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Unverified Commit bad3a974 authored by rrueger's avatar rrueger
Browse files

Algebra II: Lecture 11. Add proof of primitve element theorem

parent 57ff244f
......@@ -7,7 +7,7 @@ extension $E/F$ not Galois, but $E$ is contained in some field $\hat{E}$ such
that $\hat{E}/F$ \textit{is} Galois we can apply some known results to arrive at
the
\begin{theorem}[Primitve Element Theorem]
\begin{theorem}[Primitve Element Theorem]\label{thm:pet}
Let $E/F$ be a finite extension, with F a field with characteristic 0. Then
there exists some $\alpha$ in $E$ such that $E = F(\alpha)$.
\end{theorem}
......@@ -58,7 +58,7 @@ generates the ideal that is $\ker (\chi)$.
\section{Characteristic 0}
\begin{theorem}
\begin{theorem}\label{thm:existence_of_galois}
Consider a finite field extension $E/F$ and suppose $F$ has characteristic 0.
Then there exists some $\hat{E}$ that contains $E$ with $\hat{E}/F$ being
Galois.
......@@ -126,8 +126,66 @@ $F(r) = E$ is also fixed by $\sigma$ and thus not \textbf{only} $F$ is fixed,
meaning the extension is not Galois by the very first definition of a Galois
extension.
\section{Primitive element theorem}
\noindent
We are now in a position to prove the primitive element theorem~\ref{thm:pet}.
\begin{proof}
To that end, let $E/F$ be a finite extension $\Char (F) = 0$. As the
extension is finite, we assume $E = F(\alpha, \beta)$. Cases in which more
generators are needed are them proven by induction. From
Theorem~\ref{thm:existence_of_galois}, we have some $\hat E$ with $E \subseteq
\hat E$ and $\hat E /F$ Galois. By the Galois correspondence there can only be
finitely many subextensions $\hat E / K / F$, hence only finitely many
subextensions $E / K /F$. We can now consider the extensions over the
subfields
%
\begin{figure}[H]
\centering
\begin{tikzpicture}
\matrix(m)[%
matrix of math nodes,
row sep=5em,
column sep=3em,
minimum width=1em,
]{%
& & E & & \\
F(\alpha + \beta) & F(\alpha + 2\beta) & \cdots & F(\alpha + k \beta) & \cdots \\
& & F & & \\
};
\path[-stealth]
(m-2-1) edge [-] node [above] {$\quad\cong$} (m-2-2)
(m-2-2) edge [-] node [above] {$\cong$} (m-2-3)
(m-2-3) edge [-] node [above] {$\cong$} (m-2-4)
(m-2-4) edge [-] node [above] {$\cong\quad$} (m-2-5)
(m-3-3) edge [-] (m-2-1)
(m-3-3) edge [-] (m-2-2)
(m-3-3) edge [-] (m-2-3)
(m-3-3) edge [-] (m-2-4)
(m-3-3) edge [-] (m-2-5)
(m-1-3) edge [-] (m-2-1)
(m-1-3) edge [-] (m-2-2)
(m-1-3) edge [-] (m-2-3)
(m-1-3) edge [-] (m-2-4)
(m-1-3) edge [-] (m-2-5)
;
\end{tikzpicture}
\end{figure}
%
\noindent
for every $k \in \Z$. This generates infinitely many extensions, so they can't
all be distinct. There is some $F(\alpha + m \beta) = F(\alpha + k\beta)$. So
$\alpha + m\beta, \alpha + k\beta \in F(\alpha + m\beta)$, and thus their
difference must be too $(m-k) \beta \in F(\alpha + m\beta)$. As $F$ has
characteristic $0$, the prime field is isomorphic to $\Q$, and thus $(m-k) \in
F$. We can conclude $\beta \in F(\alpha + m\beta)$ and thus $\alpha \in
F(\alpha + m\beta)$. Finally, $E = F(\alpha, \beta) = F(\alpha + m \beta)$.
\end{proof}
\section{Skew Fields and Wedderburn's Theorem}
\begin{definition}[Skew Field, Division Algebra]
......
......@@ -252,6 +252,7 @@
% }}}
% Algebra {{{
\DeclareMathOperator{\Char}{Char}
\DeclareMathOperator{\Obj}{Obj}
\DeclareMathOperator{\Orb}{Orb}
\DeclareMathOperator{\Stab}{Stab}
......
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