diff --git a/galois-theory-notes.pdf b/galois-theory-notes.pdf index baea76a..666843f 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 de55989..d14df2c 100644 --- a/galois-theory-notes.tex +++ b/galois-theory-notes.tex @@ -73,6 +73,11 @@ {\renewcommand\theinnerlemma{#1}\innerlemma} {\endinnerlemma} +\newtheorem{innerprop}{Proposition} +\newenvironment{prop}[1] +{\renewcommand\theinnerprop{#1}\innerprop} +{\endinnerprop} + \newtheorem{innercor}{Corollary} \newenvironment{cor}[1] {\renewcommand\theinnercor{#1}\innercor} @@ -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} @@ -144,7 +154,7 @@ The isomorphism $\frac{K[t]}{} \longrightarrow K(\alpha)$ can be chosen to map $t$ to $\alpha$. \end{thm} -\begin{cor}{5.13} +\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} @@ -231,10 +241,20 @@ \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 $$f(t) = k \cdot (t- \alpha_1) \cdot \ldots \cdot (t - \alpha_n)$$ @@ -276,8 +296,8 @@ 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 $\tau: M \longarrow L$.\\ +\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} @@ -308,6 +328,17 @@ 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 $\tau: L \longrightarrow M$.\\ @@ -588,7 +619,7 @@ \subsection{Chapter 14} -\begin{defn}{14.1} +\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 @@ -656,7 +687,7 @@ Consider the series of $G$ given by combining the two previous series: $$ - 1 &= N_0 \triangleleft N_1 \triangleleft \ldots \triangleleft N_r = N = G_0 \triangleleft G_1 \triangleleft \ldots \triangleleft G_r = G + 1 = N_0 \triangleleft N_1 \triangleleft \ldots \triangleleft N_r = N = G_0 \triangleleft G_1 \triangleleft \ldots \triangleleft G_r = G $$ the quotients are either \begin{itemize} @@ -668,6 +699,139 @@ 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 \section{Tools}