diff --git a/galois-theory-notes.pdf b/galois-theory-notes.pdf index 572fd8d..baea76a 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 67b61f0..de55989 100644 --- a/galois-theory-notes.tex +++ b/galois-theory-notes.tex @@ -73,7 +73,7 @@ {\renewcommand\theinnerlemma{#1}\innerlemma} {\endinnerlemma} -\newtheorem{innercor}{Lemma} +\newtheorem{innercor}{Corollary} \newenvironment{cor}[1] {\renewcommand\theinnercor{#1}\innercor} {\endinnercor} @@ -134,6 +134,24 @@ 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} + 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 +194,15 @@ 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} @@ -190,7 +214,184 @@ 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{lemma}{8.18} + Let $q \in L$. The minimal polynomial of $q$ over $K$ \emph{splits} into linear factors over L. +\end{lemma} + +\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} + $L:K$ normal, $K \subseteq M \subseteq L$. Let $\tau$ any $K$-monomorphism $\tau: M \longarrow L$.\\ + 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{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} + +[Chapter 13 is basically a full example. More examples can be found at section \ref{ex:galoisgroups}] + \subsection{Detour: Isomorphism Theorems} \begin{thm}{i.1}(\emph{First Isomorphism Theorem}) \label{1stisothm} @@ -387,6 +588,17 @@ \subsection{Chapter 14} +\begin{defn}{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 @@ -411,7 +623,7 @@ (*): 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: 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. @@ -422,7 +634,7 @@ The series clearly exists, so now we show that the quotients are Abelian, so that $G/N$ is soluble: $$ - \frac{G_{i+1} N}{G_i N} = \frac{G_{i+1}(G_i N)}{G_i N} \stackrel{*}{\cong} + \frac{G_{i+1} N}{G_i N} = \frac{G_{i+1}(G_i N)}{G_i N} \stackrel{(*)}{\cong} \frac{G_{i+1}}{G_{i+1} \cap (G_i N)} \cong \frac{G_{i+1}/G_i}{(G_{i+1} \cap (G_i N))/G_i} $$ @@ -675,6 +887,8 @@ $s(6) = 1+2+3+6 = 12$; 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} +TODO \newpage