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\begin{document}

\noindent
{\large \bf Additional notes \#7.}

\bigskip

\noindent
{\bf Examples and Non-Examples.}

\bigskip

\noindent
{\bf Normal Extensions.}
\begin{enumerate}
\item[(E)] Examples.
\begin{enumerate}
\item[(1)] ${\BF}<\overline{\BF}$ for any field $\BF$.
(Justification: $\overline{\BF}$ is the splitting field for
${\vr F} = {\BF}[x]$.)
\item[(2)] ${\BQ}<{\BQ}[\omega]$ where $\omega$ is a root of
unity.
(Justification: ${\BQ}[\omega]$ is the splitting field for
${\vr F} = \{m_{\omega,{\BQ}}(x)\}$, since each root of this polynomial
is a power of $\omega$.)
\end{enumerate}
\item[(N)] Non-Example.
\begin{enumerate}
\item[(1)] ${\BQ}<{\BQ}[\sqrt[3]{2}]$.
(Justification: ${\BQ}[\sqrt[3]{2}]$ is the field obtained
from ${\BQ}$ by adjoining one of the three distinct 
roots $\alpha, \beta$ or $\gamma$ of the irreducible 
polynomial $x^3-2$. If this extension was normal, then
it would contain $\alpha$ and $\beta$, hence
$\omega = \alpha/\beta$. This element
satisfies $\omega^3 = 2/2 = 1$ but $\omega\neq 1$.
Thus $\omega$ is a root of $x^3-1=(x-1)(x^2+x+1)$ but not $x-1$.
Since $x^2+x+1$ is irreducible over $\BQ$, it follows
that $m_{\omega,{\BQ}}(x) = x^2+x+1$. This implies that
${\BQ}<{\BQ}[\omega]$ is an extension of degree $2$, so it
cannot be a subextension of ${\BQ}<{\BQ}[\sqrt[3]{2}]$.
The conclusion is that ${\BQ}[\sqrt[3]{2}]$ contains one
but not all of the roots of $x^3-2$.)
\end{enumerate}
\end{enumerate}

\bigskip

\noindent
{\bf Separable Extensions.}
\begin{enumerate}
\item[(E)] Examples.
\begin{enumerate}
\item[(1)] Any extension of fields of characteristic $0$. (Justification:
Let ${\BF}<{\BE}$ be an extension of fields of characteristic
$0$, $\alpha\in {\BE}-{\BF}$, and $m(x) = m_{\alpha,{\BF}}(x)$ be the minimal
polynomial of $\alpha$. Then ${\rm deg}(m)>1$,
so ${\rm deg}(m')\geq 1$. If $m$ has a multiple root,
then $m$ and $m'$ have a common root. This implies that
$d = \gcd(m,m')\in {\BF}[x]$
has degree at least one and $d(x)|m(x)$. This is impossible,
since $m(x)$ is irreducible. Hence $m$ has no multiple roots.)
\item[(2)] Any extension ${\BF}<{\BE}$ where 
${\BF}_p<{\BF}<{\BE}<\overline{{\BF}_p}$. (Justification: If not,
then ${\BE}$ would contain an element $\alpha$ that is inseparable
over ${\BF}$. The minimal polynomial of $\alpha$ would be
of the form $m_{\alpha,{\BF}}(x) = f(x^{p^n})$ with 
$f$ separable and $n\geq 1$. If the degree of $f$ is $m$,
then $[{\BF}[\alpha]:{\BF}] = mp^n$. Since $\alpha$ satisfies
$x^{p^n}-\alpha^{p^n}\in {\BF}[\alpha^{p^n}][x]$, and
$\alpha^{p^n}$ satisfies $f(x)\in {\BF}[x]$, it follws
that $[{\BF}[\alpha]:{\BF}[\alpha^{p^n}]] = p^n$
and $[{\BF}[\alpha^{p^n}]:{\BF}] = m$. In particular, this
argument shows that ${\BF}[\alpha^{p^n}]$ is a proper subfield
of ${\BF}[\alpha]$. If ${\sf f}$ is the Frobenius endomorphism,
then ${\sf f}^n({\BF}[\alpha])\subseteq {\BF}[\alpha^{p^n}]$,
so ${\sf f}$ maps ${\BF}[\alpha]$ into a proper subfield.
This is impossible, since 
the Frobenius endomorphism is an automorphism of any algebraic
extension of ${\BF}_p$.)
\end{enumerate}
\item[(N)] Non-Example.
\begin{enumerate}
\item[(1)] ${\BF} = {\BF}_p(t^p)<{\BF}_p(t) = {\BE}$ where $t$ is
transcendental over ${\BF}_p$. (Justification: (a) $t\not\in
{\BF}$, since if $t = f(t^p)/g(t^p)$ then 
$tg(t^p) = f(t^p)$, and comparing degrees on both
sides of the equality leads to a contradiction.
(b) $t$ is a root of $x^p-t^p\in {\BF}[x]$, clearly.
(c) The polynomial $x^p-t^p$ factors completely 
in ${\BE}[x]$ as $(x-t)^p$, since the characteristic is $p$.
(d) $x^p-t^p\in {\BF}[x]$ is irreducible. If not,
then there exists a factorization $x^p-t^p = f(x)g(x)$
with $f, g\in {\BF}[x]$. In view of the 
factorization over ${\BE}$, 
$(x-t)^p = x^p-t^p = f(x)g(x)$,
it follows that $f(x) = u(x-t)^k$ and $g(x) = u^{-1}(x-t)^{p-k}$
for some $u\in {\BF}_p$ and some 
$k$ strictly between $0$ and $p$.
But then the constant term $t^k$ of $f$ is in ${\BF}$.
Since $t^p\in {\BF}$, and for some $a, b\in {\BZ}$
we have $ap+bk = 1$, we get that 
$t = (t^p)^a(t^k)^b\in {\BF}$, which contradicts (a).

Now, (a)--(d) imply that $t$ is purely inseparable 
over ${\BF}$.)
\end{enumerate}
\end{enumerate}

%\bigskip
%
%\noindent
%{\bf Galois Extensions.}
%\begin{enumerate}
%\item[(E)] Examples.
%\begin{enumerate}
%\item[(1)] 
%\end{enumerate}
%\item[(N)] Non-Examples
%\begin{enumerate}
%\item[(1)] 
%\end{enumerate}
%\end{enumerate}


\end{document}

