\documentclass[12pt]{letter}
\usepackage{amsmath,amsfonts}
\textwidth 7.0in\oddsidemargin -0.25in
\textheight 10.0in\topmargin -1.3in
\pagestyle{empty}

\def \zz {\mathbb{Z}}
\def \qq {\mathbb{Q}}
\def \ee {\epsilon}
\def \ww {\omega}
\def \tt {\sqrt[3]{2}}

\begin{document}
{\bf Math 222: An Introduction to Galois Groups}

{\bf Problem.} Determine when a polynomial with integral coefficients
is solvable by radicals.

{\bf Galois' Idea:} To a polynomial $P(x)$ which is irreducible over the
rational numbers $\qq$, associate a certain finite group $Gal(P(x))$.
And translate the solvability problem to one in terms of groups.

{\bf STEP 1:} Splitting Fields -- Given an irreducible polynomial, we
first consider the smallest field which contains all the zeros.  The
construction is analogous to that of $GF(p,P(x))$.

{\bf Example 1.0.} Let $P(x) = x^5 - 1 = (x-1)(x^4 + x^3 + x^2 + x + 1)
= (x-1)f(x)$.  Then $f(x)$ is irreducible over $\qq$ with zeros $\ee$,
$\ee^2$, $\ee^3$, $\ee^4$ as previously denoted.  i.e.\ the non-trivial
fifth roots of unity. Let
$$
\qq(\ee) = \{a_0 + a_1\ee + a_2\ee^2 + a_3\ee^3: a_i \in \qq\}
$$
be a field under ``polynomial'' multiplication.  Notice that $\ee^4$ and
all higher powers can be written in terms of smaller powers.  The same
arguments as for Galois Fields shows that this will always give a
field.  When working over $\zz_p$, we saw that all the zeros were
necessarily in the Galois field.  In this example, this is also clearly
true, but this need not be true in general.

{\bf Example 2.0.} Consider $q(x) = x^3 - 2$ which is irreducible over
$\qq$. Let $\tt$ denote the real cube root of 2. Recall that the zeros
are $\tt$, $\tt\ww$, and $\tt\ww^2$ where
$\ww$ and $\ww^2$ were the non-trivial cube roots of 1 and satifsy $x^2
+ x + 1 = 0$.  Then set
$$
\qq(\tt) = \{a_0 + a_1\tt + a_2(\tt)^2: a_i \in \qq\}.
$$
Since $(\tt)^3 = 2$, we don't need any further powers.  Although this is
a field, it clearly does not contain the other zeros.  So, one has to
throw in powers of $\tt\ww$ as well as products with $\tt$ and get a
bigger field.  This turns out to be the field
$$
\qq(\tt,\ww) = \{a_0 + a_1\tt + a_2\ww + a_3(\tt)^2 + a_4\tt\ww +
a_5\ww^2 + a_6(\tt)^2\ww + a_7\tt\ww^2 + a_8(\tt)^2\ww^2: a_i \in \qq\}.
$$

{\bf STEP 2:} Groups of Automorphims -- The Galois group of the
polynomial is a group of automorphisms of the splitting field.  Recall
the notion of an automorphism of a field from the ring handout -- it's a
one-to-one, onto map, which preserves both addition and multiplication.
The group structure will be function composition. But, we only take
certain automorphisms.

{\bf Example 1.1.} We define $Gal(f(x))$ to be the set of all
automorphisms of the field $\qq(\ee)$ which ``fix'' $\qq$.  Recall, that
the field $\qq$ is a subfield of $\qq(\ee)$ and the assumption is that
we want automorphisms $\phi: \qq(\ee) \to \qq(\ee)$ for which $\phi(a) =
a$ for all $a \in \qq$.

{\bf Question:} Where can $\ee$ go under such a $\phi$?  Remember that
$\ee^5 = 1$ and hence $\phi(\ee^5) = \phi(1) = 1$ by the assumption. But,
since $\phi$ preserves multiplication, we must have $\phi(\ee^5) =
\phi(\ee)^5$.  Hence $\phi(\ee)^5 = 1$.  In other words $\ee$ must go to
one of the other zeros!  Since the map must be one-to-one $\ee$ cannot
go to 1 and so can go to $\ee$, $\ee^2$, $\ee^3$, or $\ee^4$.
Similarly, if you take any of the other zeros, one sees that they too
must go to a zero.  It turns out that the automorphisms we're looking for
simply correspond to certain {\em permutations} of the zeros!

In this case, everything is determined simply by $\phi(\ee)$. For
example, $\phi(\ee^2) = \phi(e)^2$ and so on.  As there are four choices
for $\phi(\ee)$ we get precisely four automorphisms and a group of order
four.  With a little work one can check that $Gal(f(x)) \simeq
(\zz_4,+)$.

{\bf General Facts:} For an irreducible polynomial $P(x)$ of degree $n$
over $\qq$, the Galois Group $Gal(P(x))$ is a subgroup of $S_n$, thought
of as the set of permutations of the $n$ zeros of $P(x)$.

{\bf Example 2.1.} Return to $q(x) = x^3 - 2$.  Here there are three
zeros, so the Galois Group is a subgroup of $S_3$.  In this case,
everything is determined by $\phi(\tt)$ and $\phi(\ww)$.  Now, $\tt$ can
go to any of the three zeros $\tt$, $\tt\ww$, or $\tt\ww^2$, while $\ww$
must go to either $\ww$ or $\ww^2$.  Hence there are $3\cdot2 = 6$
possible automorphisms and so $Gal(q(x))$ must be all of $S_3$.

{\bf Question:} What's the point?

Once we've constructed these groups, there's an amazingly beautiful
theorem of Galois which relates the structure of the groups to the
structure of these ``splitting fields''.  Indeed, the following is only
a glimpse of the relationship.

{\bf STEP 3: Galois' Theorem.} Let $P(x)$ be an irreducible polynomial
over
$\qq$ and $G = Gal(P(x))$ be the associated Galois group.  Then $P(x)$
is solvable by radicals if and only if there exists a chain of subgroups
$$
G = H_0 \supset H_1 \supset H_2 \supset \dots \supset H_n = \{1_G\}
$$
such that
\begin{itemize}
\item[(1)] $H_{i+1}$ is {\em normal} in $H_i$ for each $0 \leq i \leq n
-1 $.
\item[(2)] For each $0 \leq i \leq n - 1$, $|H_i|/|H_{i+1}| = p_i$ for
some prime $p_i$.  Equivalently, the quotient group $H_i/H_{i+1}$ is
isomorphic to $(\zz_{p_i},+)$.
\end{itemize}
Moreover, let $p_i$ be the largest index appearing in the chain. At
least one of the zeros of $P(x)$ has degree $p_i$ over the integers.

{\bf Example 1.2.} Consider $f(x) = x^4 + x^3 + x^2 + x + 1$.  We saw
that $Gal(f(x)) \simeq (\zz_4,+)$.  Indeed, we have the normal chain
$$
\zz_4 = \{0, 1, 2, 3\} \supset \{0, 2\} \supset \{0\}.
$$
Here the index is 2 at both steps.  Hence all the solutions are degree 2
algebraic and hence constructible as we previously claimed.

{\bf Example 2.2.} Consider $q(x) = x^3 - 2$. We saw that $Gal(q(x))
\simeq S_3$.  One can check that there is only one normal subgroup:
$$
S_3 \supset \langle (123) \rangle \simeq \zz_3 \supset \{Id\}.
$$
At the first step the index is 2 and at the second it's 3. So, the
solutions are algebraic, but at least one is degree 3.  Notice that if
one of the solutions was constructible, we could easily construct the
others and so all three must have degree 3.

{\bf Example 3.} Consider $g(x) = 3x^5 - 15x + 5$.  Using Eisenstein's
Criterion one can check that this is irreducible over $\qq$.  Moreover,
using a little Calculus, one can see that there are three distinct real
roots and two complex roots. Under these conditions, with a little group
theory, one can show that the
Galois Group must be all of
$S_5$.  We pointed out in class that $A_5$ was normal in $S_5$ since it
has index 2.  In fact, this is the only normal subgroup of $S_5$.

But, $A_5$ is a so-called {\em simple} group -- it has NO non-trivial
normal subgroups.  See Proposition 10.11.  This is proved by playing with
3-cycles.  Indeed it uses one of our homework problems (8.4 \#13) in
which we showed that every even permutation could be written as a
product of 3-cycles.

Hence, the only normal chain we can form in this case is
$$
S_5 \supset A_5 \supset \{Id\}.
$$
The first step has index 2 which is ok, but the second step has index
60, which is not prime!!  Hence, by Galois' Theorem at least one of the
zeros is not solvable by radicals of any degree!!
\end{document}

