To receive notifications about scheduled maintenance, please subscribe to the mailing-list gitlab-operations@sympa.ethz.ch. You can subscribe to the mailing-list at https://sympa.ethz.ch

Unverified Commit b82b7a1e authored by rrueger's avatar rrueger
Browse files

Algebra II: Lecture 11. Add section on characteristic p

parent c6c57797
\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}
Markdown is supported
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment