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\centerline{Mathematics 202 -- Seminar in Algebra 2}
\centerline{An ``Abstract'' of Galois Theory, and}
\centerline{Solvability by Radicals}
\centerline{April 18, 2008}
\vskip 10pt
\noindent
The following is a compendium of the ``high points'' in 
\vskip 10pt
\begin{itemize}
\item the logical
development of Galois Theory for fields of characteristic
zero, and 
\item the theorem that there exist quintic polynomials
$f(x) \in \Q[x]$ whose roots cannot be expressed in terms of radicals. 
\end{itemize}
\vskip 10pt
\noindent
These are summarized in outline form, the 
later ones with ``telegraphic'' proofs showing how they follow from 
the earlier ones.  Since the results of this subject are 
highly interrelated, this summary may be more helpful in piecing together
what implies what than the textbook presentation or the class notes, which 
include lots of examples, comments, etc. so are more discursive.
\vskip 10pt

\begin{enumerate}
\item[(0)]  Terminology and Notation:  For a finite extension $K$ over $F$:

\begin{enumerate}
\item[(a)]  An $F$-{\it automorphism\/} of $K$ is a 1-1, onto, ring 
homomorphism $\sigma : K \to K$ such that $\sigma(a) = a$ for all 
$a \in F$.
\item[(b)]  ${\rm Gal}_F K$ is set of all $F$-automorphisms of $K$, 
which is a group under function composition, called the {\it Galois group 
of $K$ over $F$.\/}
\item[(c)]  For any subgroup $H$ of ${\rm Gal}_F K$, 
$E_H = \{a \in K : \sigma(a) = a, {\rm\ all\ } \s \in H\}$, 
called the {\it fixed field\/} of $H$.  (It is a field, because
it is closed under differences, products, and inverses in $K$.)
\end{enumerate}

\item[(1)]  (Consequence of the ``Key Corollary'' of Theorem 10.7): 
{\it Let $F(u)$ contain another root $v$ of the minimal polynomial
of $u$ over $F$.
Then there exists an $F$-automorphism $\sigma$ of $F(u)$ with
$\sigma(u) = v$.  }

\item[(2)]  (The induction step of the proof of the 
Primitive Element Theorem) {\it Let $[K:F] < \infty$ 
and let $K$ be separable
over $F$. $\a, \b$ be elements of $K$ (hence both algebraic over $F$).  
Then there exists a $\g \in K$ such that $F(\a,\b) = F(\g)$.}  
(Idea of the proof:  For well-chosen
$c \in F$, we can find a $\g$ of the form $\g = \a + c\b$.)

\item[(3)] {\it If $K$ is any finite-dimensional
extension of $F$ of characteristic zero, then $K = F(\g)$ 
for some $\g \in K$.}

\item[(4)]  (Lemma for the proof of Theorems 11.7 and 11.8)
{\it Let $H$ be a subgroup of ${\rm Gal}_F K$, and let $E = E_H$.  
Then }
\begin{enumerate}
\item[(a)]  {\it If $\b = \b_1 \in K$, and $\{\b_1,\ldots,\b_t\}$ is the 
are the distinct images of $\beta$ under the elements of $H$, 
then $f(x) = (x-\b_1)\cdots(x-\b_t)$ 
is in $E[x]$.  (In fact $f(x)$ is irreducible in $E[x]$, though
we did not say this in class).}  
\item[] ({\bf Proof:}
The coefficients of $g$ are the elementary symmetric polynomials in
the $\b_i$, hence are fixed by all permutations of the $\b_i$, including
those induced by $H$.  Hence they are in the fixed field $E_H = E$.
Let $p(x)$ be the minimal polynomial of $\b_1$ over $E$.  
Then $p(\b_1) = 0$, so $p(\s(\b_1)) = 0$
for all $\s \in H$.  Hence $f(x)$ divides $p(x)$ in $E[x]$.  But this
implies $f = p$; since $p$ is irreducible, so is $f$.)
\item[(b)]  $[K:E] = |H|$.  
\item[] ({\bf Proof:} Apply the proof of (a) to $\b = u$ satisfying
$K = E(u)$, using the Primitive Element Theorem.  We get
$\deg p(x) = [K:E] = |H|$.)
\end{enumerate}

\item[(5)] (A Corollary of 4.)  {\it If $K$ is any finite extension of $F$, 
$|{\rm Gal}_F K|$ divides the degree $[K:F]$. } (This follows from 4 part b, 
by letting $E$ be the fixed field of ${\rm Gal}_F K$: $E$ is a subfield
of $K$ containing $F$, {\it possibly larger\/} than $F$.  By 4, part b,
$[K:E] = |{\rm Gal}_F K|$.  But $[K:E][E:F] = [K:F]$, so 
$|{\rm Gal}_F K|$ divides $[K:F]$.)

\item[(6)]  A \emph{Galois} extension is a normal, separable,
finite dimensional extension field $K$ of $F$.  If $F$ and $K$
have characteristic zero, this is equivalent to saying that
$K$ is the splitting field of some polynomial in $f(x) \in F[x]$
(Theorem 10.15).  

\item[(7)] If $K$ is a Galois extension of $F$ and 
$E$ is any intermediate field -- $F \subseteq E \subseteq K$, then 
$K$ is also a Galois extension of $E$ (the polynomial $f \in F[x]$
is also in $E[x]$.)

\item[(8)]  (The Fundamental Theorem of Galois Theory -- Theorem 11.11)  
{\it Let $K$ be a Galois extension of $F$.}
\begin{enumerate}
\item[(a)] {\it There is a one-to-one, 
inclusion reversing correspondence between the set of intermediate fields
$F\subseteq E \subseteq K$ and the set of subgroups 
${\rm Gal}_F K \supseteq H \supseteq \{id\}$.  The correspondence is given by 
$E \mapsto {\rm Gal}_E K$ and $H \mapsto E_H$.  Moreover, if $E = E_H$, then
$[E:F] = [G:H]$, where $G = {\rm Gal}_F K$.}
\item[(b)]  {\it The subfield $E$ is normal over $F$ if and only if its corresponding
subgroup $H = {\rm Gal}_E K$ is a normal subgroup of ${\rm Gal}_F K$.  Moreover,
if $E$ is Galois over $F$, then ${\rm Gal}_F E$ is isomorphic
to the quotient group ${\rm Gal}_F K/{\rm Gal}_E K$. }
\item[] The proof of the direct implication is based on the observation
that if $E$ is normal over $F$, any element $\sigma \in {\rm Gal}_F K$
must satisfy $\sigma(E) \subseteq E$.  Therefore we have a restriction 
mapping
\begin{eqnarray*}
\rho : {\rm Gal}_F K &\to & {\rm Gal}_F E\\
           \sigma &\mapsto& \left.\sigma\right|_E,
\end{eqnarray*}
which is a group homomorphism.  It is onto because of the ``Key Corollary''
of Theorem 10.7 (see 10.14).  The kernel is ${\rm Gal}_E K$, which 
is therefore a normal subgroup of ${\rm Gal}_F K$.  The isomorphism
\[
{\rm Gal}_F E \simeq {\rm Gal}_F K/{\rm Gal}_E K
\]
is then a consequence of the First Isomorphism Theorem for groups.
\end{enumerate}

\item[(10)]  (Root Towers, I) An polynomial $f(x) \in F[x]$ with
$F \subset {\bf C}$ is said to be {\it solvable by radicals\/}  
if there is a ``root tower'' 
\[
F = F_0 \subset F_1 \subset \cdots \subset F_n,
\]
where $\a \in F_n$, and for each $i$, $F_i = F_{i-1}(\b_i)$
where $\b_i^{n_i} \in F_{i-1}$ for some integer $n_i$, and $F_n$
contains the splitting field of $f(x)$.    

\item[(11)] {\it Let $F$ be a subfield of ${\bf C}$ containing 
the primitive $p$th root of 1, denoted $\z_p$.  Let $a$ be an element of $F$
which is not a $p$th power in $F$.  Then the splitting field $K$ of 
$x^p - a$ over $F$ has degree $p$ over $F$ and ${\rm Gal}_F K$ is cyclic
of order $p$.}  
\item[] ({\bf Proof}:  Let $\a$ be any one root of $x^p - a$, 
and let $K = F(\a)$.
The other roots of $x^p - a$ are $\z_p \a, \ldots, \z_p^{p-1}\a$.
Since $\z_p \in F$, all these roots are in $K$, so $K$ is the splitting 
field.  By 6, $K$ is Galois over $F$.  If $\s \in {\rm Gal}_F K$ is 
not the identity, then $\s(\a) = \z_p^i\a$ for some $i$.  It is easy 
to see that $\s$ generates a subgroup of ${\rm Gal}_F K$ isomorphic to 
the cyclic group $\Z_p$.  Since $K = F(\a)$, and $\a$ is a root
of a polynomial of degree $p$ over $F$, $[K:F] \le p$.  Hence
$[K:F] = p = |{\rm Gal}_F K|$ and ${\rm Gal}_F K$ is cyclic as claimed.)

\item[(12)]  {\it Let $p$ be a prime and let $\z_p$ be as in 11.  Then} 
\begin{enumerate}
\item[(a)] {\it $\qzp$ is Galois over $\Q$ and ${\rm Gal}_\Q \qzp$ is 
cyclic of order $p - 1$.} 
\item[] ({\bf Proof:} The minimal polynomial
of $\z_p$ over $\Q$ is $\phi_p(x) = x^{p-1} + \cdots + x + 1$.  
The roots of $\phi_p$ are just the powers $\z_p^i$ for $i = 1,\ldots, p-1$.  
Hence $\qzp$ is the splitting field of $\phi_p$ over $\Q$, and 
we're done by 6.  The Galois group ${\rm Gal}_\Q \qzp$ is isomorphic
to the multiplicative group of the field $\Z/\langle p\rangle$, hence
is cyclic of order $p-1$.)
\item[(b)]  {\it If $F$ is any subfield of ${\bf C}$, then
$F(\z_p)$ is Galois over $F$ and
${\rm Gal}_F F(\z_p)$ is cyclic.}  (Follows since this Galois group 
is isomorphic to a subgroup of ${\rm Gal}_\Q \qzp$.)
\end{enumerate}

\item[(13)]  (Root Towers, II)  {\it If $f(x) \in F[x]$ is solvable by radicals over $F$,
then there is a root tower 
$$F = F_0 \subset F_1 \subset \cdots \subset F_n$$
as in 10 above  (for each $i$, $F_i = F_{i-1}(\b_i)$
where $\b_i^{n_i} \in F_{i-1}$ for some integer $n_i$)
such that the splitting field of $f(x)$ is contained in 
$F_n$, $F_n$ is Galois over $F$, and in addition, 
for each $i$, $F_i$ is Galois over $F_{i-1}$ with ${\rm Gal}_{F_{i-1}} F_i$ a cyclic
group.} 
\item[] ({\bf Proof:}  Since $\a$ is expressible in terms of radicals, 
there is a root tower as in 10.  Since ${\root mn \of \b} = 
{\root m \of {\root n \of \b }}$, by inserting additional fields in 
the tower, we may assume that all the $n_i$ are {\it prime}.
 Make a new root tower by first adjoining
the primitive roots of unity $\z_{n_i}$ one at a time.  Then adjoin
the $\b_i$.  Each step of this new tower falls into the categories
studied in 11 or 12.  Hence we are done.)

\item[(14)]  (Corollary of 13.)  {\it Suppose we have a root tower
as in 13.  Then $G = {\rm Gal}_\Q F_n$ has a chain of subgroups of the form:
\[
\{id\} \subset G_1 \subset G_2 \subset \cdots \subset G_n = G
\]
where for all $i$, $G_i$ is normal in $G_{i+1}$, and 
$G_{i+1}/G_i$ is cyclic.} 
\item[] ({\bf Proof:}  Let $G_i = {\rm Gal}_{F_i} F_n$.
By 7, $F_n$ is Galois over each of the intermediate fields $F_i$.  
Hence by the last part of 9, 
\[
G_{i+1}/G_i \simeq {\rm Gal}_{F_{i+1}} F_n/{\rm Gal}_{F_i} F_n
\simeq {\rm Gal}_{F_i} F_{i+1},
\]
which is a cyclic group by hypothesis.)

\item[(15)]  (Continuation) Since $F_n$ is Galois over $\Q$ and $\a \in F_n$, 
the splitting field $K$ of $\a$ is contained in $F_n$, it is Galois
over $\Q$ by 6, and ${\rm Gal}_\Q K = {\rm Gal}_\Q F_n/{\rm Gal}_K F_n$.  As a result, 
$H = {\rm Gal}_\Q K$ also has chain of subgroups of the form in 14:
$$\{e\} \subset H_1 \subset H_2 \subset \cdots \subset H_n = G$$
where for all $i$, $H_i$ is normal in $H_{i+1}$, and 
$H_{i+1}/H_i$ is cyclic.  Groups with this property are said to 
be {\it solvable groups\/}.

\item[(16)]  {\it The symmetric group $S_5$ is not solvable.}  
\item[] ({\bf Proof}:  We try to construct a chain of subgroups as in 14 or 15.
Each normal subgroup must be a union of conjugacy classes in $S_n$.
By examining the sizes of the conjugacy classes, we see that the only
possible nontrivial normal subgroup in $S_5$ is the alternating
group $A_5$.  But $A_5$ is simple
(it has no normal subgroups except the identity subgroup and $A_5$).
Hence the only chains of subgroups 
$\{e\} \subset H_1 \subset H_2 \subset \cdots \subset H_n = S_5$
with $H_i$ normal in $H_{i+1}$ for all $i$ are 
are $\{ () \} \subset A_5 \subset S_5$ and $\{ () \}  \subset S_5$.
Since $A_5$ and $S_5$ are not cyclic, $S_5$ is not solvable.)

\item[(17)]  (The ``punch line'') Let $f(x) = 2x^5 - 10x + 5 \in \Q[x]$.
By the Eisenstein Criterion with $p = 5$, $f$ is irreducible.  
Let $K$ be the splitting field of $f$.  The Galois 
group ${\rm Gal}_F K$ permutes the five roots of $f$, so it is 
isomorphic to a subgroup of $S_5$.
By calculus, you can see that $f$ has exactly three real roots
and one pair of complex conjugate roots.  If we number the two 
complex roots as $\a_1,\a_2$ and the three real roots as 
$\a_3,\a_4,\a_5$, complex conjugation induces an automorphism 
of the splitting field $K$ of $f$ over $\Q$ giving the permutation
$(12)$ on the roots.  Moreover, since $[\Q(\a_i): \Q] = 5$,
$5$ divides $[K:F] = |{\rm Gal}_F K|$.  By the Sylow theorems, 
${\rm Gal}_F K$ must contain an element of order 5.  But
the only permutations of the 5 roots that have order 5 are the 
5-cycles.  It is easy to see that if a subgroup of $S_5$ contains
a 5-cycle and a 2-cycle, then it is all of $S_5$.  But then
by 16, ${\rm Gal}_F K$ is not a solvable group.  So by the reasoning 
leading up to 15, the roots of $f$ are not expressible by radicals(!)
\end{enumerate}
\end{document}
