$\forall0\neq f \in\frac{K[t]}{<m>},~~ \exists f^{-1}$ 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}
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 +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 $L$.
(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 $\geq1$ 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}
$L:K$ 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 $L$.
