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\begin{document}

\chapter*{September 7:}

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

\bigskip

\begin{itemize}
\item
Uniqueness of Splitting Fields
\item 
The Fundamental Theorem of Algebra
\end{itemize}

\section{Uniqueness of Splitting Fields}

Earlier we saw that there was a splitting field for
any polynomial.  We now prove uniqueness.  Curiously,
it turns out that we have to state and prove a slightly
stronger result, for the sake of making an induction
argument go smoothly.

\begin{theorem}
If $\phi \colon F \to \tilde F$  is a field isomorphism and
$E$ and $\tilde E$  are splitting fields of polynomials $f \in F[x]$
and $\tilde f \in \tilde F[x]$, respectively,  then
there exists an isomorphism $\Phi \colon E \to \tilde E$
such that $\Phi(x) = \phi(x)$ for all $x$ in~$F$.
\end{theorem}

\noindent {\bf Proof:} By induction on $[E:F]$.
If $[E:F]=1$, then $E=F$ and $\tilde E=\tilde F$, so $\phi$ itself is the
desired mapping.  

Assume that the theorem is true for any extension of fields
of degree less than~$n$.  Let $p(x)$ be an irreducible factor 
of~$f(x)$, let $\alpha$ be a zero of $p(x)$ in $E$, and let $\beta$
be a zero of $\phi(p(x))$ in $\tilde E$.  

By the Uniqueness of Root Fields Theorem, there
exists an isomorphism $\Phi_1$ from $F(\alpha)$ to 
$\tilde F(\beta)$ that
agrees with $\phi$ on $F$ and carries $\alpha$ to~$\beta$.  

So $f(x)=(x-\alpha)g(x)$ for some $g(x) \in F(\alpha)[x]$.  
Then $E$ is a splitting field for $g(x)$
over $F(\alpha)$ and $\tilde E$  is a splitting field for $\phi_1(g(x))$
over $\tilde F(\beta)$.  The degree of $g$ is less than the degree of
$f$, so $[E:F(a)] < [E:F] = n$ and thus, by the induction
assumption, there
exists an isomorphism $\Phi$ from $E$ to $\tilde E$  that agrees
with $\Phi_1$ on $F(\alpha)$ and therefore with $\phi$ on~$F$.

\section{Fundamental Theorem of Algebra}

There is a surprisingly simple proof of the Fundamental
Theorem of Algebras that uses the existence of splitting fields,
the theorem on symmetric polynomials proved last time, the
fact that square roots exist in the complex numbers, and the
fact that a polynomial of odd degree with real coefficients
has a real root.

More precisely, we prove:

\begin{theorem} 
Any polynomial of positive degree in $\C[x]$ has a root in $\C$.
\end{theorem}

From this one easily deduces that any polynomial in $\C[x]$ factors
into linear factors, i.e., that $\C$ is a splitting field
for any $f \in \C[x]$, i.e., that $\C$ is ```algebraically closed.''

\begin{proof}
Let $f$ be a polynomial of degree~$n>0$.  
Without loss of generality, we can assume that $f$ has
real coefficients; indeed, the polynomial
$f \overline f$ has real coefficients, and if it has a
complex root then so does~$f$.

Write the degree~$n$ as a power of 2 times an odd number:
$n = 2^e m$.  We prove that a polynomial $f$ with real coefficients
has a complex root by induction on~$e$.

The base case is $ e= 0$, i.e., the degree~$n$ is odd.
In this case the result is well-known from the Intermediate
Value Theorem in calculus: for large positive and negative~$x$,
$f(x) $ has opposite signs, and is hence zero somewhere
in between.

Now assume that $e > 0$ and that the result is true for
all real polynomials whose degree has a smaller power of 2 than~$2^e$.

Choose a splitting field~$E$ for~$f$ over the real numbers, so that
$$f(t) = a_0 \prod_{i=1}^n (t- \alpha_i)$$
for $a_0 \in \R$ and $\alpha_i \in E$.
For a real number~$\lambda$ consider the polynomial
$$g(t) = \prod_{i<j} (t - \alpha_i-\alpha_j -\lambda\alpha_i\alpha_j).$$
The product is over all pairs of roots of~$f$, so the degree of~$g$
is
$$\frac{n(n-1)}{2} = 2^{e-1}m(n-1).$$
The coefficients of~$g$ are symmetric functions of the $\alpha_i$
and, by the symmetric functions theorem, are polynomials in the
elementary symmetric functions of the $\alpha_i$ and are therefore
real numbers.

By our induction assumption, we know that $g$ has a complex root!
Thus for every $\lambda$ there is a pair $i,j$ such that
$$\alpha_i+\alpha_j + \lambda \alpha_i \alpha_j \quad \in \quad \C.$$
Since there are infinitely many $\lambda$ and only finitely
many pairs $i,j$ it follows that there are distinct 
$\lambda$, $\lambda'$ that give rise to the same pair.  From
the previous equation and the equation
$$\alpha_i+\alpha_j + \lambda' \alpha_i \alpha_j \quad \in \quad \C.$$
we deduce that $\alpha_i \alpha_j \in \C$ (subtract the equations and
divide by $\lambda-\lambda'$).  It follows by subtracting the
real number $\lambda \alpha_i \alpha_j$ that $\alpha_i + \alpha_j$
is complex.

Therefore the polynomial
$$(t-\alpha_i)(t-\alpha_j) = t^2 - (\alpha_1+\alpha_2) t +
\alpha_1 \alpha_2$$
has complex coefficients.  By the quadratic formula it has a root
in~$\C$ and we are finished.
\end{proof}

\end{document}
