### 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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