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\begin{document}

\chapter*{September 12:}

\begin{center}
{\Huge \bf Math 431 Class Lecture Notes}
\end{center}

\bigskip

\begin{itemize}
\item
Finite fields
\item 
Automorphisms of fields
\item 
Separable Extensions
\end{itemize}


\section{Finite fields}

The proof of the cyclicity of $F^*$ for finite~$F$ was
finished  --- see the class notes from September 10 for details.
This enables us to answer several questions about finite fields.

Suppose that $F$ is a finite field.  Then we know that the number
of elements in~$F$ is a power of a prime, say $q = p^n$, so
that $[F:\Fp] = n$.

The multiplicative group $F^*$ has order $q-1$ so every nonzero
element~$u$ of~$F$ satisfies $u^{q-1} = 1$ and it follows that
every element of~$F$ satisfies $u^q = u$.  The polynomial $x^q-x$
has $q$ roots in~$F$, and therefore $F$ is a splitting field
for this polynomial.

Since splitting fields exist are unique, we see that there is
a finite field with $q$ elements and that it is unique up to
isomorphism.

If $u$ is a generator of the cyclic group $F^*$ then $F = \Fp(u)$.
The minimal polynomial for~$u$ has degree~$n$, and this proves
that there is an irreducible polynomial in $\Fp$ of degree~$n$.

\begin{remark}
With  a little more work, one can show that $x^q-x$ is the product
of all irreducible polynomials in $\Fp[x]$ whose degree divides~$n$.
\end{remark}

\Ex
Over $\F_2$, $$x^4-x = x^4+x =  x(x+1)(x^2+x+1)$$
so there are two irreducible polynomials of degree~1 and one
of degree~2.  Therefore the polynomial $x^{16}$ must factor as
$$x^{16}+x = x(x+1)(x^2+x+1)f(x) g(x)h(x)$$
where $f$, $g$, and~$h$ are irreducible of degree~4.
A little work shows that those irreducible quartics are
$x^4+x+1$, $x^4+x^3+1$, $x^4+x^3+x^2+x+1$.
\Eex

\section{Automorphisms of fields}

If $E$ is an extension of~$F$ define
$$\Aut_{F}(E) := \{\sigma : E \rightarrow E \quad | \quad 
\sigma(x) = x \forall x \in F \}.$$

\Ex
If $F = \Q$ and $E = \Q(i)$ then
$$\Aut_F(E) = \{1, \mbox{complex conjugation}\}.$$
\Eex

\Ex
If $F = \Q$ and $E = \Q(\sqrt[3]{2})$ then
$\Aut(E) = \{1\}$.  Indeed, any automorphism maps a root of
$x^3-2$ to a root of the same polynomial, and there is a unique
root of that polynomial in~$E$.
\Eex

This is an instance of a general ``lifting'' result that comes up
over and over again.

\begin{theorem} If $\phi \colon F \to F'$ is an isomorphism of
fields, $f(x) \in F[x]$, $E$ is a root
field of~$F$, $f'$ is the polynomial obtained by applying 
$\phi$ to each coefficient of~$f$, then any lifting of $\phi$ to
a mapping from $E$ to an extension field $E'$ of $F'$ must map
$\alpha$ to a root of $f'$.  Thus there are at most $\degg(f)$
such extensions. \end{theorem}

The proof of the theorem is immediate.  If $E = F(\alpha)$,
and
$$f(\alpha) = \sum a_i \alpha^i = 0$$
then applying an extension $\Phi$ to this equation gives
$$ \sum \phi(a_i) \Phi(\alpha)^i = f'(\Phi(\alpha)) = 0.$$
Thus the proof of the theorem is shorter than the statement.

(And, as seen in class, the diagram is even shorter.)

\section{Separable Extensions}
         
From the above general fact, it is clear that the number of
embeddings from a root field $F(\alpha)$ into an extension~$E$
of~$F$ depends on how many roots the minimal polynomial of
$\alpha$ has in~$E$.  An important technical point comes up:
does the polynomial have distinct roots in its splitting field?

\begin{defin} A polynomial $f \in F[x]$ is {\bf separable}
if its roots are distinct in its splitting field.
An element of an extension field is separable over~$F$ if
its minimal polynomial is separable.
An extension field is separable if and only if all of its
elements are separable.
\end{defin}

In characteristic 0, or over finite fields, any irreducible
polynomial, and
hence any element and any extension, is separable.

\begin{theorem}
If $F$ is finite or has characteristic 0, then any irreducible
polynomial is separable.
\end{theorem}

If $f$ has a multiple root then $(x-\alpha)^2$ divides $f$,
and hence $x-\alpha$ divides $f'$.  Thus $f$ and $f'$ have
a common factor, which is impossible if $f$ is irreducible
and the characteristic is 0 (so that the degree of $f'$ is
one less than the degree of~$f$).  Conversely, any root
$\alpha$ of the gcd of $f$ and $f'$ gives a double factor
of~$f$.

\begin{example}
If $F = \F_p(t)$ is the field of rational functions
$r(t)/s(t)$ in an indeterminate~$t$, then the root
$\alpha$ of the polynomial $f(x) = x^p-t$ in a root
field $F(\alpha)$ is not separable.  Indeed, $f$ is
irreducible, and is therefore the minimal
polynomial of~$\alpha$, but
$$x^p-t = (x-\alpha)^p$$
so that the root field is the splitting field, and
the minimal polynomial has repeated roots.  
\end{example}

One shows that any inseparable polynomial in characteristic~$p$
has the shape $f(x) = g(x^p)$, so that $f' = 0$.
If the field is finite, where
the Frobenius automorphism is surjective, then
$g(x^p) = h(x)^p$ for a suitable~$h$, so that if $f$ is
inseparable then it is not irreducible.

An easy argument using the ``lifting'' principle in the preceding
section shows that if $E$ is a splitting field over~$F$ then
$$\#\Aut_F(E) \le [E:F] \; .$$
With more work, one can show that this is also true for any
finite extension $E/F$ of fields.

\begin{defin}
A finite extension $E/F$ is a {\bf galois} extension if equality
holds, i.e., if
$$\#\Aut_F(E) = [E:F] \; .$$
\end{defin}

By using the lifting theorem one can show that this holds if $E$
is the splitting field of a separable polynomial.  
A nontrivial result from galois theory says that in fact any
galois extension has the form.

\begin{theorem}
If $E/F$ is a galois extension then $E$ is the splitting field
of a separable polynomial $f \in F[x]$.
\end{theorem}

\end{document}
       
