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\begin{document}

\chapter*{August 31:}

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

\bigskip

\begin{itemize}
\item Quotient rings
\item Root fields
\item Splitting fields
\end{itemize}

\section{Quotient rings}

Several questions have been raised, in class and 
in connection with the homework, that come
down to how to interpret quotient rings.  If~$R$ is a
ring and~$I$ is an ideal then the quotient ring~$R/I$ is
the of equivalence classes in~$R$ under the equivalence
relation
$$x \sim y \quad \mbox{\rm if and only if } \quad x-y \in I.$$
Often these equivalence classes are most easily understood
by choosing convenient or natural ``representatives'' from
each equivalence class.

\Ex It is often convenient to think of $\Z/n\Z$ as the
set 
$$\{0,1, \cdots , n-1 \}$$ 
with addition and
multiplication ``modulo~$n$'' rather than a more unwieldy
set of $n$ infinite sets.  If $x$ is any integer then there
is a unique integer~$r$ so that
$$x = qn+r, \quad 0 \le r < n$$
and this shows that every integer is in the equivalence class
of a unique $r$, $0 \le r < n$.
\Eex

\Ex
If $f$ is a nonzero polynomial then
$F[x]/(f)$ is a set of equivalence classes.  The division
algorithm
$$g = qf + r$$
shows that there is a unique element of each equivalence class
whose degree is less than $n := \degg(f)$ and it is customary
to think of the quotient as the set of all polynomials
$$a_0 + a_1 x + \cdots + a_{n-1}x^{n-1}$$
of degree less than $n$ with coefficients in~$F$.
\Eex

\section{Root fields}

If $E$ is an extension of~$F$ and $\alpha$ is an element of
$E$ then last time we found a minimal polynomial $m_\alpha(x)$
that is monic, has $\alpha$ as a root, and is a divisor of
any polynomial that has $\alpha$ as a root.  The polynomial
$m_\alpha$ is irreducible and 
$$F(\alpha) \simeq F[x]/m_\alpha.$$
Finally, the powers of~$\alpha$, $\alpha^k$, $0 \le k < n$ form
a basis for the extension $F(\alpha)/F$.

Now suppose that we start with an irreducible polynomial $f(x)$
and we want to construct an extension field in which $f$ has
a root.  This is easy!  Namely, the irreducibility of $f$ implies
that the quotient ring
$$ F[x]/(f) $$
is actually a field, and
the equivalence class containing~$x$ is a root of~$f$, 
tautologically.

\begin{defin}
If $f \in F[x]$ is irreducible then a \defn{root field} for f is a
field $E$ containing a root $\alpha$ of $f$ such that
$E=F(\alpha)$.
\end{defin}

\begin{theorem}
Root fields exist, and they are unique, i.e. if $E$ and $E'$ are
root fields with roots $\alpha$ and $\alpha'$, respectively, then
there exists an isomorphism $\phi : E \rightarrow E'$ such that
$\phi(\alpha) = \alpha'$, and $\phi(x)=x$ for all $x \in F$.
\end{theorem}

Indeed, suppose that $E$ is a root field.  Map~$F[x]$ to $E$
by evaluating polynomials at~$\alpha$.
Since $E = F(\alpha)$ we see that
$$F[x]/(f) \simeq F(\alpha) = E.$$
Since any root field is isomorphic to $F[x]/(f)$ it follows
that any two root fields are isomorphic, and from the proof we
see that there is an isomorphism with the stated property.

\Ex
There are three root fields of $f(x) = x^3-2$ inside the
complex numbers.  Namely, let
$$f(x) = x^3-2 = (x-\alpha_1)(x-\alpha_2)(x-\alpha_3)$$
where $\alpha_1 = \sqrt[3]{2}$ is the real cube root of~$2$,
$\alpha_2 = \omega \alpha_1$, $\alpha_3 = \omega^2 \alpha_1$,
where $\omega = (-1+\sqrt{3}i)/2 = \exp(2\pi i/3)$ is a third
root of unity.  The field $\Q(\alpha_1)$ is a subfield of the
real numbers and is therefore not equal to either $\Q(\alpha_2)$
or $\Q(\alpha_3)$, neither of which are contained in the
real numbers.

This leaves open the possibility that $\Q(\alpha_2)$
and $\Q(\alpha_3)$ are the same field.  Any field containing
both $\alpha_2$ and $\alpha_3$ would contain their quotient
$\omega = \alpha_3/\alpha_2$.  However, the minimal polynomial
of $\omega$ is easily checked to be $m_\omega(x) = x^2+x+1$,
so $\Q(\omega)$ is of degree~2 over~$\Q$.  The equation
$$3 = [\Q(\alpha_2):\Q] = [\Q(\omega):\Q][\Q(\alpha_2):\Q(\omega)]$$
is nonsensical, and this contradiction shows that
$\Q(\alpha_2)$ and $\Q(\alpha_3)$ are distinct.
\Eex

\section{Splitting fields}

\begin{defin}
If $f(x)$ is any nonzero polynomial in $F[x]$, then an
extension~$E$ of $F$ is said to be a splitting field of~$f$
over~$F$ if $f(x)$ factors into linear factors in $E[x]$,
and no proper subfield of~$E$ has this property.
\end{defin}

\begin{theorem}
Splitting fields exist and are unique, up to isomorphism.
\end{theorem}

We'll prove existence by induction on the degree, and leave
uniqueness (and some cool applications) until next time.

Let $n = \degg(f)$ be the degree of a polynomial $f(x)\in F[x]$.
The case $n = 1$ of the existence part of the theorem is trivial.

Now suppose that all polynomials over any field with degree 
less then $n$ have splitting fields.  
Choose $g(x) \in f[x]$, of positive degree, that divides $f$ and is
irreducible over $F$. Let $E$ be
the root field of $g$, and let $\alpha$ be a root of $g$ in~$E$.

Now, over $E$, $f$ has at least one linear factor, namely $f(x) =
(x - \alpha)f_{1}(x)$ for some $f_{1} \in E[x]$.  Now $\deg(f_{1}) <
n$ so by the induction hypothesis, there exists a splitting field
of $f_{1}$, call it $E'$.  Since $f_1$ splits in $E'[x]$ it
follows that $f$ splits into linear factors in~$E'$.
The smallest field containing the roots of~$f$ is a splitting
field.  (In fact, as checked in class, $E'$ is the smallest
field containing the roots.)

\end{document}
