(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}
$\forall0\neq f \in\frac{K[t]}{<m>},~~ \exists f^{-1}$ iff $m$ is irreducible in $K[t]$.\\
Then $\frac{K[t]}{<m>}$ 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 $\frac{K[t]}{<m>}$.\\
The isomorphism $\frac{K[t]}{<m>}\longrightarrow K(\alpha)$ 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 $\frac{K[t]}{<m>}$.
\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 $L$.
$L:K$ 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
$$f(t)= k \cdot(t-\alpha_1)\cdot\ldots\cdot(t -\alpha_n)$$
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 $\geq1$ 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$.
(*): 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:
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
$$
the quotients are either
\begin{itemize}
\item$\frac{N_{i+1}}{N_i}$, Abelian
\item$\frac{G_{i+1}}{G_i}$, isomorphic to $\frac{G_{i+1}/N}{G_i/N}$,
which is Abelian.
\end{itemize}
\end{enumerate}
Therefore, the quotients are always Abelian; hence $G$ is soluble.
\end{proof}
\begin{defn}{14.5}
$G$ is \emph{simple} if it's nontrivial and it's only normal subgroups are $1$ and $G$.
\end{defn}
\begin{thm}{14.6}\label{14.6}
A \emph{soluble} group is \emph{simple} iff it is cyclic of prime order.
\end{thm}
\begin{thm}{14.7}\label{14.7}
if $n \geq5$, then $\mathbb{A}_n$ is simple.
\end{thm}
\begin{cor}{14.8}
$\mathbb{S}_n$ is not soluble if $n \geq5$.
\end{cor}
\begin{proof}
if $\mathbb{S}_n$ were soluble, then $\mathbb{A}_n$ would be soluble by Theorem \ref{14.1}(i), and simple by Theorem \ref{14.7}, hence of prime order by Theorem \ref{14.6}.
But observe: $|\mathbb{A}_n|=\frac{1}{2}(n!)$ is not prime if $n \geq5$.\\
Thus $\mathbb{S}_n$ is not soluble if $n \geq5$.
\end{proof}
\begin{lemma}{14.14}\label{14.14}
if $A$ finite and abelian group with $p \biggr\vert |A|$ ($p$ prime), then $A$ has an element of order $p$.
\end{lemma}
\begin{proof}
\begin{enumerate}[i.]
\item if $|A|$ prime and Abelian $\Longrightarrow$ then $A$ is cyclic.
Since $p \vert |A| ~~ \Longrightarrow ~~ \exists! ~ B \subseteq A$ such that $|B|=p$, where $B = <b>$ with $ord(b)=p$. So the lemma is proven.
\item if $|A|$ non-prime:
take $M \subseteq A$ with $|M|=m$, $m$ maximal. Then
\begin{enumerate}[a.]
\item if $p|m ~~ \Longrightarrow ~~ \exists! ~ B'=<b'>,~ b' \in A$ with $|B'|=p$ and $ord(b')=p$.
\item if $p \nmid m$:
Let $b \in A \not M$ and $B=<b>$.\\
Then $MB \supseteq M$, and by maximility must be $MB=A$.
By the 1st Isomorphism Theorem (\ref{1stisothm}),
$$|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 $Z(G)\neq1$.
