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Unverified Commit fa49c7c8 authored by rrueger's avatar rrueger
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Algebra II: Add Lecture 1

parent a243e3e7
\chapter{Splitting Field Extensions}
\section{Field Automorphisms and Permutations}
\begin{definition}[Field Automorphism, Fixed field]
A \textbf{field automorphism}, often just ``automorphism'', on a field, is a
field isomorphism from a field to itself. Furthermore, if a automorphism
$\sigma$ restricted onto a subfield $F$ is the identity, then we say that $F$
is a \textbf{fixed field} of $\sigma$. Equivalently, we say that the $\sigma$
\textbf{fixes} $F$. Clearly, the set of all field automorphisms form a group.
The automorphism group on a field $E$ is denoted $\Aut (E)$.
\end{definition}
\begin{definition}[Automorphism group of a field extension]
Let $E/F$ be a field extension, then $\Aut(E/F) \subseteq \Aut(E)$ denotes the
set of automorphisms on $E$ that fix $F$.
\end{definition}
\begin{proposition}
Let $F$ be a field, and $f \in F[X]$ with $R = \set{r_1, \ldots, r_n}$ the set
of its roots. Further, let $E = F(R)$ be the splitting field of $f$ over $F$.
Then $\Aut(E/F)$ is isomorphic to some subgroup of $\Sym (R)$.
\end{proposition}
\begin{proof}
With $f(X) := \sum _{i=1} ^k f_i X^i$ and noting that for $\sigma \in
\Aut(E/F)$
%
\begin{align*}
0
\overset{\text{aut.}}{=} \sigma(0)
= \sigma (f(r_i))
= \sum _{j=1} ^k \sigma (f_j r_i^j)
= \sum _{j=1} ^k \sigma (f_j) \sigma(r_i^j)
= \sum _{j=1} ^k f_j {\sigma(r_i)}^j
= f (\sigma (r_i))
\end{align*}
%
we see that $\sigma (r_i)$ is again a root of $f$. Meaning $\sigma (R)
\subseteq R$. As $\sigma$ is an automorphism, it is injective, and since $R$
finite, $\sigma$ is then also surjective onto $R$. Thus, restricted onto $R$,
$\sigma$ is a permutation. This makes the map $\Phi \colon \Aut(E/F) \to
\Sym(R); \sigma \to \sigma |_R$ a group homomorphism. As the image of a group
under a group homomorphism is a subgroup, this concludes our proof. In fact,
$\Phi$ is also injective: as $E = F(R)$, the behaviour of $\sigma$ on $E$ is
uniquely determined by action on $R$.
\end{proof}
\begin{corollary}
If $\sigma \in \Aut(E/F)$ as in the setting above, then $\sigma$ acts on $E$
by permuting the roots of $f$.
\end{corollary}
\begin{remark}
It is worth noting that not every permutation of roots induces an
automorphism. See Example~\ref{ex:not_every_perm}.
\end{remark}
\section{Cardano's Formula}
In the lecture, we derived Cardano's formula for cubic polynomials. The details
won't be included here.
......@@ -14,7 +14,7 @@ automorphisms is bounded by the dimension of the extension: 2.
% @todo: Adjust colloquial tone.
% @todo: Restructure sentences, somewhat confusing.
\subsection{$\Q(\sqrt[3]{5})$ over $\Q$}
\subsection{$\Q(\sqrt[3]{5})$ over $\Q$}\label{ex:not_every_perm}
We claim that the only automorphism that fixes $\Q$ is $\set{\id}$. Why? Well,
a field isomorphism is uniquely defined by the image of a basis. The minimal
......
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