The isomorphism $\frac{K[t]}{<m>}\longrightarrow K(\alpha)$ 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 $L:K$.
$L:K$ is radical if $L=K(\alpha_1, \ldots, \alpha_m)$ where for each $j=1, \ldots, m$, $\exists~ n_j$ such that $\alpha_j^{n_j}\in K(\alpha_1, \ldots, \alpha_{j-1})~~(j\geq1)$
\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
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 $n \geq5$, then $\mathbb{A}_n$ is simple.
\end{thm}
\begin{cor}{14.8}
$\mathbb{S}_n$ is not soluble if $n \geq5$.
\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 $n \geq5$.\\
Thus $\mathbb{S}_n$ is not soluble if $n \geq5$.
\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 $B = <b>$ 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 $p|m ~~ \Longrightarrow ~~ \exists! ~ B'=<b'>,~ b' \in A$ with $|B'|=p$ and $ord(b')=p$.
\item if $p \nmid m$:
Let $b \in A \not M$ and $B=<b>$.\\
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 $j \geq2$.
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 $x\neq1$, so $Z(G)\neq1$.
