{enumerate}[a.] + \item does the maximum occour at a point where $Df > 0$?\\ + No, because if $Df > 0$, then $f$ is increasing, but it can not increase since we're at its maximum point. + \item does the maximum occour at a point where $Df < 0$?\\ + No, because if $Df < 0$, then $f$ is deccreasing, which means that at our left it was larger, but we're at a maximum point, so a contradiction. + \end{enumerate} + Same with minimus.\\ + $\Longrightarrow$ Hence, since $Df \nless 0$ and $Df \ngtr 0$, and $Df$ exists, then $Df=0$. + \end{enumerate} + \\ + Thus, $f$ must have extrema (either max or min or both), and at that extrema $Df$ must be zero. +\end{proof} + +Consequence of Rolle's Theorem: + +\begin{cor}{}(Zero separation property) + Between any two distinct consecutive zeros of $f$, there lies at least one zero of $Df$. +\end{cor} +\begin{eg}{} + If $f(t)$ has zeros $t_1, t_2$, with $t_1 < t_2$, and $f$ is derivable, then by Rolle'z theorem: + + $\exists c \in (t_1, t_2)$ such that $Df(c)=0$. + + Hence, the zeros of $Df$ \emph{separate} the zeros of $f$. +\end{eg} \newpage