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\begin{document}

\chapter*{August 29:}

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

\bigskip

\begin{itemize}
\item Fields, Characteristic, Prime Field
\item Degree of a Field Extension
\item Minimal Polynomials
\item Lemma
\end{itemize}

\section{The characteristic of a fields}

A field is a set~$F$ together with two binary operations called addition
and multiplication (written with standard algebraic notation and conventions)
such that: 
\begin{itemize}
\item
Addition is commutative, associative, has an identity, and has inverses.
\item
Multiplication is commutative, associative, and has an identity; elements
not equal to the additive identity have multiplicative inverses.
\item The additive and multiplicative identities are distinct.
\item Multiplication distributes over addition, i.e., $x(y+z) = xy+xz$ for 
all $ x,y,z \in F$.
\end{itemize}

The additive and multiplicative identities in a field~$F$
will usually be denoted
$0$ and~$1$ respectively, though on occasion they will be denoted
$0_F$ and $1_F$ if it is useful to emphasize that they are in a
specific field~$F$.

Let $F$ be a field.  Then there is a canonical ring homomorphism
$h\colon \Z \to F$, basically defined by taking $1$ to~$1_F$.
More precisely we define
$$h(n) = n_F := 1_F + \cdots + 1_F, \quad (n \quad {\rm times})$$
for nonnegative~$n$ (i.e., $h(n)$ is defined by recursion) and take
$h(n) = - h(-n) $ for $n < 0$.

There are two things that can happen.  The kernel of~$h$ might
be trivial.  By the isomorphism theorem
``domain/kernel $\simeq$ image'' it follows that
the image is isomorphic to~$\Z$.
Since~$F$ ``contains'' integers $r$ and~$s$ it contains
their quotient $r/s$, and $F$ contains
(an isomorphic copy of) the rational numbers~$\Q$.
In this case the field is said to be of \defn{characteristic}~0,
written $\chart(F) = 0$, and the \defn{prime field} of~$F$, which
is the smallest field contained in~$F$, is~$\Q$.

On the other hand the kernel of~$h$ might be a nonzero ideal in~$\Z$.
(An ideal in a ring is a nonempty subset closed under addition, and 
multiplication by arbitrary elements of the ring.)  Any ideal in
the ring of integers~$\Z$ is ``principal'', i.e., the set of all
multiples of a fixed integer:
$$(n) := n\Z = \{ xn : x \in \Z\}.$$
(Any ring with this property is said to be a principal
ideal domain, PID; the proof that~$\Z$ is a PID is given in an Appendix
below.)  If $h(n) = 0$ and $n = rs$ then
$$h(n) = h(r)h(s) = 0$$
and since $F$ is a field we conclude that $h(r) = 0$ or $h(s) = 0$.
It follows that that $\ker(h)$ is of the form $(p)$ where~$p$ is
a prime.  

By the aforementioned isomorphism theorem for ring maps, the image of~$h$
is isomorphic to the domain modulo the kernel, i.e.,
$$\im(\Z) \simeq \Z/p\Z.$$
The image is (isomorphic to) the field $\Fp := \Z/p\Z$
with~$p$ elements.  
In this case we say that $F$ has characteristic~$p$,
written $\chart(F) = p$, and that the prime field of~$F$ is~$\Fp$.

\section{The degree of a field extension}

Suppose that a field~$F$ is a subfield of a field~$E$.  We say
that~$E$ is an \defn{extension} of~$F$.  If we restrict the
multiplication map on~$E$ to multiplication of elements of~$E$
by elements of~$F$, then the resulting operation
$$\times \colon F \times E \to E$$
gives~$E$ the structure of a vector space over~$F$ (where addition of
``vectors'' is just the underlying field operation of addition on~$E$;
the vector space axioms are immediate consequences of the field axioms
on~$E$).

The dimension of this vector space is the \defn{degree} of the extension,
written
$$[E:F] := \dim_F(E).$$
An extension is \defn{finite} if its degree is finite.  In that case
one can choose a basis $x_1, \cdots , x_n$ consisting of elements
of~$E$ such that every element of~$E$ has a unique representation in
the form
$$x = a_1 x_1 + \cdots + a_n x_n$$
where the $a_i$ lie in the field~$F$.

\Ex
The field~$\C$ is an extension of the real numbers~$\R$.
If we forget how to multiply complex numbers by complex numbers, and
merely remember how to multiply complex numbers by real numbers, we
get the two-dimensional vector space $\R^2$; thus $[\C:\R] = 2$.
\Eex

\Ex
If $F$ is a finite field, then its prime field is $\Fp$ for some prime~$p$,
and $F$ is a finite-dimensional vector space over~$\Fp$.  From the
representation
$$x = a_1 x_1 + \cdots + a_n x_n$$
we see that $F$ has $p^n$ elements, where $n = [E:F]$, since each of
the $a_i$ can be chosen arbitrarily in~$\Fp$.
\Eex

A direct proof (see homework!) shows that if $E$ is an extension
of~$F$ and~$E'$ is an extension of~$F$ then $$[E':F] = [E':E] [E:F].$$

\section{Minimal polynomials}

Let $E$ be an extension of~$F$, and let $\alpha$ be an
element of~$E$.

Then there is a natural ring homomorphism, $h$, from the ring
$F[x]$ of polynomials with coefficients in~$F$ to the field~$E$
defined by being the identity on~$F$ and taking the indeterminate
$x$ to $\alpha$.
Thus
$$h(f) = h(\sum a_i x^i) = \sum a_i h(x)^i = \sum a_i \alpha^i$$
and $h$ is just the ``evaluate at~$\alpha$ map.

The kernel of~$h$ is an ideal in~$F[x]$.  The ring $F[x]$ is a PID
(see appendix below).  There are two cases.

If $\ker(h) = \{0\}$ is trivial, then $\alpha$ is said to be
\defn{transcendental} over~$F$.

\Ex
By a highly nontrivial theorem, $\pi$ is transcendental over~$\Q$
and $[\R:\Q]$ is infinite.  Similarly, $e$ is transcendental over~$\Q$.
\Eex

If, on the other hand, $\ker(f)$ is nontrivial then
$$\im(f) \simeq F[x]/(f(x))$$
where the kernel of~$h$ is the principal ideal $(f)$ of all
multiples of a polynomial~$f$.  The polynomial~$f$ can be chosen
canonically if we require it to be the unique element of the kernel
that is monic (the coefficient of its highest degree term is~$1$); 
THEN $f$ is said to the the \defn{minimal} polynomial
of~$\alpha$ over~$F$, and we will write this polynomial as
$m_\alpha(x) \in F[x]$.

In this case $\alpha$ is said to be \defn{algebraic} over~$F$.

Moreover, in order for $F[x]/(m_\alpha)$ to be a field, the minimal
polynomial~$m_\alpha$ has to be irreducible (analogous to
the earlier proof that the kernel of the map from $\Z$ to a field
is generated by a prime).

It is easy to see that if $n = \deg(m_\alpha)$ then
$1, \alpha, \alpha^2, \cdots \alpha^{n-1}$ is a basis for $\im(h)$ over~$F$.
(Hint: given an element $g(\alpha)$ of~$\im(h)$, divide $m_\alpha$ into
$g(x)$ to get a quotient and a remainder
$$g(x) = q(x) m_\alpha(x) + r(x)$$
and then evaluate at~$\alpha$.)

The smallest subfield of~$E$ that contains $F$ and~$\alpha$ is the
set of all rational functions $f(\alpha)/g(\alpha)$, where $g$ isn't
divisible by~$m_\alpha$; this field is usually denoted $F(\alpha)$.
However, by the remarks above on the basis of the field we see that
the set of polynomials of degree less than $n = \deg(m_\alpha) $
is a field, and we conclude that
$$F(\alpha) = \left \{\frac{f(\alpha)}{g(\alpha)}\right \}
= F[\alpha] = \{f(\alpha) : \deg(f) < n\} .$$

\Ex
$\C = \R(i)$ and $m_i(x) = x^2+1$.
\Eex

\Ex
If $F = \F_2$ and $m_\alpha(x) = x^3+x+1$ then $\F_2(\alpha)$ is
a field with 8 elements.
\Eex

\section{Appendix: Two lemmas}

\lemma
Any ideal in $\Z$ is principal, i.e. an ideal~$I$ has the
form 
$$I = (n) = n\Z$$
for some integer~$n$.

\smallskip

\noindent {\bf Sketch of Proof}: If $I$ is nontrivial let~$n$ be its
least positive element.   Then $I$ contains the set of all multiples
$(n)$ of~$n$.  On the other hand, if $x$ is any element of~$I$ then
by dividing $x$ by~$n$ we get
$$x = qn+r, \quad 0 \le r < n.$$
then $r = x-qn$ is in~$I$.  By the definition of~$n$ as the least
positive element, we have~$r = 0$ and therefore~$x$ is in~$(n)$ as
desired.

\lemma
Any ideal in $\F[x]$ is principal.
\smallskip

\noindent{\bf Sketch}: If an ideal~$I$ is nontrivial then 
it contains an element~$f$
of least possible degree.  Divide $f$ into an arbitrary element~$g$ of~$I$
$$g = qf+r$$
and reason that $I$ consists exactly of the multiples of~$f$ i.e.,
$I = (f)$ as desired.
\end{document}
