Commit fa49c7c8 by rrueger

### 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 ... ...
Markdown is supported
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!