... ... @@ -15,46 +15,48 @@ the However, before we prove this, we will take a step back and investigate in which situations such an $\hat{E}$ exists. \section{Characteristic 0} \begin{siderules} \begin{definition}[Reminder Algebra I, Characteristic] The characteristic, $p$, of a field is given by the number of times the multiplicative neutral element must be added to itself before we arrive at the additive neutral element. If no such $p$ exists, we say the field has characteristic 0. \begin{align*} \underset{p \text{~times}}{\underbrace{1+1+1+\cdots}} = 0 \end{align*} \end{definition} \begin{example}~ \begin{itemize} \item The characteristic of $\F _{p^n}$ is $p$. \item The characteristics of $\R, \C, \Q$ are $0$. \end{itemize} \end{example} This definition becomes more natural when considering it in the following manner: let $\chi \colon \Z \to k$ be a homomorphism, mapping $n \mapsto nl$ for $l$ some unit of $k$. Then $\ker (\chi) = (p)$ for $p$ zero or a prime (This is a result from Algebra I. It is a direct consequence of $\Z$ being a PID and $k$ an integral domain). If $p \neq 0$, then $p$ is clearly the characteristic as defined before, since if $x \in (p) = \ker (\chi)$, then $x = pl$ as $(p)$ is a prime ideal. Then % \section{Reminder on Characteristics} \begin{definition}[Reminder Algebra I, Characteristic] The characteristic, $p$, of a field is given by the number of times the multiplicative neutral element must be added to itself before we arrive at the additive neutral element. If no such $p$ exists, we say the field has characteristic 0. \begin{align*} pl = 0 = pll^{-1} = p1 = \underset{p \text{~times}}{\underbrace{1+1+1+\cdots}} \underset{p \text{~times}}{\underbrace{1+1+1+\cdots}} = 0 \end{align*} % Assigning a field characteristic zero, when no such $p$ exists at first seemed somewhat arbitrary, but now is quite canonical, as it simply is the $p$ that generates the ideal that is $\ker (\chi)$. \end{siderules} \end{definition} \begin{example}~ \begin{itemize} \item The characteristic of $\F _{p^n}$ is $p$. \item The characteristics of $\R, \C, \Q$ are $0$. \end{itemize} \end{example} This definition becomes more natural when considering it in the following manner: let $\chi \colon \Z \to k$ be a homomorphism, mapping $n \mapsto nl$ for $l$ some unit of $k$. Then $\ker (\chi) = (p)$ for $p$ zero or a prime (This is a result from Algebra I. It is a direct consequence of $\Z$ being a PID and $k$ an integral domain). If $p \neq 0$, then $p$ is clearly the characteristic as defined before, since if $x \in (p) = \ker (\chi)$, then $x = pl$ as $(p)$ is a prime ideal. Then % \begin{align*} pl = 0 = pll^{-1} = p1 = \underset{p \text{~times}}{\underbrace{1+1+1+\cdots}} \end{align*} % Assigning a field characteristic zero, when no such $p$ exists at first seemed somewhat arbitrary, but now is quite canonical, as it simply is the $p$ that generates the ideal that is $\ker (\chi)$. \section{Characteristic 0} \begin{theorem} Consider a finite field extension $E/F$ and suppose $F$ has characteristic 0. ... ... @@ -124,6 +126,8 @@ $F(r) = E$ 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} \section{Skew Fields and Wedderburn's Theorem} \begin{definition}[Skew Field, Division Algebra] ... ...