\setcounter{chapter}{10} \chapter{Galois Theory without Galois Extensions} When dealing with Galois extensions, it at some point becomes inevitable to ask: ... ... @@ -64,6 +62,9 @@ which situations such an $\hat{E}$ exists. Galois. \end{theorem} \begin{proof} \end{proof} %\begin{proof} % We will utilise the splitting field of a polynomial with distinct roots in % this proof as by the characterisation of Galois extensions over normal ... ... @@ -102,3 +103,22 @@ which situations such an $\hat{E}$ exists. % roots, as otherwise, by previous argumentation, some factor would divide them. % However, they could be the same. If so omit one of them in the product. %\end{proof} \section{Characteristic p} In the case of $F$ having characteristic $p \neq 0$, there is no such $\hat E$ in general. To show this, we pose the following counterexample: Let $F = \F _p(X)$. Note, this field is infinite. Moreover, let $E = F(r)$ for $r$ a root of $p(Z) = Z^p - X$. We immediately now notice that $p'(Z) = 0$, so our previous line of argument doesn't work. Since $r$ is a root, $0 = p(r) = r^p - X$ implying $X = r^p$ so $P(Z) = Z^p - r^p$, and as we are in characteristic $p$ we have that $p(Z) = Z^p - r^p = {(Z - r)}^p$. This means that $E$ is the splitting field over a polynomial with repeated roots. We know that the automorphism group of $E/F$ can only permute the roots, and as there is only one root, $\Aut (E/F)$ is trivial. The extension is not Galois. \section{Skew Fields and Wedderburn's Theorem}