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\bec {\bf GALOIS THEORY NOTES}\\ Prof. J. Spencer \\ Spring 2014 \ece
We deal here entirely with subfields of the complex numbers. We
say $\ah$ is {\em expressible by radicals} if there is a tower
of fields
\beq \label{tower} Q=K_0\subset K_1 \subset \ldots \subset K_s \eeq
with $\ah\in K_s$ and $K_i=K_{i-1}(\beta_i)$ where $\beta_i^{n_i}\in K_{i-1}$.
\begin{theorem}\label{biggie}
{\tt Galois's Big Result:} There are algebraic $\ah$ which are
not expressible by radicals.
\end{theorem}
Suppose we have (\ref{tower}). First we'll massage the tower a bit.
Let $D=lcm(n_1,\ldots,n_s)$ and set
$\eps=e^{2\pi i/D}$. We'll throw in the roots of unity we need first
by setting $K_1=K_0(\eps)$ and then add the $\beta_i$. So we now have
a tower (call it the intermediate tower)
\beq \label{tower1} Q=K_0\subset K_1 \subset \ldots \subset K_{s+1} \eeq
with $\ah\in K_{s+1}$, $K_1=Q(\eps)$ and, each other extension by some
$n$-th root of an element we already have.
Now suppose we had $K_i=K_{i-1}(\beta)$ with $\beta^n=\ah\in K_{i-1}$.
Let $g(x)$ be the minimal polynomial of $\ah$ in $Q[x]$ and let the
roots of $g(x)$ be $\ah=\ah_1,\ldots,\ah_u$. (It may be that $\ah\in Q$
in which case this step will not be necessary.) Now we expand the
tower. We extend in turn by the $n$-th roots of each of the $\ah_i$.
Formally, lets set $K_{i0}=K_{i-1}$ and for $1\leq j\leq u$ set
$K_{ij}=K_{i,j-1}(\beta)$ where $\beta^n = \ah_j$. So the part of
the tower $K_{i-1}\subset K_i$ becomes longer:
$K_{i0}\subset K_{i1}\subset \ldots \subset K_{iu}$. (Yes, we may have
added extra elements here.) Call this the final tower.
\noi {\tt Example:} The original tower is \[ Q\subset Q(\sqrt{2})\subset
Q(\sqrt{2},(68+11\sqrt{2})^{1/5}) \] Here $n_1=2,n_2=5$ so $D=10$. Set
$\eps=e^{2\pi i/10}$. Now the intermediate tower
\[ Q\subset Q(\eps)\subset Q(\eps,\sqrt{2})\subset
Q(\eps,\sqrt{2},(68+11\sqrt{2})^{1/5}) \] But further $68+11\sqrt{2}$
and $68-11\sqrt{2}$ satisfy the same minimal polynomial. So we expand the
last part to \[
Q(\eps,\sqrt{2})\subset
Q(\eps,\sqrt{2},(68+11\sqrt{2})^{1/5})\subset
Q(\eps,\sqrt{2},(68+11\sqrt{2})^{1/5},(68-11\sqrt{2})^{1/5}) \]
giving the final tower.
\begin{claim}\label{finalnormal} Let $M$ be the final field in the final tower.
($M=Q(\eps,\sqrt{2},(68+11\sqrt{2})^{1/5},(68-11\sqrt{2})^{1/5})$ in
the example.) Then $M$ is normal over $Q$.
\end{claim}
\noi {\tt Proof:}
We have various polynomials $g_i(x)$, $2\leq i\leq s+1$, for which we
are adding an $n_i$-th root of all of its roots.
Set $h_i(x)=g_i(x^{n_i})$. This $h_i$
has as its roots all the $n_i$-th roots of all the
roots of $g_i$. Now we claim $M$ is the splitting field of
$(x^D-1)h_2(x)\cdots h_{s+1}(x)$. Well, we put the roots of
$x^D-1$ (namely, $1,\eps,\eps^2,\ldots,\eps^{D-1}$) into $M$ on
the first step. Further, we added an $n_i$-th root of each $\ah_{ij}$.
But the $n_i$-th roots of unity (as $n_i|D$) were already in the
field so when we added one $n_i$-th root we added them all.
\noi {\tt Example:}
\[ M=Q(\eps,\sqrt{2},(68+11\sqrt{2})^{1/5},(68-11\sqrt{2})^{1/5}) \]
as defined above is the splitting field of
\[ (x^{10}-1)(x^2-2)((x^5-68)^2-242) \]
as $68+11\sqrt{2}$ has minimal polynomial over $Q$: $(x-68)^2=(11\sqrt{2})^2=242$.
Let us renumber so that
\beq \label{towerfinal}
Q=K_0\subset K_1\subset\ldots\subset K_s = M \eeq
is the final tower.
\begin{claim}\label{normalover}
$K_{i+1}$ is normal over $K_i$.
\end{claim}
\noi {\tt Proof:} For $i=0$ we have $Q(\eps)$ normal over $Q$, as it
is the splitting field of $x^D-1$. Otherwise $K_{i+1}=K_i(\beta)$
where $\beta^n=\ah \in K_i$ for some $n$ and $n|D$. Consider
the polynomial $x^n-\ah$. (This is a polynomial in $K_i[x]$.)
Its splitting field over $K_i$ is an extension of $K_i$ by {\em all}
its roots. Its roots are $\beta,\beta\gamma,\beta\gamma^2,\ldots,\beta\gamma^{n-1}$
where $\gamma=e^{2\pi i/n}$. But $n|D$ and we added the $D$-th roots of unity
(and hence the $n$-th roots of unity) at the very start so all $\gamma^j\in K_i$.
So adding $\beta$ to $K_i$ gives us all the other roots for free!
\noi {\tt Warning:} We do not say that $K_{i+1}$ is normal over $Q$. Indeed,
if a big field is normal over an intermediate field which is normal over a
small field it does {\em not} follow that the big field is normal over the
intermediate field.
\noi Now we apply the Galois Correspondence to tower (\ref{towerfinal}) to
give the {\em Galois tower}
\beq \label{towergroup}
\Gamma(M:Q) = \Gamma_0 \supset \Gamma_1 \supset \ldots \supset \Gamma_s=\{e\}
\eeq
where $\Gamma_i=\Gamma(M,K_i)$ is the set of automorphisms of $M$ which perserve $K_i$.
Lets ``review" some group theory. Let $H$ be a subgroup of $G$.
\begin{theorem}\label{groups}
$H$ is a normal subgroup of $G$ if and only if for
all $\tau\in G$ and $\sig\in H$ we have $\sig^{-1}\tau\sig\in H$. Given
that, the factor group $G/H$ is Abelian if and only if for all $\sig_1,\sig_2\in
G$, $\sig_1\sig_2\sig_1^{-1}\sig_2^{-1} \in H$.
\end{theorem}
\noi {\tt Proof:} The first is the definition. For the second $G/H$ is Abelian means all
$(\sig_1H)(\sig_2H)=(\sig_2H\sig_1H)$ so that all $\sig_1\sig_2H=\sig_2\sig_1H$
so that all $\sig_1\sig_2(\sig_2\sig_1)^{-1}H=H$ so that all
$\sig_1\sig_2(\sig_2\sig_1)^{-1}\in H$.
This has an application that shall be important later:
\begin{theorem}\label{psi} Let $\psi:G\ra H$ be a homomorphism. Let $G_1\supset G_2$
be subgroups of $G$ and set $H_1=\psi(G_1), H_2=\psi(H_2)$.
\ben
\item
If $G_2$ is a normal
subgroup of $G_1$ then $H_2$ is a normal subgroup of $H_1$.
\item If further $G_1/G_2$ is Abelian then $H_1/H_2$ is Abelian
\een
\end{theorem}
\noi {\tt Proof:} First suppose $G_2$ is a normal subgroup of $G_1$.
Let $h_2\in H_2, h_1\in H_1$. There exist $g_2\in G_2$,
$g_1\in G_1$ with $\psi(g_2)=h_2$ and $\psi(g_1)=h_1$.
By normality $g_1^{-1}g_2g_1\in G_2$. Applying $\psi$,
$h_1^{-1}h_2h_1\in H_2$.
Second, suppose $G_1/G_2$ is Abelian. Let $\kappa,\lam\in H_1$.
There exist $\mu,\nu\in G_1$ with $\psi(\mu)=\kappa$, $\psi(\nu)=\lam$.
From Theorem \ref{groups}, $\mu\nu\mu^{-1}\nu^{-1}\in G_2$. Applying
$\psi$, $\kappa\lam\kappa^{-1}\lam^{-1}\in H_2$. Thus $H_1/H_2$
is Abelian.
Now comes an important general result. It connects the notion of a normal
extension with the group theoretic idea of a normal subgroup.
\begin{theorem}\label{normalnormal}
Let $Q\subset K \subset L \subset M$ with $M$ normal
over $Q$ and $L$ normal over $K$. Then $\Gamma(M,L)$ is
a {\em normal subgroup} of $\Gamma(M,K)$. Further, the
factor group $\Gamma(M:K)/\Gamma(M:L)$ is isomorphic
to $\Gamma(L:K)$
\end{theorem}
\noi {\bf Proof:} We define a map $\Psi: \Gamma(M:K)\ra \Gamma(L:K)$
by letting $\Psi(\sig)$ denote the {\em restriction} of $\sig$ to $L$.
As $L:K$ is normal an isomorphism of $L$ over $K$ must send $L$ to $L$
and so the restriction is indeed an automorphism of $L$ over $K$.
It is a homomorphism as the restriction of the product is the product
of the restrictions. We further claim it is surjective. Let $\tau: L\ra L$
be an automorphism over $K$. By the Galois Extension Theorems proven in
class we may extend $\tau$ to some $\sig$ with domain $M$. As $M$ is normal
over $K$ this $\sig$ will be an automorphism of $M$ over $K$, that is,
$\sig\in \Gamma(M:K)$. The restriction of $\sig$ to $L$ then takes us back
to $\tau$. Now we employ one of the fundamental theorems in Algebra I. Given
a surjective homomorphism from, say, $G_1$ to $G_2$, $G_2$ is isomorphic
to $G_1/H$ where $H$ is the kernal of the homomorphism. In our case the
kernal of $\Psi$ is those $\sig\in \Gamma(M:K)$ whose restriction to $L$
is the identity, and this is precisely $\Gamma(M:L)$. As the kernal of $\Psi$,
$\Gamma(M:L)$ is a normal subgroup of the domain $\Gamma(M:K)$ and the
range $\Gamma(L:K)$ is isomorphic to the factor group
$\Gamma(M:K)/\Gamma(M:L)$ as claimed.
\par Applying Theorem \ref{normalnormal} to (\ref{towerfinal},\ref{towergroup}) we have that
$\Gamma_{i+1}$ is a normal subgroup of $\Gamma_i$.
({\tt Warning:} A normal subgroup of a normal subgroup is not necessarily
a normal subgroup.)
We have set up our final tower (\ref{towerfinal}) so that each extension is a quite
simple one.
\begin{claim}
For $1\leq i \leq s$, the Galois group $\Gamma(K_i:K_{i-1})$
is Abelian.
\end{claim}
\noi {\tt Proof:} For $i=1$ we have the Galois group of $Q(\eps)$ over $Q$.
In class we examined this when $\eps=e^{2\pi i/p}$ and $p$ is prime. Here
$\eps=e^{2\pi i/D}$ and the situation is more complicated. We can say
that all automorphisms $\sig$ are determined by $\sig(\eps)$ and that
we must have $\sig(\eps)=\eps^j$ for some $j$. (The complicated part is
which $j$ actually give automorphisms, but that won't concern us here.)
If $\sig(\eps)=\eps^j$ and $\tau(\eps)=\eps^k$ then $\sig\tau(\eps)=
\tau\sig(\eps)=\eps^{ij}$ and so $\sig\tau=\tau\sig$. For $i>1$ we have
$K_i=K_{i-1}(\beta)$ where $\beta^n=\ah\in K_{i-1}$. Let $\gamma=e^{2\pi i/n}$
so that $\gamma\in K_{i-1}$. An automorphism $\sig\in \Gamma(K_i:K_{i-1})$
is determined by $\sig(\beta)$ and $\sig(\beta)$ must be an $n$-th root
of $\ah$ so $\sig(\beta)=\beta\gamma^j$ for some $0\leq j\leq n-1$.
As $\gamma\in K_{i-1}$ we must have $\sig(\gamma)=\gamma$.
(Again, the complicated part is
which $j$ actually give automorphisms, but that won't concern us here.)
If $\sig(\beta)=\beta\gamma^j$ and $\tau(\beta)=\beta\gamma^k$ then
$\sig\tau(\beta)=\tau\sig(\beta)=\beta\gamma^{i+j}$ and so $\sig\tau=\tau\sig$.
To review: If $\ah$ is expressible by radicals we can find a (final) tower
with $\ah\in M$. We will have $M$ normal over $Q$. Further each $K_i$ in the tower
is normal over its predecessor $K_{i-1}$. The crucial conditions are on the
Galois Tower
\[ \Gamma(M:Q) = \Gamma_0 \supset \Gamma_1 \supset \ldots \supset \Gamma_s=\{e\} \]
We have for each $1\leq i\leq s$ that $\Gamma_i$ is a {\em normal}
subgroup of $\Gamma_{i-1}$ {\em and} that the quotient group $\Gamma_i/\Gamma_{i-1}$
is Abelian.
We give the group conditions a name.
\begin{definition}\label{defsolvable}
A finite group $G$ is called {\em solvable}
if there exists a tower
\beq\label{solvable} G = \Gamma_0 \supset \Gamma_1 \supset \ldots \supset
\Gamma_s=\{e\} \eeq
so that for each $1\leq i\leq s$ that $\Gamma_i$ is a {\em normal}
subgroup of $\Gamma_{i-1}$ {\em and} that the quotient group $\Gamma_i/\Gamma_{i-1}$
is Abelian.
\end{definition}
There is a lot of group theory about solvable groups. The key point is that
some finite groups are solvable and some are not. This is the key to getting
a necessary condition for an algebraic $\ah$ to be expressible by radicals.
We'll need a useful result:
\begin{theorem}\label{a1} Let $G$ be a solvable group and let $\psi:G\ra H$ be a surjective
homomorphism. Then $H$ is a solvable group.
\end{theorem}
\noi {\tt Proof:} Applying $\psi$ to the tower (\ref{solvable}) gives
\beq\label{xxx} H = H_0 \supset H_1 \supset \ldots \supset
H_s=\{e\} \eeq
with $H_j=\psi(G_j)$. From Theorem \ref{psi} this tower has the same
properties and so $H$ is solvable.
Let $\ah$ be an algebraic number with minimal polynomial $p(x)$,
let $\ah=\ah_1,\ldots,\ah_n$ be the roots of $p(x)$, and let
$L=Q(\ah_1,\ldots,\ah_n)$ be the splitting field of $p(x)$ over $Q$.
If $\ah$ is expressible by radicals there is a (final) tower $M$ with
$\ah\in M$. Since $M$ is normal over $Q$, all $\ah_1,\ldots,\ah_n\in M$
so that $L\subset M$.
\begin{theorem}\label{yes}
Under the above conditions $\Gamma(L:Q)$ is a
solvable group.
\end{theorem}
\noi {\tt Proof:} The group $\Gam[M:Q]$ is solvable. Define
$\psi: \Gam[M:Q]\ra \Gam[L:Q]$ be letting $\psi(\sig)$ be the
{\em restriction} of $\sig$ to $L$. As $L:Q$ is normal,
$\psi(\sig)$ does indeed send $L$ to $L$ and is in $\Gam[L:Q]$.
The restriction of a product is the product of the restrictions
so $\psi$ is a homomorphism. When $\tau\in \Gam[L:Q]$ the
Galois Extension Theorems given in class show that it may be
extended to a $\sig\in \Gam[M:Q]$. Then the restriction
$\psi(\sig)=\tau$. Thus $\psi$ is a surjective homomorphism.
From Theorem \ref{a1}, $\Gam[L:Q]$ is solvable.
\noi {\tt Conclusion:} We express the Theorem in contrapositive form.
Let $\ah$ be an algebraic number and let $L$ be the splitting field
of its minimal polynomial.
\bec IF $\Gamma(L:Q)$ IS {\em NOT} SOLVABLE \\ then \\
$\ah$ is {\em not} expressible by radicals \ece
It turns out that the full symmetric group $S_5$ is not solvable. It
also turns out that there are irreducible quintic polynomials $p(x)$
whose Galois Group is $S_5$. For those polynomials, the quintic
equation $p(x)=0$ cannot be solved by radicals!
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