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\bec{\Large\bf NOTES ON GALOIS THEORY}\\ {\large\bf Prof. Joel Spencer} \ece
{\tt A Special Note:} Galois Theory involves the study of arbitrary fields
and fields can come in many different guises. However, throughout these
notes we shall restrict ourselves to fields whose elements are complex
numbers. That is, all of our fields $F$ (even when we forget to mention
it!) will have $F\subset C$.
\section{Galois Basics}
\begin{define}\label{def1} Let $F\subset K,K'$, all fields. We
say $\sig:K\ra K'$ is an isomorphism {\em over $F$}
if
\ben
\item $\sig$ is a bijection from $K$ to $K'$
\item $\sig(a+b)=\sig(a)+\sig(b)$ for all $a,b\in K$
\item $\sig(ab)=\sig(a)\sig(b)$ for all $a,b\in K$
\item $\sig(c)=c$ for all $c\in F$
\een
\end{define}
We say that the elements of $F$ are fixed by $\sig$.
Note, however, that it is acceptible that other elements
(not in $F$) are also fixed by $\sig$.
The most important case is when $K=K'$.
\begin{define} Let $F\subset K$, both fields. We
say $\sig:K\ra K$ is an automorphism {\em of $K$ over $F$}
if
\ben
\item $\sig$ is a bijection from $K$ to itself
\item $\sig(a+b)=\sig(a)+\sig(b)$ for all $a,b\in K$
\item $\sig(ab)=\sig(a)\sig(b)$ for all $a,b\in K$
\item $\sig(c)=c$ for all $c\in F$
\een
\end{define}
Now we come to the object of study.
\begin{define}\label{defgaloisgroup} Let $F\subset K$, both fields. We
define the Galois Group of $K$ over $F$, denote
$\Gamma(K:F)$ as follows. The elements are all of
the automorphims of $K$ over $F$. The group
operation is concatenation. That is, given two
automorphisms $\sig,\tau$ we define their product
$\sig\tau$ by
\beq (\sig\tau)(a) = \tau(\sig(a)) \eeq
We define the inverse $\sig^{-1}$ by
\beq (\sig^{-1})(b) \mbox{ is that } a \mbox{ such that } \sig(a)=b \eeq
\end{define}
{\tt Exercise:} Show that $\sig\tau\in \Gamma(K:F)$. That is,
given that $\sig,\tau$ satisfy the four properties above
show that $\sig\tau$ also satisfies the four properties above.
{\tt Exercise:} Show that $\sig^{-1}\in \Gamma(K:F)$. That is,
given that $\sig$ satisfies the four properties above
show that $\sig^{-1}$ also satisfies the four properties above.
{\tt Example:} Consider $\Gamma(C:Re)$, the automorphisms of the
complex numbers $C$ over the real numbers $Re$. We claim that
complex conjugation, defined by (for $a,b$ real)
\beq\label{conj} \sig(a+bi) = a-bi \eeq
satisfies the four properties above. For example, the third property
states that, setting
\[ (a+bi)(c+di)=e+fi \]
one has
\beq\label{prod} (a-bi)(c-di)=e-fi \eeq
\\ {\tt Exercise:} Check that $\sig$, as defined by (\ref{conj}) satisfies all
four properties.
\\ There is another element of $\Gamma(C:Re)$. The identity! The map $e:K\ra K$
given by $e(\ah)=\ah$ for all $\ah\in K$ is {\em always} an element of $\Gamma(K:F)$.
\begin{theorem}\label{CoverRe} The {\em only} elements of $\Gamma(C:Re)$ are complex conjugation $\sig$
and the identity $e$. \end{theorem}
{\tt Proof:} Let $\tau\in \Gamma(C:Re)$ and consider $\tau(i)$. As $i^2+1=0$
\beq \label{a1} 0 = \tau(0) = \tau(i^2+1) = \tau(i)^2+1 \eeq
That is, denoting $\tau(i)$ by $z$, $z$ must satisfy $z^2+1=0$. There are only two
possibilities for $z$, either $z=i$ or $z=-i$. Further, the value of $z=\tau(i)$
determines the entire map $\tau$. This is because any complex number $\ah$ can be
written $\ah=a+bi$ with $a,b$ real and so
\beq \label{a2} \tau(\ah) = \tau(a)+\tau(b)\tau(i) = a+bz \eeq
as $\tau$ fixes all real numbers. When $z=i$ we have $\tau(\ah)=\ah$ so that $\tau$
is the identity. When $z=-i$ we have $\tau(a+bi)=a-bi$ and so $\tau$ is complex
conjugation $\sig$.
We therefore have $\Gamma(C:Re)=\{e,\sig\}$. The identity acts as the identity of the
group and
\beq\label{a3} \sig^2(a+bi) = \sig(\sig(a+bi)) = \sig(a-bi) = a+bi \eeq
so that $\sig^2=e$. We have a group on two elements. We can further write
\beq\label{a4} \Gamma(C:Re)\cong (Z_2,+) \eeq
by mapping $e$ to $0$ and $\sig$ to $1$.
{\tt Caution:} An expression such as $\sig^3(\ah)$ does {\em not} mean the cube
of $\sig(\ah)$ but rather the result of applying $\sig$ three times to $\ah$,
in this case, $\sig(\sig(\sig(\ah)))$. To say, for example, that $\sig^3=e$,
would be to say that $\sig(\sig(\sig(\ah)))=\ah$ for {\em all} $\ah$.
The proof ideas in Theorem \ref{CoverRe} can be greatly generalized.
\begin{theorem}\label{polysat}
Let $F\subset K,K'$, all fields.
Let $\sig:K\ra K'$ be an isomorphism over $F$ as given by
Definition \ref{def1}. Let $\ah\in K$ and let $\ah$
be a root of some $p(x)\in F[x]$. That is, we may
write
\[ p(x)=a_0+a_1x+\ldots+a_nx^n \in F[x] \]
with the coefficients in $F$. Set $\beta=\sig(\ah)$
Then $\beta$ is a root of $p(x)$.
\end{theorem}
{\tt Proof:} As
\[ 0=p(\ah)=a_0+a_1\ah+\ldots+a_n\ah^n \]
we apply $\sig$ to both sides and (noting, critically,
that as $a_i\in F$, $\sig(a_i\ah^i)=\sig(a_i)\sig(\ah)^i = a_i\beta^i$)
\[ 0 = \sig(0)= a_0+a_1\beta+\ldots+a_n\beta^n \]
as desired.
\begin{theorem}\label{b1}
Let $F\subset K,K'$, all fields. Assume $K=F(\ah_1,\ldots,\ah_s)$.
Let $\sig:K\ra K'$ be an isomorphism over $F$ as given by
Definition \ref{def1}. Then $\sig$ is determined by the values of
$\sig(\ah_1),\ldots,\sig(\ah_s)$.
\end{theorem}
{\tt Proof:} For any mononomial $\kappa=c\ah_1^{m_1}\cdots \ah_s^{m_s}$ with
$c\in F$, the value of $\sig(\kappa)$ is determined by
\[ \sig(\kappa) = c\sig(\ah_1)^{m_1}\cdots \sig(\ah_s)^{m_s} \]
Any polynomial $\lam$ in $\ah_1,\ldots,\ah_s$ is the sum of monomials and
hence $\sig(\lam)$ is determined. When $K$ is an extension of $F$ of finite
dimension (which is pretty much all we look at) every $\lam\in K$ is such
a polynomial and so $\sig$ is determined on $K$. But even in the general
case every $\lam\in K$ can be written as the quotient $\lam=\lam_1/\lam_2$
of polynomials and hence $\sig(\lam)=\sig(\lam_1)/\sig(\lam_2)$ is still
determined.
Here is a powerful consequence.
\begin{theorem}\label{finitegalois}
Let $F\subset K$, both fields, and assume only that $[K:F]$ is finite.
Then the Galois Group $\Gamma[K:F]$ (as given by Definition \ref{defgaloisgroup})
is finite.
\end{theorem}
{\tt Proof:} Suppose $n=[K:F]$.
Write $K=F(\ah_1,\ldots,\ah_s)$ for some $\ah_1,\ldots,\ah_s$.
(One way to do this is to take a basis $\ah_1,\ldots,\ah_n$ for $K$ over $F$. Later,
Theorem {e5}, we'll find a more efficient way but we don't need that now.)
Let $\sig\in \Gamma[K:F]$ and set
$\beta_i=\sig(\ah_i)$ for each $i$.
Each $\ah_i$ satisfies some polynomial
$p_i(x)\in F[x]$ of degree at most $n$. From theorem \ref{polysat}, $\beta_i$
satisfies the same polynomial. But we know that a polynomial of degree at most
$n$ can have at most $n$ roots so there are at most $n$ choices for $\beta_i$.
These choices, from Theorem \ref{b1}, determine $\sig$ on all of $K$.
{\tt A Cautionary Note:} Suppose that in Theorem \ref{finitegalois} we have
$K=F(\ah_1,\ldots,\ah_s)$. Each $\sig(\ah_i)$ must be one of a finite number
of choices. {\em However,} not all choices necessarily give a good $\sig$.
For example, suppose we wrote $K=Q(\sqrt{2},\sqrt{3},\sqrt{6})$. For
$\sig\in \Gamma[K:Q]$ we must have $\sig(\sqrt{2})=\pm \sqrt{2}$, $\sig(\sqrt{3})=\pm \sqrt{3}$,
$\sig(\sqrt{6})=\pm \sqrt{6}$. But we don't get all eight choices. If, say,
$\sig(\sqrt{2})=-\sqrt{2}$, $\sig(\sqrt{3})=+ \sqrt{3}$, then we must have (as $\sig$ sends
products to products) $\sig(\sqrt{6})=-\sqrt{6}$.
Now we can give a wide class of examples for which the Galois Group is determined.
\begin{theorem}\label{b3}
Let $F\subset K$, both fields, and assume that $K=F(\ah)$. Let $p(x)\in F[x]$
be the minimal polynomial for $\ah$. Suppose further that $\beta\in K$ is also
a root of $p(x)$. Then there is an automorphism $\sig$ of $K$ over $F$ with
$\sig(\ah)=\beta$.
\end{theorem}
{\tt Proof:} Set $n$ to be the degree of $p(x)$. Then
\[ K = \{a_0+a_1\ah+\ldots+a_{n-1}\ah^{n-1}: a_0,\ldots,a_{n-1}\in F\} \]
As $p(x)$ is a minimal polynomial for $\ah$ it must be irreducible (over $F$)
and hence it must be a minimal polynomial for $\beta$ as well. Thus
$[F(\beta):F]=n$. But as $\beta\in F(\ah)$, $F(\beta)\subseteq F(\ah)$. As
the dimensions over $F$ are the same we deduce that $F(\beta)=K$. Thus we
can write
\[ K = \{a_0+a_1\beta+\ldots+a_{n-1}\beta^{n-1}: a_0,\ldots,a_{n-1}\in F\} \]
We define $\sig$ by
\[ \sig(a_0+a_1\ah+\ldots+a_{n-1}\ah^{n-1}) = a_0+a_1\beta+\ldots+a_{n-1}\beta^{n-1} \]
The key point is that products are sent to products. Write $p(x)=x^n+b_{n-1}x^{n-1}+\ldots+b_0$.
In multiplying elements of the form
$a_0+a_1\ah+\ldots+a_{n-1}\ah^{n-1}$ we use the reduction
$\ah^n=-b_{n-1}\ah^{n-1}-\ldots-b_0$. As $\beta$ has the same minimal polynomial
in multiplying elements of the form
$a_0+a_1\beta+\ldots+a_{n-1}\beta^{n-1}$ we use the same reduction
$\beta^n=-b_{n-1}\beta^{n-1}-\ldots-b_0$.
\begin{theorem}\label{b4}
Let $F\subset K$, both fields, and assume that $K=F(\ah)$. Let $p(x)\in F[x]$
be the minimal polynomial for $\ah$. Let $\ah_1=\ah, \ah_2,\ldots,\ah_s$ be all
the roots of $p(x)$ in $K$.
Then the Galois Group $\Gamma[K:F]$ (as given by Definition \ref{defgaloisgroup})
will have precisely $s$ elements $\sig_1=e,\sig_2,\ldots,\sig_s$.
\end{theorem}
{\tt Proof:} As $K=F(\ah)$, $\sig$ is determined (Theorem \ref{b1}) by $\sig(\ah)$
which must be (Theorem \ref{polysat}) one of $\ah_1,\ldots,\ah_s$. From
Theorem \ref{b3} each of these give a valid $\sig_i\in \Gamma[K:F]$.
\section{Three Examples}
In all our examples the ground field shall be $Q$ and the extension field will be a
subfield of the complex numbers $C$.
We take as basic that the {\em only} nonzero rational numbers $c$ for which $\sqrt{c}\in Q$
are those positive $c$ for which each prime factor $p$ appears an even number of times.
In particular, $\sqrt{2},\sqrt{3},\sqrt{3/2}$ are all irrational.
{\tt Example I:} $K=Q(\sqrt{2},\sqrt{3})$.
As $\sqrt{2}$ has minimal polynomial $x^2-2$, $Q(\sqrt{2})$ has basis $1,\sqrt{2}$ over $Q$.
Now we need a simple result:
\begin{theorem}\label{sqrt3}
$\sqrt{3}\not\in Q(\sqrt{2})$
\end{theorem}
{\tt Proof:} If it were we would have
\[ \sqrt{3} = a+b\sqrt{2} \]
with $a,b\in Q$. Squaring both sides
\[ 3 = a^2+2b^2+2ab\sqrt{2} \]
As $1,\sqrt{2}$ is a basis the coefficient of $\sqrt{2}$ would need be zero.
That is, $2ab=0$. So either $a=0$ or $b=0$.
\ben
\item $b=0$: Then $\sqrt{3}=a\in Q$, contradiction.
\item $a=0$. Then $\sqrt{3}=b\sqrt{2}$ so $\sqrt{3/2}=b\in Q$, contradiction.
\een
From Theorem \ref{sqrt3} and that $\sqrt{3}$ satisfies a quadratic
(namely, $x^2-3$) over $Q(\sqrt{2})$, $1,\sqrt{3}$ is a basis for $Q(\sqrt{2},\sqrt{3})$
over $Q(\sqrt{2})$ and so $1,\sqrt{2},\sqrt{3},\sqrt{6}$ is a basis for $K$ over $Q$.
That is, we may write
\beq\label{a11}
K= Q(\sqrt{2},\sqrt{3}) = \{ a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}: a,b,c,d\in Q\}
\eeq
and every $\ah\in K$ has a {\em unique} expression in this form.
Now we turn to the Galois Group $\Gamma(K:Q)$. Any $\sig\in K$ has
$\sig(\sqrt{2})=\pm \sqrt{2}$ and
$\sig(\sqrt{3})=\pm \sqrt{3}$ which gives four ({\tt Caution:} this is $2$
{\em times} $2$) possibilities. The value of $\sig$ on $\sqrt{2},\sqrt{3}$
determines the value on all of $K$.
The four elements of the Galois Group
are $e,\sig_1,\sig_2,\sig_3$ where
\[ e(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \]
\[ \sig_1(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6} \]
\[ \sig_2(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= a+b\sqrt{2}-c\sqrt{3}-d\sqrt{6} \]
\[ \sig_3(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6} \]
{\tt Exercise:} Check that these really are automorphisms, that they are
bijections (easy!) that send sums to sums (easy!) and products to products
(a bit messier!). This will actually come out of more general stuff later.
What does the group $\Gamma(K:Q)$ look like. The identity is the identity,
no problem. Any $\sig$ when squared gives the identity. For example
\[ \sig_1^2(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= \sig_1(a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6})
=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \]
We can also see this by noticing that either $\sig(\sqrt{2})=\sqrt{2}$ or
$\sig(\sqrt{2})=-\sqrt{2}$ but in either case $\sig^2(\sqrt{2})=\sqrt{2}$ and
similarly $\sig^2(\sqrt{3})=\sqrt{3}$ so that $\sig^2=e$. We also calculate
that if you multiply any two of $\sig_1,\sig_2,\sig_3$ in either directtion
you get the other third one. For example
\[ \sig_2(\sig_1(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= \sig_2(a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6})
=a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6} \]
We have the Klein Vierergruppe, aka the Fourgroup, which is isomorphic to $Z_2\times Z_2$.
Actually, there are only two groups with four elements (up to isomorphism, of course), the
cyclic group $Z_4$ and the Vierergruppe $Z_2\times Z_2$ so once it isn't the first it
{\em must} be the second!
{\tt Example II:} $K=Q(\eps)$ with $\eps = e^{2\pi i/5}$.
$\eps$ satisfies $x^5-1=0$ and, as $\eps\neq 1$, it satisfies $p(x)=0$ with
\beq\label{minpolyeps}
p(x) = \frac{x^5-1}{x-1} = x^4+x^3+x^2+x+1
\eeq
This is irreducible (one can show this by replacing $x$ by $x+1$ giving
$x^4+5x^3+10x^2+5x+5$ and using Eisenstein's criterion) so $[K:Q]=4$ and we write
\beq \label{basiseps}
K = \{ a + b\eps + c\eps^2+d\eps^3: a,b,c,d\in Q \}
\eeq
The minimal polynomial (\ref{minpolyeps}) has roots $\eps,\eps^2,\eps^3,\eps^4$.
From Theorem \ref{b4} the Galois Group $\Gamma(K:Q)$ consists
of four automorphism which we shall label $\sig_1,\sig_2,\sig_3,\sig_4$. They
are determined by their values on $\eps$ and we shall let $\sig_j$ be that
automorphism with $\sig_j(\eps)=\eps^j$. Note that $\sig_1$ is the identity
and $\sig_4$ is our old friend, complex conjugation.
What is the product $\sig_j\sig_k$. Lets see what it does to $\eps$.
\beq \label{prod1}
(\sig_j\sig_k)(\eps)= \sig_j(\sig_k(\eps))= \sig_j(\eps^k)= \sig_j(\eps)^k = (\eps^j)^k = \eps^{jk} \
\eeq
Hmmmm, so it looks like $\sig_j\sig_k = \sig_{jk}$. But we only have four automorphisms. What does
it mean to say $\sig_3\sig_3=\sig_9$. The key is that $\eps^5=1$ so that we can reduce $\eps^{jk}$
by reducing $jk$ modulo $5$. As $\eps^9=\eps^4$ we have $\sig_3\sig_3=\sig_4$. So we {\em can} and
{\em do} say
$\sig_j\sig_k = \sig_{jk}$ with the understanding that $jk$ is computed modulo $5$. With this we
have
\beq\label{iso1} \Gamma(K:Q) \cong Z_5^* \eeq
where we associate $\sig_j$ with $j$. Finally $Z_5^*\cong (Z_4,+)$ (the cyclic group, not the
Vierergruppe) by associating $1,2,3,4$ with $0,1,3,2$ respectively.
{\tt Example III:} $K=Q(2^{1/3},\omega)$ with $\omega = e^{2\pi i/3}$.
The polynomial
\beq \label{minpoly3} p(x)=x^3-2 \eeq
is irreducible (by Eisenstein's criterion, or simply that, as a cubic, it has
no rational roots) and its roots are $\ah,\beta,\gamma$ where for convenience we
write
\beq\label{ahbetagam}
\ah=2^{1/3},\beta=2^{1/3}\omega,\gam=2^{1/3}\omega^2
\eeq
Any field that contains $2^{1/3},\omega$ contains
$\ah,\beta,\gam$
and any field that contains
$\ah,\beta,\gam$
contains $2^{1/3},\omega$ so we may also write $K=Q(\ah,\beta,\gam)$,
so $K$ is the {\em splitting field} (to be defined later in Definition \ref{splittingdef}) of $p(x)$ over $Q$.
A basis for $Q(\ah)$ over $Q$ is $1,\ah,\ah^2$.
As all elements of $Q(\ah)$ are real,
$\omega\not\in Q(\ah)$. As $\omega$ satisfies the quadratic $1+x+x^2=0$ over
$Q(\ah)$, a basis for $K=Q(\ah,\omega)$ over $Q(\ah)$ is $1,\omega$. Hence
a basic for $K$ over $Q$ is $1,\ah,\ah^2,\omega,\omega\ah,\omega\ah^2$ and we
can write
\beq\label{3basis}
K= \{a+b\ah+c\ah^2+d\omega+e\omega\ah+f\omega\ah^2:a,b,c,d,e,f\in Q \}
\eeq
and each $\zeta\in K$ has a unique such representation.
What are the automorphisms $\sig\in\Gamma(K:Q)$? As $K=Q(\ah,\beta,\gam)$, $\sig$
is determined by its values on $\ah,\beta,\gam$. Further, as $\ah,\beta,\gam$
satisfy the same irreducible (\ref{minpoly3}), $\sig$ of any of them must be one
of them. Further, as $\sig$ must be a {\em bijection}, $\sig$ cannot send
two of $\ah,\beta,\gam$ to the same value and hence $\sig$ must be a permutation
on $\ah,\beta,\gam$. This gives that there are {\em at most} six automorphisms
and that $\Gamma(K:Q)$ is isomorphic to a subgroup of $S_3$, the full symmetric
group on three elements, here $\ah,\beta,\gam$.
Actually, {\em all} permutations of $\ah,\beta,\gam$ yield automorphisms of
$K$ and so
\beq\label{IIIgal} \Gamma(K:Q) \cong S_3 \eeq
This actually will follow from some general stuff but we can give an idea
here. Two automorphisms are easy, the identity (we {\em always} have the
identity) and complex conjugation $\sig$. We do have to check that
complex conjugation is a bijection from $K$ to itself. As $\sig(\ah)=\ah$ and
$\sig(\omega)=\omega^2\in K$ it sends $K$ to $K$ and since $\sig^2=e$ it
must be a bijection. (That is, $\sig^{-1}(\zeta)=\sig(\zeta)$.) This
$\sig$ corresponds to the permutation that keeps $\ah$ fixed and transposes
$\beta,\gam$.
Here is another $\tau\in \Gamma(K:Q)$. Generate it by setting $\tau(\ah)=\beta$
and $\tau(\omega)=\omega$. Then
\beq\label{whitherbeta}
\tau(\beta)=\tau(\ah\omega)=\tau(\ah)\tau(\omega)=\beta\omega=\gamma
\eeq
and
\beq\label{whithergamma}
\tau(\gam)=\tau(\beta\omega)=\tau(\beta)\tau(\omega)=\gamma\omega=\ah
\eeq
so it cycles $\ah$ to $\beta$ to $\gam$ back to $\ah$. With the representation
of (\ref{3basis})
\beq\label{tauzeta}
\tau(a+b\ah+c\ah^2+d\omega+e\omega\ah+f\omega\ah^2)=
a+b\omega\ah + c\omega^2\ah^2+d\omega+e\omega^2\ah+f\ah^2
\eeq
In this form (but, like I said, there are other approaches) one need show
$\tau$ is bijective (pretty easy), that $\tau(\zeta_1+\zeta_2)=\tau(\zeta_1)+\tau(\zeta_2)$ (quite
easy), and that $\tau(\zeta_1\zeta_2)=\tau(\zeta_1)\tau(\zeta_2)$ (lengthy, unless you use some
tricks).
Indeed, here is another approach to show that $\tau$ is indeed an automorphism from $K$ to $K$.
Consider the intermediate field $L=Q(\omega)$. We first claim that $p(x)$ given by (\ref{minpoly3})
is irreducible over $L$. As it is a cubic, if it reduced it would have a root in $L$. So either
$\ah,\beta=\ah\omega,\gam=\ah\omega^2$ would be in $L$. As $\omega\in L$ if any of $\ah,\beta,\gam$
were in $L$ then all three would be in $L$, in particular $\ah\in L$. But then $L$ would have $\omega$
and $\ah$ and so would be $Q(\ah,\omega)=K$. As $[L:Q]=2\neq 6=[K:Q]$ that cannot happen. Now
$\ah,\beta$ have the same minimal polynomial in $L[x]$ and $K=L(\ah)=L(\beta)$, so, from the previous notes, there
does exist an automorphism $\tau$ of $K$ that preserves $L$ and has $\tau(\ah)=\beta$. As it preserves $L$ we also
have $\tau(\omega)=\omega$. Since $\tau$ preserves $L$ it certainly preserves the smaller $Q$ and so
$\tau\in\Gamma(K:Q)$.
Once we have $\sig,\tau\in \Gamma(K:Q)$ we have that $\Gamma(K:Q)$ is a subgroup of $S_3$ that
contains an element $\tau$ of order three and an element $\sig$ of order two. From $\tau$ it
must have at least three elements, from $\sig$ it can't have exactly three elements, so it
has more than three elements, so it has all six elements, it is all of $S_3$.
\section{Splitting Fields and Normal Extensions}
Here we are usually dealing with a ``ground field" $F$ and an extension field $K$.
Throughout we will only consider finite extensions $K:F$. In most examples
$F$ is the field of rational numbers $Q$. While other examples will be considered,
one may well think about $F$ as $Q$ in the first reading.
\begin{define} We say $f(x)$ completely splits into linear factors in $K[x]$ if we
may write
\beq\label{one} f(x) = \prod_{i=1}^r (x-\ah_i)^{m_i} \eeq
\end{define}
\begin{define}\label{splittingdef} We say $K$ is the splitting field of a polynomial $f(x)\in F[x]$
over $F$ if
\ben
\item $f(x)$ completely splits into linear factors in $K[x]$.
\item $K=F(\ah_1,\ldots,\ah_r)$, where the $\ah_i$ are the roots of $f(x)$ in $K$.
That is, $K$ may be generated from $F$ using the roots of $f(x)$ and has nothing more.
\een
\end{define}
Our goal is the following result, but we shall first require preliminaries interesting
in their own right.
\begin{theorem}\label{bigsplit}
Suppose $K\subset C$ is the splitting field of $f(x)\in F[x]$ over $F$.
Suppose $g(x)\in F[x]$ is irreducible (over $F$) and suppose further that there
is an $\beta\in K$ with $g(\beta)=0$. Then $g(x)$ completely splits into linear
factors in $K[x]$.
\end{theorem}
{\tt Remark:} The condition that $K$ is a subfield of the complex numbers isn't totally
necessary but it somewhat simplifies the presentation and this is the main case we shall
use.
\begin{define}
Let $\sig:K_1\ra K_2$ be an isomorphism over $F$.
Let $h(x)=h_0+h_1x+\ldots+h_wx^w \in K_1[x]$. Then $\sig h$ is that polynomial
achieved by applying $\sig$ to all of the coefficients. That is,
\[ (\sig h)(x) = \sig(h_0)+\sig(h_1)x+\ldots+ (\sig h_w)x^w \]
We note $(\sig h)(x)\in K_2[x]$.
\end{define}
\begin{theorem}\label{aa1} If $c(x)=a(x)b(x)$ in $K_1[x]$ then $(\sig c)(x)=(\sig a)(x)(\sig b)(x)$ in
$K_2[x]$
\end{theorem}
{\tt Proof:} Immediate.
\begin{theorem}\label{aa2} $p(x)\in K_1[x]$ is irreducible in $K_1$ if and only if
$(\sig p(x))\in K_2[x]$ is irreducible in $K_2$.
\end{theorem}
{\tt Proof:} If $p(x)=a(x)b(x)$ in $K_1[x]$ then $(\sig p)=(\sig a)(\sig b)$ in $K_2[x]$. Conversely we may apply
the isomorphism $\sig^{-1}$ so if $(\sig p)(x)=a(x)b(x)$ in $K_2[x]$, $p(x)=(\sig^{-1}a)(x)(\sig^{-1}b(x))$ in
$K_1[x]$.
\begin{theorem}\label{aa3} Let $f(x)\in F[x]$ be irreducible over $F$.
Let $\sig:K_1\ra K_2$ be an isomorphism over $F$.
Let
$f(x)=p_1(x)\cdots p_l(x)$
be the factorization of $f(x)$ into irreducible factors in $K_1[x]$.
Then
$f(x)=(\sig p_1)(x)\cdots (\sig p_l)(x)$
is the factorization of $f(x)$ into irreducible factors in $K_2[x]$.
\end{theorem}
{\tt Proof:} As $f(x)\in F[x]$, $(\sig f)(x)$ is $f(x)$. From Theorem \ref{aa1},
$f(x)=(\sig p_1)(x)\cdots (\sig p_l)(x)$ is a factorization and from Theorem
\ref{aa2} the factors are irreducible in $K_2[x]$.
Now we shall {\em extend} an isomorphism to a larger structure.
\begin{theorem}\label{aa4}{\tt Isomorphism Extension Theorem.}
Let $\sig:K_1\ra K_2$ be an isomorphism over $F$.
Let $p(x)\in K_1[x]$ be irreducible and let $\ah$ be a root of $p$.
Let $\beta$ be a root of $(\sig p)(x)$. Then we may extend $\sig$
to an isomorphism $\sig^+ K_1(\ah)\ra K_2(\beta)$ by setting
$\sig^+(\ah)=\beta$.
\end{theorem}
{\tt Proof:} Set $n$ equal the degree of $p(x)$ so we may write
\[ K_1(\ah) = \{ c_0+c_1\ah+\ldots+ c_{n-1}\ah^{n-1}: c_0,\ldots,c_{n-1}\in K_1\} \]
and
\[ K_2(\beta) = \{ d_0+d_1\beta+\ldots+ d_{n-1}\beta^{n-1}: d_0,\ldots,d_{n-1}\in K_2\} \]
Then $\sig^+$ is defined by
\[
\sig(c_0+c_1\ah+\ldots+ c_{n-1}\ah^{n-1})= d_0+d_1\beta+\ldots+ d_{n-1}\beta^{n-1} \]
where $d_j=\sig(c_j)$. Write
$p(x)=x^n-a_{n-1}x^{n-1}-\ldots-a_0$.
To check that $\sig^+$
preserves products we basically (formally, we'd need to write more here)
have to look at
$\ah^n= a_{n-1}\ah^{n-1}+\ldots+a_0$.
We have
$(\sig p)(x)=x^n-b_{n-1}x^{n-1}-\ldots-b_0$.
where $b_j=\sig(a_j)$. So
$\beta^n= b_{n-1}\beta^{n-1}+\ldots+b_0$ and indeed $\sig(\ah^n)=\beta^n$.
{\tt Example:} Let $F=Q$, $\theta=2^{1/3}$, $\omega=e^{2\pi i/3}$, $\eta=\theta\omega$. Define
$\sig: Q(\theta)\ra Q(\eta)$ by $sig(\theta)=\eta$. Take $p(x)=x^2+x+1$ so that in this
case $\sig p$ is $p$. Take the root $\omega$ for both sides. Now we extend $\sig$
to $\sig^+:Q(\theta,\omega)\ra Q(\eta,\theta)$ be setting $\sig^+(\omega)=\omega$.
Applying Theorem \ref{aa4} repeatedly we get:
\begin{theorem}\label{aa4e}{\tt Full Isomorphism Extension Theorem.}
Let $\sig:K_1\ra K_2$ be an isomorphism over $F$.
Let $f(x)\in K_1[x]$ have complex roots $\ah_1,\ldots,\ah_s$.
Let $\beta_1,\ldots,\beta_s$ denote the complex roots of $(\sig f)(x)\in K_2[x]$.
Then $\sig$ may be extended to an isomorphism
$\sig^{++}:K_1(\ah_1,\ldots,\ah_s)\ra K_2(\beta_1,\ldots,\beta_s)$
\end{theorem}
Proof: Let $p_1(x)\in K_1[x]$ be the irreducible polynomial of $\ah_1$
over $K_1$. Let $\beta_1$ be a root of $(\sig p_1)(x)$. From
Theorem \ref{aa4} we extend $\sig$ to $\sig^+: K_1(\ah_1)\ra K_2(\beta_1)$.
Continue in this fashion extending each of the $a_i$. (When $a_i$ is
already in the field you don't do anything.) At the end you have
a $\sig^{++}$ with domain $K_1(\ah_1,\ldots,\ah_s)$ But the $\sig^{++}(\ah_i)$
must be the roots of $(\sig f)(x)$ so they must be the $\beta_1,\ldots,\beta_s$
giving the desired extension.
Now we prove Theorem \ref{bigsplit}.
Let $\ah_1,\ldots,\ah_r$ denote the complex roots of $f(x)$ and let
$\beta_1=\beta,\beta_2,\ldots,\beta_s$ denote the complex roots of $g(x)$. If the
theorem fails we can assume, without loss of generality, that $\beta_1\in K$ and
$\beta_2\not\in K$. Set $K_1=F(\beta_1)$, $K_2=F(\beta_2)$. As $\beta_1,\beta_2$
have the same minimal polynomial $g(x)$ over $F$ we find an isomorphism
$\sig:K_1\ra K_2$ which preserves $F$ and has $\sig(\beta_1)=\beta_2$.
Now from Theorem \ref{aa4e} we extend $\sig$ to the roots of $f(x)$. As
$f(x)\in F[x]$, $(\sig f)(x)=f(x)$ so both $f(x)$ and $(\sig f)(x)$ have
the roots $\ah_1,\ldots,\ah_r$. So the extended $\sig^{++}$ permutes
these roots. That is, $\sig^{++}$ is an isomorphism from
$K_1(\ah_1,\ldots,\ah_r)$ to $K_2(\ah_1,\ldots,\ah_r)$.
Whats wrong with this? Well, remember that $K_1=F(\beta_1)$ with $\beta_1\in F(\ah_1,\ldots,\ah_r)$
so that $K_1(\ah_1,\ldots,\ah_r)=K_1(\ah_1,\ldots,\ah_r)$.
But $K_2=F(\beta_2)$ with $\beta_2\not\in F(\ah_1,\ldots,\ah_r)$
and so $K_2(\ah_1,\ldots,\ah_r)= F(\ah_1,\ldots,\ah_r,\beta_2)$ is a nontrivial extension of
$K_1(\ah_1,\ldots,\ah_r)$. But isomorphisms over $F$ preserve dimension over $F$ (a nice exercise!)
and the Tower Theorem would give
$[F(\beta_2,\ah_1,\ldots,\ah_r):F]$ to be strictly bigger than
$[F(\beta_1,\ah_1,\ldots,\ah_r):F]$, a contradiction.
Now that we have proven Theorem \ref{bigsplit} we give an important definition that distinguishes
certain kind of field extensions.
\begin{definition}\label{normal} Suppose $F\subset K$ are subfields of $C$ with $K$ a finite
extension of $F$. We say that the extension
$K:F$ is {\em normal} if the following hold.
\ben
\item\label{y1} There is an $f(x)\in F[x]$ with $K$ the splitting field of $f(x)$ over $F$.
\item\label{y2} Every $g(x)\in F[x]$ which is irreducible (over $F$) and has a root in $K$
completely splits into linear factors in $K[x]$.
\een
When this occurs we often say that $K$ is a normal extension of $F$.
\end{definition}
\begin{theorem}\label{easy1}
The two conditions in Definition \ref{normal} are equuivalent. That is, either
one implies the other.
\end{theorem}
{\tt Proof:} We've already done the hard part. Theorem \ref{bigsplit} gives that
condition \ref{y1} implies condition \ref{y2}. Now assume condition \ref{y2}.
As $[K:F]$ is finite write $K=(\ah_1,\ldots,\ah_s)$ for some finite number of
$\ah_1,\ldots,\ah_s$. For each $\ah_i$ let $p_i(x)\in F[x]$ be its irreducible
polynomial over $F$.
\par We claim $K$ is the splitting field of $f(x)$ where we
set $f(x)$ to be the product
$p_1(x)\cdots p_s(x)$. By Condition \ref{y2} {\em all} of the roots of each
$p_i(x)$ are in $F$ and so the extension of $F$ by all of the roots of $f(x)$
(that is, all of the roots of each $p_i(x)$) is still inside of $F$. But the
roots include $\ah_1,\ldots,\ah_s$ so the extension must include $K(\ah_1,\ldots,\ah_s)$
which is all of $F$. That is, the extension of $F$ by all of the roots of $f(x)$
is precisely $K$, giving the claim.
{\tt Some Examples:}
\ben
\item $K=Q(2^{1/3})$ is {\em not} a normal extension of $Q$ as the polynomial $x^3-2\in Q[x]$
(irreducible by Eisenstein's criterion)
has one root in $K$ but its other roots are not in $K$.
\item $K=Q(\sqrt{2},\sqrt{3})$ is a normal extension of $Q$ as the polynomial $(x^2-2)(x^2-3)$
has roots $\sqrt{2},-\sqrt{2},\sqrt{3},-\sqrt{3}$ and extending $Q$ by these four roots gives
precisely $K$.
\item $K=Q(2^{1/3},\omega)$ (with $\omega=e^{2\pi i/3}$) is a normal extension of $Q$ as the
polynomial $x^3-2\in Q[x]$ has roots $2^{1/3},2^{1/3}\omega,2^{1/3}\omega^2$ and extending
$Q$ by these three roots gives precisely $K$.
\item Let $\eps=e^{2\pi i/5}$. Then $K=Q(\eps)$ is a normal extension of $Q$ as the polynomial
$(x^5-1)/(x-1)=x^4+x^3+x^2+x+1$ has roots $\eps,\eps^2,\eps^3,\eps^4$ and extending $Q$ by
these four roots gives precisely $K$.
\item Let $K=Q(2^{1/4})$ and $F=Q(2^{1/2})$. Then $K$ is a normal extension of $F$ as the
polynomial $x^2-2^{1/2}\in F[x]$ has roots $2^{1/4},-2^{1/4}$ and extending $F$ by these
two roots gives precisely $K$.
\item Let $K=Q(2^{1/4})$. Then $K$ is {\em not} a normal extension of $Q$ as the polynomial
$x^4-2\in Q[x]$ (irreducible by Eisenstein's criterion) has two roots $2^{1/4},-2^{1/4}$ in
$K$ but the other two roots $2^{1/4}i, -2^{1/4}i$ are not in $K$. (One reason why $2^{1/4}i\not\in K$
is that all elements of $K$ are real.)
\een
{\tt A Cautionary Note:} The last two examples emphasize that when we talk about a normal extension
we are talking about {\em two} fields, that $K$ is normal over $F$. Further, consider the tower
$Q\subset F\subset K$, with $F=Q(2^{1/2})$ and $K=Q(2^{1/4})$. Then $F$ is a normal extension of $Q$
as it is an extension of $Q$ by the two roots of $x^2-2$. We've seen that $K$ is a normal extension
of $F$. But it is {\em not} true (as we just saw) that $K$ is a normal extension of $Q$. That is,
we do {\em not} have a transitive property for normality, just because blip is normal over blop
which is normal over blunk we {\em cannot} deduce that blip is normal over blunk.
While we have to be careful about towers of fields, the following is useful and easy.
\begin{theorem}\label{cc1} {\tt The Middle Normal Theorem}
Let $K\subset L subset F$ be fields and {\em assume} $F:K$ is a normal field extension.
Then $F:L$ is a normal field extension.
\end{theorem}
{\tt Proof:} From Definition \ref{normal}, condition \ref{y1},
there is an $f(x)\in F[x]$ with $K$ the splitting field of $f(x)$ over $F$. That is, $f$ splits
entirely in $K[x]$ with roots $\ah_1,\ldots,\ah_r\in F$ and $K=F(\ah_1,\ldots,\ah_r)$. But now we can
simply consider $f(x)$ as a polynomial in $L[x]$. It still splits entirely in $K[x]$ with
roots $\ah_1,\ldots,\ah_r$. As $F\subset L$ we have
$F(\ah_1,\ldots,\ah_r)\subset L(\ah_1,\ldots,\ah_r)$ and since $\ah_1,\ldots,\ah_r\in F$ and $L\subset F$,
$L(\ah_1,\ldots,\ah_r)\subset K$ so that $L(\ah_1,\ldots,\ah_r)=F$ and so the $F$ is a normal extension
over $L$ by the same Definition \ref{normal}, condition \ref{y1} and the same $f(x)$.
{\tt Caution:} Under the assumptions of Theorem \ref{cc1} we do {\em not} necessarily hafve $L:K$ a
normal extension.
Here is a nice property of normal field extensions that say, somehow, that they are nailed down.
\begin{theorem}\label{fixF} Let $K:F$ be a normal field extension. Let $K'$ be a field and $\sig:K\ra K'$
an isomorphism over $F$. (Recall, this means $\sig(c)=c$ for all $c\in K$.) Then $K'=K$.
\end{theorem}
{\tt Proof:} We can write $K=F(\ah_1,\ldots,\ah_s)$. Then $K'=K(\sig(\ah_1),\ldots,\sig(\ah_s))$.
For each $i$, $\ah_i$ and $\sig(\ah_i)$ satisfy the same irreducible polynomial $p_i(x)\in F[x]$.
As $K:F$ is normal this means $\sig(\ah_i)\in K$. Thus $K'\subset K$. Similarly, going backward
with $\sig^{-1}$, $K\subset K'$ and so $K=K'$.
{\tt A Cautionary Note:} Theorem \ref{fixF} does not say that each element of $K$ is fixed by
$\sig$. Indeed, $\sig$ can move around the elements of $K$ but the set of elements remains the
same.
\begin{theorem}\label{biggernomal} Let $K:F$ be a finite field extension. Then there is an
extension $K\subset K^+$ so that $K^+$ is a normal field extension of $F$.
\end{theorem}
{\tt Proof:} As $[K:F]$ is finite we can write $K=F(\ah_1,\ldots,\ah_r)$ for some finite
number of $\ah$'s. Let $p_i(x)$ be the minimal polynomial for $\ah_i$ in $F[x]$. Set
$K^+$ to be the splitting field for the product $f(x)=p_1(x)\cdots p_r(x)$. As a splitting
field it is a normal extension of $F$ and it contains $\ah_1,\ldots,\ah_r$ and therefore $K$.
\section{Fields to Groups and Back Again}
Let us fix some finite extension $F\subset K$ of subfields of $C$
and set $G$ to be the Galois Group $\Gamma(K:F)$. We will be
interested in intermediate fields $L$, that is, $F\subset L \subset K$,
and in subgroups $H$ of $G$.
We will describe first a mapping from
fields $L$ to groups $H$
\begin{definition}\label{stardef} Let $F\subset L \subset K$ be an
intermediate field. We define $L^*$, a subgroup of $G$, by
\beq \label{stareq}
L^* = \{ \sig\in G: \sig(\ah)=\ah \mbox{ for all } \ah\in L \}
\eeq
That is, $L^*$ is those automorphisms of $L$ which fix all elements
of $L$.
\end{definition}
Why is $L^*$ a subgroup of $G$? Well, suppose $\sig,\tau$ were two
automorphims of $K$ over $F$. Then, as we have discussed before,
so is $\sig\tau$. But further, if $\sig(\ah)=\ah$ and $\tau(\ah)=\ah$ for all
$\ah\in L$ then
\[ (\sig\tau)(\ah)= \sig(\tau(\ah)) = \sig(\ah) = \ah \]
for all $\ah\in L$ and so $\sig\tau\in L^*$. Similarly $\sig^{-1}\in L^*$.
Finally the identity $e\in L^*$ as $e$ fixes all elements.
We now will describe first a mapping from groups $H$ to fields $L$.
\begin{definition}\label{dagdef} Let $H \subset G$ be a
subroup of the Galois Group. We define $H^{\dag}$, an intermediate
field, by
\beq \label{dageq}
H^{\dag} = \{ \ah \in K: \sig(\ah)=\ah \mbox{ for all } \sig \in H \}
\eeq
That is, $H^{\dag}$ is those elements of $K$ which are fixed by all automorphisms
$\sig\in H$.
\end{definition}
Set $L=H^{\dag}$. Why is $L$ an intermediate field? First of all, as all
automorphisms $\sig\in \Gamma$ fix all elements $c\in F$, any element $c\in F$
will be fixed by all $\sig\in H$, so that $F\subset L$. Now suppose $\ah,\beta\in L$
and take any $\sig\in H$. As $\sig(\ah)=\ah$ and $\sig(\beta)=\beta$,
we must have $\sig(\ah+\beta)=\sig(\ah)+\sig(\beta)=\ah+\beta$. Thus $\ah+\beta\in L$
and similarly $\ah\beta, -\ah, \ah^{-1}\in L$.
{\tt An Extended Example:} Take ground field $F=Q$ and extension $K=Q(\sqrt{2},\sqrt{3})$.
The four elements of the Galois Group $G$
are $e,\sig_1,\sig_2,\sig_3$ where (as done earlier)
\[ e(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \]
\[ \sig_1(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6} \]
\[ \sig_2(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= a+b\sqrt{2}-c\sqrt{3}-d\sqrt{6} \]
\[ \sig_3(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6} \]
$G$ is the Vierergruppe. There are five subgroups (we count the trivial ones
here) of $G$:
\[ \{e\}, H_1=\{e,\sig_1\}, H_2=\{e,\sig_2\}, H_3=\{e,\sig_3\}, \mbox{ and } G
\mbox{ itself.} \]
\par Lets start in the ``middle" with $H_1=\{e,\sig_1\}$. What is $H_1^{\dag}$.
That is,
which $\ah=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ are fixed by all $e,\sig_1$.
Well, $e$ is the identity so it fixes everything so we can ignore it. If we think
of $\sig_1(\ah)=\ah$ as an (easy!) equation it is true precisely when $b=d=0$.
So $\ah\in H_1^{\dag}$ if and only if we can write $\ah=a+c\sqrt{3}$. That is,
$H_1^{\dag}= Q(\sqrt{3})$.
Similarly, for $\ah\in H_2^{\dag}$ the necessary and
sufficient condition is that $c=d=0$ so $\ah=a+b\sqrt{2}$ and $H_2^{\dag}=Q(\sqrt{2})$.
Similarly, for $\ah\in H_3^{\dag}$ the necessary and
sufficient condition is that $b=c=0$ so $\ah=a+d\sqrt{d}$ and $H_3^{\dag}=Q(\sqrt{6})$.
\par The case $\{e\}^{\dag}$ is always the same, since the identity preserves everything
$\{e\}^*=K$. Finally, what about $G^{\dag}$. That is,
which $\ah=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ are fixed by all $e,\sig_1,\sig_2,\sig_3$.
To be fixed by $\sig_1$ forces $b=d=0$, to be fixed by $\sig_2$ forces $c=d=0$, to be
fixed by $\sig_3$ is now redundant as it forces $b=c=0$. So all of $b,c,d$ must be
zero but $a$ can be an arbitrary rational and so $G^*=Q$.
\par Now set
\[ L_1=Q(\sqrt{2}), L_2=Q(\sqrt{3}), L_3=Q(\sqrt{6}) \]
and consider
the groups associated (via $*$) with the fields $Q,L_1,L_2,L_3,K$. The easiest
is $Q^*=G$, which is to say that all $\sig\in G$ fix all $\ah\in Q$ which is
true as $G$ was {\em defined} as all automorphisms $\sig$ of $K$ which fix
all $\ah\in Q$.
\par How about $L_1^*$? Clearly $e\in L_1^*$ as $e$ fixes everything.
Also $\sig_2\in L_1^*$ as $\sig_2(a+b\sqrt{2})=a+b\sqrt{2}$. But
$\sig_1,\sig_3\not\in L_1^*$ as they send $\sqrt{2}$ to $-\sqrt{2}$.
So $L_1^*=\{e,\sig_2\}$
\par How about $L_2^*$? Clearly $e\in L_2^*$ as $e$ fixes everything.
Also $\sig_1\in L_2^*$ as $\sig_1(a+c\sqrt{3})=a+c\sqrt{3}$. But
$\sig_2,\sig_3\not\in L_2^*$ as they send $\sqrt{3}$ to $-\sqrt{3}$.
So $L_2^*= \{e,\sig_1\}$
\par How about $L_3^*$?
Clearly $e\in L_2^*$ as $e$ fixes everything.
Also $\sig_3\in L_3^*$ as $\sig_3(a+d\sqrt{6})=a+d\sqrt{6}$. But
$\sig_1,\sig_2\not\in L_3^*$ as they send $\sqrt{6}$ to $-\sqrt{6}$.
So $L_3^*= \{e,\sig_3\}$
\par Finally, how about $K^*$.
Clearly $e\in K^*$ as $e$ fixes everything. But the other $\sig_1,\sig_2,\sig_3$
do not fix everything and so are not in $K^*$. Thus $K^*=G$.
\par We can put this all in tabular form.
\bec
\begin{tabular}{rr}
Field & Group \\ \hline
$K=Q(\sqrt{2},\sqrt{3})$ & $\{e\}$ \\
$L_1=Q(\sqrt{2})$ & $H_2=\{ e,\sig_2\}$ \\
$L_2=Q(\sqrt{3})$ & $H_1=\{ e,\sig_1\}$ \\
$L_3=Q(\sqrt{6})$ & $H_3=\{ e,\sig_3\}$ \\
$Q$ & $G=\{e,\sig_1,\sig_2,\sig_3\}$
\end{tabular}
\ece
We see we have a one-to-one correspondence. We can go from fields to groups by applying
$*$. And we can go from groups to fields by applying $\dag$. And $\dag$ and $*$ are
inverses as maps, if we apply one and then the other we get back where we started.
Does this always work? No. But it works {\em in important cases} and that will be
the substance of the major theorem of Galois Theory. Indeed, not to keep you in
suspence, here is that theorem.
The normal extensions are precisely those extensions for which
the correspondence works.
\begin{theorem}\label{bigkahuna}{\tt The Galois Correspondence Theorem}
Let $F\subset K$ be subfields of $C$ with $K:F$ a {\em normal} extension.
Set $G=\Gamma(K:F)$.
Then there is a bijection between the intermediate fields
$L$, meaning that $F\subset L\subset K$ and the subgroups
$H$ of $G$. (We include $L=F$, $L=K$ as intermediate fields
and we include $\{e\}$ and $G$ itself as subgroups.) The
bijection is given by $*$ and $\dag$ as previously defined.
That is, $H=L^*$ if and only if $L=H^{\dag}$. Thus
\beq\label{d10}
(L^*)^{\dag} = L \mbox{ and } (H^{\dag})^*=H
\eeq
Furthermore, the
correspondence {\em reverses} containment, making $L$ bigger
makes $H=L^*$ smaller and making $H$ bigger makes $L=H^{\dag}$
smaller.
The field $F$ is associated with all of $G$ while the field
$K$ is associated with $\{e\}$. Setting $n=[K:F]$ we have
$n=|G|$.
Further the sizes are connected, when $H=L^*$
\beq \label{size1}
[K:L] = |H|
\eeq
or, equivalently,
\beq \label{size2}
[L:F] = |G/H|
\eeq
\end{theorem}
{\tt A Cautionary Note:} In the listing for $K=Q(\sqrt{2},\sqrt{3})$ above as $G$
has only four elements it doesn't take too much work (try it!) to show that
$\{e\},H_1,H_2,H_3,G$ are the {\em only}
subgroups. It is not at all clear that we have listed {\em all} of the subfields
of $K$. How do we know there isn't some other weird intermediate field between $Q$
and $K$? After all, these are infinite sets so we can't try everything. It will turn
out that from Galois Theory we will be able to show that the above list gives all of
the intermediate fields.
{\tt A Cautionary Example:} Let the ground field $F=Q$ and the extension field
$K=Q(2^{1/3})$. Any automorphism $\sig:K\ra K$ must send $2^{1/3}$ to a root of
$x^3-2$ but $2^{1/3}$ is the {\em only} root of $x^3-2$ in $K$, as the other roots
are not real. Thus we must have $\sig(2^{1/3})=2^{1/3}$ and so $\sig$ must be the
identity. That is, $G=\Gamma(K:Q)=\{e\}$. As $[K:Q]=3$ there are no intermediate
fields except for $Q$ and $K$ themselves. So $Q^*=\{e\}$ and $K^*=\{e\}$. As
every element is fixed by $e$, $\{e\}^*=K$. So in this case we do {\em not} get
a bijection between subgroups of the Galois Group and intermediate fields.
For {\em any} extension $K:F$ the following ``easy" result is one part of
(\ref{d10}), the main part of the Galois Correspondence Theorem, Theorem
\ref{bigkahuna}.
\begin{theorem}\label{d1} Let $F\subset K$ be subfields of $C$. Then for
any intermediate field $L$
\beq\label{d2} L\subset (L^*)^{\dag} \eeq
and for any subgroup $H$
\beq\label{d3} H\subset (H^{\dag})^* \eeq
\end{theorem}
{\tt Proof:}
$L^*$ is those automorphisms $\sig$ such that $\sig(\ah)=\ah$ for
all $\ah\in L$. That is, all $\sig\in L^*$ fix all $\ah\in L$. That is, all
$\ah\in L$ are fixed by all $\sig\in L^*$ and hence all $\ah\in L$ belong
to $(L^*)^{\dag}$.
Similarly, $H^{\dag}$ is those $\ah\in K$ such that $\sig(\ah)=\ah$ for all
$\sig\in H$. That is, all $\sig\in H$ fix all $\ah\in H^{\dag}$. That is,
all $\sig\in H$ are in $(H^{\dag})^*$.
\section{Fields to Groups and Back Again II}
Let us fix some finite extension $F\subset K$ of subfields of $C$
and set $G$ to be the Galois Group $\Gamma(K:F)$. However, we
now assume $K$ is a Normal Extension of $F$. Recall that we
have already defined the map $*$ from intermediate fields to
subgroups and the map $\dag$ from subgroups to intermediate
fields.
\begin{theorem} Let $F\subset K$ be subfields of $C$ with $K$ a
Normal extension of $F$ and set $G$ to be the Galois Group $\Gamma(K:F)$.
Then for any intermediate field $L$
\beq\label{e1} (L^*)^{\dag} = L \eeq
\end{theorem}
{\tt Proof:} We already know $L\subset (L^*)^{\dag}$. Now suppose $\beta\in K$
and $\beta\not\in L$. Our goal is to show $\beta\not\in (L^*)^{\dag}$. Recall
that as $K$ is a normal extension of $F$, $K$ is a normal extension of $L$.
Let $p(x)$ be the minimal polynomial for $\beta\in L[x]$ and let $\beta_1$ be
another root of $p(x)$. As $K$ is a normal extension of $L$, $\beta_1\in K$.
Thus there is an isomorphism $\sig: L(\beta)\ra L(\beta_1)$ which fixed $L$
and has $\sig(\beta)=\beta_1$. Applying the Full Isomorphism Extension Theorem
(Theorem \ref{aa4e})
we extend $\sig$ to an isomorphism $\sig^{++}$ with domain $K$. But as
$\sig^{++}$ fixes $L$ and $K$ is normal over $L$, the range of $\sig^{++}$ must
be $K$. That is, $\sig^{++}$ is an automorphism of $K$ which fixes all $\ah\in L$
but does not fix $\beta$. So $\beta\not\in (L^*)^{\dag}$. End of Proof.
This has a perhaps surprising followup.
\begin{theorem}\label{e2} Let $F\subset K$ be subfields of $C$ with $K$ a
normal extension of $F$. Then there are only {\em finitely many}
intermediate fields $L$.
\end{theorem}
{\tt Proof:} From Theorem \ref{e2}, $L$ is determined by $L^*$ but as
$G=\Gamma(K:F)$ is finite there can be only finitely many subgroups
$H$, only finitely many possible $L^*$.
\begin{theorem}\label{e3}
Let $K$ be a finite extension of $F$, both
subfields of $C$.
Then there are only {\em finitely many}
intermediate fields $L$.
\end{theorem}
{\tt Proof:} Extend $K$ to $K^+$ so that $K^+$ is a normal extension of
$F$. From Theorem \ref{e2} there are only finitely many intermediate
fields between $F$ and $K^+$ and thus only finitely many intermediate
fields between $F$ and the smaller $K$.
\begin{theorem}\label{e4} Let $F$ be a subfield of $C$ and $\ah,\beta\in C$,
both algebraic over $F$. Then there exists $\gam\in C$ with
\beq\label{twotoone} F(\gam)=F(\ah,\beta) \eeq
\end{theorem}
{\tt Proof:} As $\ah,\beta$ are algebraic over $F$, $F(\ah,\beta)$ is a
finite extension of $F$. Now for each integer $i$ set $F_i=F(\ah+i\beta)$.
Each of these are subfields of $F(\ah,\beta)$ but by Theorem \ref{e3}
there are only finitely many such subfields so there must be $i\neq j$
with $F_i=F_j$. Thus $F_i$ contains $\ah+i\beta$ {\em and} $\ah+j\beta$.
But then it contains $\ah = \frac{1}{j-i}j(\ah+i\beta)-i(\ah+j\beta)$ and
$\beta = \frac{1}{j-i}((\ah+j\beta)-(\ah+i\beta))$. Thus $F_i$ must be
all of $F(\ah,\beta)$ and so we can take $\gam=\ah+i\beta$.
\begin{theorem}\label{e5} {\tt Single Generator Theorem.}
Let $K$ be a finite extension of $F$, both subfields of $C$.
Then there is an element $\gam\in K$ such that $K=F(\gam)$.
\end{theorem}
{\tt Proof:} We claim that for any $\ah_1,\ldots,\ah_r\in C$, all
algebraic over $F$, there exists a $\gam\in C$ with
$F(\gam)=F(\ah_1,\ldots,\ah_r)$. This comes from repeatedly
applying Theorem \ref{e4} to replace two of the generators by one.
(Formally we apply induction on $r$.) Now as $K$ is a finite extension
of $F$ we can write $K=F(\ah_1,\ldots,\ah_r)$ for some finite set of
$\ah$'s and then replace them by a single $\gam$.
\section{Toward the Galois Extension Theorem}
OK, lets see where we are. We have a map $*$ from fields to
groups and a map $\dag$ from groups to fields (more precisely,
intermediate fields $L$, $F\subset L\subset K$, and subgroups
$H\subset G$. We know that if we apply $*$ and then $\dag$
we get back to the same $L$. This gives a bijection between
intermediate fields $L$ and subgroups $H$ which may be expressed
as $H=L^*$ though, a priori, there may be $H$ that are not of that
form.
Now we look at the size relationship in the Galois
Correspondence Theorem \ref{bigkahuna}. We make strong use of the Single Generator
Theorem \ref{e5}.
\begin{theorem}\label{f1} Let $K$ be a normal extension of $F$ and
let $L$ be an intermediate field. Set $r=[K:L]$ Then $L^*$ consists of
precisely $r$ automorphisms.
\end{theorem}
{\tt Proof:} We know from the Middle Normal Theorem that $K$ is a
normal extension of $L$, which will allow us basically to ignore $F$.
From the Single Generator Theorem \ref{e5} write $K=L(\ah)$. Let $p(x)$
be the minimal polynomial in $L[x]$ with $\ah$ as a root so that
$p(x)$ has degree $r$. In $C$ let $\ah=\ah_1,\ah_2,\ldots,\ah_r$
denote the roots of $p(x)$. As $K:L$ is normal, $\ah_1,\ldots,\ah_r\in K$.
As $[L(\ah_i):L]=r$ we must have all $L(\ah_i)=K$. For each $1\leq i\leq r$
as $\ah,\ah_i$ have the same minimal polynomial in $L[x]$ there is an
isomorphism $\sig_i:L(\ah)\ra L(\ah_i)$ given by setting $\sig_i(\ah)=\ah_i$.
These are automorphisms of $K$ fixing $L$, so elements of $L^*$. Conversely
$\sig\in L^*$ is determined by $\sig(\ah)$ and $\sig(\ah)$ must also satisfy
$p(x)$ and so must be one of $\ah_1,\ldots,\ah_r$ and so $\sig_1,\ldots,\sig_r$
are {\em all} of the automorphisms in $L^*$.
{\tt Example:} Take $K=Q(\sqrt{2},\sqrt{3})$. The Single Generator Theorem \ref{e5}
works and we can set $K=Q(\sqrt{2}+\sqrt{3})$. This is not certain a priori,
one must check that $\sqrt{2}+\sqrt{3}$ does indeed generate $K$ which takes
some linear algebra. Setting $\gam=\sqrt{2}+\sqrt{3}$ we check that
$1$,$\gam$,$\gam^2=5+2\sqrt{6}$,$\gam^3=11\sqrt{2}+9\sqrt{3}$ are indeed
linearly independent. I wasn't kidding about the Linear Algebra, here you need
to show that the vectors $(1,0,0,0)$, $(0,1,1,0)$, $(5,0,0,2)$, $(0,11,9,0)$,
representing $1,\gam,\gam^2,\gam^3$ are linearly independent.
Now set, for example, $\ol{\gam}=\sqrt{2}-\sqrt{3}$
and we want a $\sig\in \Gamma(K:Q)$ with $\sig(\gam)=\ol{\gam}$. Then
$\sig(\gam^2)=\sig(5+2\sqrt{6})=\ol{\gam}^2 = 5-2\sqrt{6}$ and
$\sig(\gam^3)=\sig(9+11\sqrt{6})=\ol{\gam}^3 = 11\sqrt{2}-9\sqrt{3}$.
$\sig$ is a linear transformation from $K$ to itself. From $\sig(\gam^2)=\ol{\gam}^2$
we deduce $\sig(\sqrt{6}=-\sqrt{6}$. Further $\gam^3 - 9\gam = 2\sqrt{2}$ and so
$\sig(2\sqrt{2})=\ol{\gam}^3-9\ol{\gam}=2\sqrt{2}$. Thus we must have $\sig(\sqrt{2})=
\sqrt{2}$ and, finally, $\sig(\sqrt{3})=\sig(\sqrt{6})/\sig(\sqrt{2}) = -\sqrt{3}$.
That is, $\sig$ is one of the four automorphisms we knew we had.
Now we have shown the size relationship in the Galois Correspondence Theorem \ref{bigkahuna}. The only
item left is to show that the correspondence between $L$ and $L^*$ gives us {\em all}
of the subgroups $H$. This will take an interesting side detour.
\section{Symmetric Functions}
Now for a change of pace which has applications to what we are doing and
is interesting by itself. Lets look at polynomials in $n$ variables
$x_1,\ldots,x_n$. In our examples we'll take $n=3$ and call the
variables simply $x,y,z$.
We'll call a polynomial symmetric if no matter how you permute the
variables you get the same thing. For example: $x^{20}+y^{20}+z^{20}$
or $x^5y^5+x^5z^5+y^5z^5$. One class is of particular interest
to us. For $1\leq i\leq n$ define the $i$-th {\em elementary symmetric
polynomial} as the sum of all of the products of $i$ distinct variables.
That is, $s_1=x+y+z$, $s_2=xy+xz+yz$, $s_3=xyz$.
Suppose
\beq f(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_0 \eeq
is a monic polynomial of degree $n$ with
roots $\ah_1,\ah_2,\ldots,\ah_n$. (When $\ah$ is a root of multiplicity $m$ just
write it $m$ times here.) Then
\beq
f(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_0 = (x-\ah_1)\cdots (x-\ah_n)
\eeq
Multiplying out the product, $x^{n-1}$ has coefficient $-(\ah_1+\ldots+\ah_n)$,
the constant coefficient is $(-1)^n\ah_1\cdots \ah_n$ and, in general, the
coefficient of $x^{n-i}$ is $(-1)^i$ times the value of the $i$-th symmetric
polynomial $s_i$ on the values $\ah_1,\ldots,\ah_n$. For $n=3$, calling the
roots $\ah,\beta,\gam$, we have
\beq a_1 = -(\ah+\beta+\gam) \eeq
\beq a_2 = (\ah\beta+\beta\gam+\ah\gam) \eeq
\beq a_3 = -(\ah\beta\gam) \eeq
\begin{theorem}\label{g1} Any symmetric polynomial in $x_1,\ldots,x_n$ can be expressed
in terms of the elementary symmetric polynomials.
\end{theorem}
This can get pretty bogged down in notation so we will first only show the
argument for $n=3$ with variable $x,y,z$. We say two monic momomials (i.e., with constant
coefficient one) have the same {\em form} if they have the same exponents with
the same multiplicities. For any monic monomial $g(x)$ we let $\ol{g(x)}$ denote the
sum of all monomials with that form. Thus for $a,b,c$ distinct nonnegative integers
\beq \ol{x^ay^bz^c}:= x^ay^bz^c+x^az^by^c+y^ax^bz^c+y^az^bx^c+z^ax^by^c+z^ay^bx^c
\eeq
We do not double count so, for example,
\beq \ol{x^2y^2z} = x^2y^2z+x^2z^2y+ y^2z^2x \eeq
(That is, we don't count, say, $y^2x^2z$ separately. This is only a technical point.)
A monoic monomial is determined by the exponents written in decreasing order, that is,
$(a_1,\ldots,a_n)$ with $a_1\geq\ldots\geq a_n\geq 0)$ or, for $n=3$, simply $(a,b,c)$
with $a\geq b\geq c \geq 0$.
Any symmetric polynomial can be expressed as a linear combination of these
so it suffices (we are now doing only the case $n=3$) to write $\ol{x^ay^bz^c}$
in terms of $s_1,s_2,s_3$. We actually get the expression by reducing to simpler
(more on that later) terms.
If $a,b,c$ are positive, simply take out a common factor of $s_3=xyz$.
If $a=b>0=c$ express
\beq \ol{x^ay^a} = (xy+xz+xy)^a - \Delta \eeq
As the other two parts are symmetric their difference
$\Delta$ is also symmetric. But all the terms in $\Delta$ have all three
variables and so they can be reduced.
{\tt Example:} With $a=b=2, c=0$:
\beq (x^2y^2+x^2z^2+y^2z^2)= (xy+xz+yz)^2 - 2xyz(x+y+z) \eeq
If $a>b>0=c$ express
\beq \ol{x^ay^b} = \ol{x^by^b}\cdot \ol{x^{a-b}} + \Delta \eeq
The polynomials $\ol{x^by^b}$ and $\ol{x^{a-b}}$, being smaller, have already
been done. Again, $\Delta$, as the difference of symmetric polynomials, is
symmetric and again all terms in $\Delta$ have all three variables and so
they can be reduced.
{\tt Example:} With $a=5$, $b=2$, $c=0$:
\beq \ol{x^5y^2}= (x^2y^2+x^2z^2+y^2z^2)(x^3+y^3+z^3)-x^2y^2z^2(x+y+z) \eeq
Finally when $a>0=b=c=0$ we express
\beq x^a+y^a+z^a = (x+y+z)^a + \Delta \eeq
Again $\Delta$ is symmetric, and consists only of terms on two or three
variables, which we have already done.
{\tt Example:} With $a=4$, $b=0$, $c=0$:
\beq x^4+y^4+z^4 = (x+y+z)^4 - 4\ol{x^3y} - 6\ol{x^2y^2} - 12xyz(x+y+z) \eeq
There is a powerful consequence.
\begin{theorem}\label{g2} Let $L$ be any subfield of $C$. Let $f(x)\in L[x]$
be a polynomial of degree $n$ with complex roots $\ah_1,\ldots,\ah_n$.
(For multiple roots we repeat the root.) Then any symmetric polynomial of
$\ah_1,\ldots,\ah_n$ is in $L$.
\end{theorem}
{\tt Proof:} From Theorem \ref{g1} we write any symmetric polynomial in
terms of the elementary symmetric polynomials and their values are $\pm$
the coefficients of $f(x)$, which are in $L$.
{\tt Example:} Take $L=Q$ (the main case we shall use) and let
$\ah,\beta,\gam$ be the roots of $f(x)=x^3+x^2+2x+1$. Consider
$\kappa=\ah^3+\beta^3+\gam^3$. We express
\beq \kappa = (\ah+\beta+\gam)^3-3(\ah^2\beta+\cdots+\gam^2\beta)-6\ah\beta\gam
\eeq
and further reduce
\beq (\ah^2\beta+\cdots+\gam^2\beta)=(\ah\beta+\ah\gam+\beta\gam)(\ah+\beta+\gam)
- 3\ah\beta\gam
\eeq
We know $\ah+\beta+\gam=-1$ and $\ah\beta+\ah\gam+\beta\gam= 2$ and
$\ah\beta\gam =-1$ so
$\ah^2\beta+\ldots+\gam^2\beta=2(-1)-3(-1)= -5$ and so
$\kappa=(-1)^3-3(-5)-6(-1)=-10$.
Lets return to Theorem \ref{g1}.
How do we turn our arguments into a rigorous proof for a general number of
variables $n$. For each $\vec{a}=(a_1,\ldots,a_n)$ with
$a_1\geq\ldots a_n\geq 0$
let $MM(\vec{a})$ denote the monic polynomial
\beq \label{g4} MM(\vec{a}) = \ol{x_1^{a_1}\cdots x_n^{a_n}} \eeq
We set $D=D(\vec{a})=a_1+\ldots a_n$ and call $D(\vec{a})$ the degree of $\vec{a}$.
(Note it is the degree of the associated monic polynomial.
The idea is to subtract off from $MM(\vec{a})$ some combination of elementary
symmetric polynomials so as to be left with simpler forms. But what do
we mean by simpler? The notion of simpler, and even more the analysis of simpler,
shall be quite complicated! We define an ordering on the possible $\vec{a}$. Let
$\vec{a}=(a_1,\ldots,a_n)$ and $\vec{b}=(b_1,\ldots,b_n)$ with
$a_1\geq\ldots a_n\geq 0$ and $b_1\geq\ldots b_n\geq 0$.
\ben
\item\label{g5} If $D(\vec{a})\neq D(\vec{b})$ then the one with
the smaller degree is called
simpler.
\item\label{g6} Now suppose $b_1+\ldots+b_n=a_1+\ldots+a_n$. Let $i$ be the
smallest index (it may be that $i=1$) with $a_i\neq b_i$. If $a_i < b_i$ we
then say $\vec{a}$ is simpler than $\vec{b}$, else $\vec{b}$ is simpler.
\een
Among the $\vec{a}$ with the same sum of coordinates, simpler can be thought
of as a lexicographical ordering of the possible vectors, thought of as
words.
We want to show that, for any $\vec{a}$, $MM(\vec{a})$ can be expressed in
terms of the elementary symmetric functions. We do this by a double induction,
first on the degree $D$ and then, amongst those of a given degree $D$, in
order of simplicity as given by (\ref{g6}) above. To start the induction, for
$D=1$ the only vector is $\vec{a}=(1,0,\ldots,0)$ and $MM(\vec{a})=s_1$.
Now suppose the result is true for all $\vec{b}$ of degree less than $D$
and for all $\vec{b}$ simpler than $\vec{a}$. If $a_n\neq 0$ we reduce
by writing
\beq\label{g7} MM(\vec{a}) = s_n\cdot MM((a_1-1,\ldots,a_n-1)) \eeq
That is, we take out the common factor of $s_n=x_1\cdots x_n$ and are left
with something of smaller degree. Now we come to the main case. We write
(we assume $a_n=0$)
\beq\label{g8}
MM(\vec{a}) = s_1^{a_1-a_2}s_2^{a_2-a_3}\cdots s_{n-1}^{a_{n-1}-a_n} + \Del
\eeq
Observe that the product on the RHS is a symmetric polynomial of degree
$D$ and so $\Del$ is a symmetric polynomial of degree $D$ and so we only have
to check that all of the forms in $\Del$ are simpler than $\vec{a}$.
We pause for an example. Consider $n=5$ and $\vec{a}=(10,7,4,0,0)$. Call
the variables $v,w,x,y,z$ for convenience. Then
(\ref{g8}) becomes
\beq \label{g9}
\ol{v^{10}w^7x^4}=(v+w+x+y+z)^3(vw+\ldots+yz)^3(vwx+\ldots + xyz)^4 + \Del
\eeq
Looking at the leftmost terms in the products on the right we get
$v^3(vw)^3(vwx)^4$ which is precisely the $v^{10}w^7x^4$ that we want.
The general term consists of three from $v,\ldots,z$, three from $vw,\ldots,yz$
and four from $vwx,\ldots,xyz$. One such term would be taking $v,v,w;vw,vw,wx,
vwx,vwx,vwy,vwy$. That gives $v^8w^8x^3y^2$ and, indeed, $(8,8,3,2,0)$ is simpler
than $(10,7,4,0,0)$.
It may be helpful to think of the forms of these monomials in terms of balls in
bins. Imagine $n$ bins labelled with the $n$ variables $x_1,\ldots,x_n$. The
monomial polynomial $MM(\vec{a})$ corresponds to placing $a_i$ balls in the
bin marked $x_i$ Here is a picture corrsponding to the $\vec{a}$ of the example.:
\begin{verbatim}
x
x
x
x x
x x
x x
x x x
x x x
x x x
x x x
- - - - -
v w x y z
\end{verbatim}
Now each term from $s_i$ consists of $i$ balls in $i$ different bins. A term in the product
consists of placing $i$ balls in $i$ different bins $a_i-a_{i+1}$ times for $1\leq i \leq n$.
In the example above we have
$v^8w^8x^3y^2$ as follows:
\begin{verbatim}
x
x
x
x x
x x
x x
x x x
x x x
x x x
x x x
- - - - -
v w x y z
\end{verbatim}
The claim is that no matter how we
place $i$ balls in $i$ different bins $a_i-a_{i+1}$ times for $1\leq i \leq n$ we
end up with a $\vec{b}=(b_1,\ldots,b_n)$ which is simpler than $\vec{a}$.
First look at $b_1$. We can have at most one ball in a bin from each placement
and so
\beq \label{g10}
b_1 \leq (a_1-a_2)+(a_2-a_3)+\ldots+ (a_{n-1}-a_n) = a_1
\eeq
Now consider the total number of balls in two bins. We get at most two balls in
the two bins from each placement except that the placement of one ball (corresponding
to the $a_1-a_2$ factors of $s_1$) gives only one ball in the two bins so
\beq \label{g11}
b_1+b_2 \leq (a_1-a_2)+2(a_2-a_3)+\ldots+ 2(a_{n-1}-a_n) = a_1 + a_2
\eeq
In general, for $1\leq j\leq n-1$ consider the total number of balls in any $j$ bins.
When $i\leq j$ and $i$ balls are placed in $i$ different bins there are at most
$i$ balls in those $j$ bins while when $i>j$ there are at most $j$ balls in thos
$j$ bins. Thus
\beq \label{g12}
\sum_{k=1}^j b_k \leq \sum_{i=1}^j i(a_i-a_{i+1}) + j\sum_{i=j+1}^{n-1} (a_i-a_{i+1})
= \sum_{k=1}^j a_k
\eeq
But (\ref{g12}), for $1\leq j\leq n-1$, implies $\vec{b}$ is simpler than (or equal to)
$\vec{a}$. If $\vec{b}\neq \vec{a}$ let $j$ denote the first coordinate for which
$b_j\neq a_j$. As $b_k=a_k$ for $k1$ be such that $st=r$.
Set $\Lam^-=\{\sig_1(\ah),\ldots,\sig_s(\ah)\}$. These $s$ elements are
distinct as if $\sig_i(\ah)=\sig_j(\ah)$ then $\sig_i^{-1}\sig_j(\ah)=\ah$
which, as $K=L(\ah)$, would imply $\sig_i^{-1}\sig_j=e$, or $\sig_i=\sig_j$.
Further, each of the $\sig\in H^-$ permutes $\Lam^-$ as for any $\sig\in H^-$,
$\sig\sig_1,\ldots,\sig\sig_s$ ranges over $\sig_1,\ldots,\sig_s$.
Therefore all symmetric polynomials
in $\Lam^-$ are fixed by all $\sig\in H^-$. But we are assumming that
$L=(H^-)^{\dag}$ so any element fixed by all $\sig\in H^-$ must be in $L$.
Let $u_1,\ldots,u_s$ denote the values of the $i$-th symmetric polynomials
on $\ah_1,\ldots,\ah_s$. So these $u_1,\ldots,u_s\in L$. But then $\ah=\ah_1$
would satisfy the polynomial $x^s-u_1x^{s-1}+\ldots + (-1)^su_s$ which would
be a polynomial in $L[x]$ of degree $s$. This contradicts that, as $K=L(\ah)$,
the minimal polynomial of $\ah$ in $L[x]$ has degree $r$.
This complete the Galois Correspondence Theorem \ref{bigkahuna}. Theorem \ref{h1} can
be restated that for any subgroup $J$ (above $J$ is $H^-$) applying $\dag$
and then $*$ gets one back to $J$. Thus $*$ and $\dag$ are inverses of
each other and give a bijection as desired.
\section{Cyclotomic Fields}
Lets return to one of our original examples: $K=Q(\eps)$ with $\eps = e^{2\pi i/5}$.
The minimal polynomial for $\eps$ in $Q[x]$ is $(x^5-1)/(x-1)$ which has roots
$\eps,\eps^2,\eps^3,\eps^4$ which all all in $K$. Thus $K$ is the splitting field
of that polynomial (over $Q$) and hence $K$ is normal over $Q$.
The Galois Group is $Z_5^*$ which is cyclic. Let $\sig\in \Gamma[K:Q]$ be determined
by $\sig(\eps)=\eps^2$. Then $\sig^2(\eps)=\eps^4$ and $\sig^3(\eps)=\eps^8=\eps^3$
and $\sig^4(\eps)=\eps^{16}=\eps$ so $\sig^4=e$. There is one nontrivial subgroup:
$H=\{e,\sig^2\}$. From the Galois Correspondence theorem \ref{bigkahuna} this means there is one
nontrivial intermediate field $Q\subset L \subset Q(\eps)$ and $L= H^{\dag}$.
As $2=|H|=[Q(\eps):L]$ we have $[L:Q]=2$, so $L$ is a quadratic extension of $Q$.
To find $L$ we look for which $\ah\in Q(\eps)$ are in $H^{\dag}$. As $e$ fixes
all elements, $\ah\in Q(\eps)$ if and only if $\sig^2(\ah)=\ah$. Writing
$\ah=a+b\eps+c\eps^2+d\eps^3$ we must have
\beq \label{i1}
\ah = \sig^2(\ah) = a + b\eps^4 + c\eps^8 + d\eps^{12} = a + b(-1-\eps-\eps^2-\eps^3) + c\eps^3+ d\eps^2
\eeq
Equating the coefficients using the basis $1,\eps,\eps^2,\eps^3$ yields the equation system:
$a=a-b$, $b=-b$, $c=d-b$, $d=c-b$ which reduces to $b=0$, $c=d$. Thus the elements of $L$ may be
uniquely written as $a+c(\eps^2+\eps^3)$. Set $\kappa=\eps^2+\eps^3$. As $L=Q(\kappa)$, $\kappa$
must satisfy a quadratic. We find it by calculating $\kappa^2=\eps^4+2+\eps = 1 - \eps^2-\eps^3$.
Then $1,\kappa,\kappa^2$ are dependent, more precisely $\kappa^2=1-\kappa$. Solving the quadratic
gives
\beq\label{i2} \kappa = \frac{-1 \pm \sqrt{5}}{2} \eeq
(The actual sign is minus, but this method doesn't tell us that.) Thus we find $L=Q(\kappa)=Q(\sqrt{5})$.
When $p$ is an odd prime we can define $K=Q(\eps)$ with $\eps=e^{2\pi i/p}$. The minimal polynomial for
$\eps$ is $p(x)=(x^p-1)/(x-1)$ which has roots $\eps,\eps^2,\ldots,\eps^{p-1}$. Again, $K$ is normal
over $Q$. The Galois Groups $\Gamma[K:Q]$ has automorphisms $\sig_i$ given by $\sig_i(\eps)=\eps^i$
for each $i\in Z_p^*$ and $\sig_i\sig_j= \sig_{ij}$ where multiplication is done modulo $p$. Thus
$\Gamma[K:Q]\cong Z_p^*$. It is known that this is a cyclic group of order $p-1$,
so $\Gamma[K:Q]\cong (Z_{p-1},+)$. This group has a unique subgroup with half the elements, namely the
multiples of $2$ (thinking of it as $Z_{p-1}$). Hence, by the Galois Correspondence Theorem \ref{bigkahuna} there is
a unique quadratic extension of $Q$ lying inside of $Q(\eps)$.
Here is a way of finding the square root in $Q(\eps)$ for general odd prime $p$. Rather than the usual basis
$1,\eps,\ldots,\eps^{p-2}$ we use the basis ({\tt Exercise:} Show this is a basis.) $\eps,\eps^2,\ldots,\eps^{p-1}$.
The Galois Group $\Gamma(Q(\eps):Q)$ consists of $\sig_i$ for $1\leq i\leq p-1$ where $\sig_i(\eps)=\eps^i$.
Associating $\sig_i$ with $i\in Z_p^*$, the group is isomorphic to $Z_p^*$. The unique subgroup $H$ of $Z_p^*$
of index $2$ (that is, size $(p-1)/2$) consists of the squares (modulo $p$). That is $H$ has the automorphisms
$\sig_{k^2}$ for $1\leq k\leq p-1$. (Each square appears twice so there are $(p-1)/2$ elements of $H$.
Now write an arbitrary element $\ah\in Q(\eps)$ with the new basis as
\beq\label{i11} \ah = \sum_{i=1}^{p-1} a_i\eps^i \eeq
For $\ah\in H^{\dag}$ we need that for each $k$ we have $\sig_{k^2}(\ah)=\ah$. That is,
\beq\label{i21} \ah = \sig_{k^2}(\ah)=\sum_{i=1}^{p-1} a_i\eps^{k^2i} \eeq
Here as $\eps^p=1$ we can consider the exponent $k^2i$ as calcuated in $Z_p$. Thus the condition becomes
\beq\label{i12} a_i = a_{k^2i} \mbox{ for all } i,k\in Z_p^* \eeq
But (\ref{i12}) just says that $a_i$ is constant over the quadratic residues and constant (maybe a different
constant) over the quadratic nonresidues. ($0$ is special and is counted neither as a quadratic residue nor
as a quadratic nonresidue.)
Let $R,N\subset Z_p^*$ denote the sets of quadratic residues and
quadratic nonresidues respectively. Set
\beq\label{i13} \kappa = \sum_{r\in R} \eps^r = \frac{1}{2}\sum_{k=1}^{p-1}\eps^{k^2} \eeq
\beq\label{i14} \lam = \sum_{r\in N} \eps^r \eeq
Then $\kappa,\lam$ form a basis for $H^{\dag}$. It is convenient to note that
\beq\label{i15} \kappa+\lam = \sum_{r=1}^{p-1} \eps^r = -1 \eeq
so \beq\label{i15a} \lam= -1-\kappa \eeq and we can replace the basis $\kappa,\lam$ with the basis $1,\kappa$. Thus
\beq \label{i16} H^{\dag} = \{ a+ b\kappa : a,b\in Q \} \eeq
is the unique quadratic extension of $Q$ inside $Q(\eps)$. Thus $\kappa$ satisfies a quadratic equation (and is not itself rational)
and one can write $\kappa=a_1+a_2\sqrt{d}$ so that the unique quardatic extension of $Q$ inside $Q(\eps)$ can be written
$Q(\sqrt{d})$.
One can also find $\kappa$ explicitly. From (\ref{i13}) we find
\beq\label{i16b} \kappa^2 = \frac{1}{4}\sum_{x,y=1}^{p-1} \eps^{x^2+y^2} \eeq
This gets into some interesting number theory. For each $r\in Z_p$ one
examines the number of solutions to the equation $x^2+y^2=r$ over $Z_p$ with $x,y\neq 0$.
Let $r,s$ both be quadratic residues. Then we can write $r=st^2$. Each solution
$x^2+y^2=r$ corresponds to a solution of $x_1^2+y_1^2=s$ by setting $x_1=xt, y_1=yt$.
We can go in the other direction, dividing a solution by $t$. Thus there is a
value, call it $R$. so that $x^2+y^2=r$ has precisely $R$ solutions for every
quadratic residue $r$. Now let $r,s$ both be quadratic nonresidues. Again we
can write $r=st^2$ and again the number solutions is the same. Thus there is value
value, call it $N$. so that $x^2+y^2=r$ has precisely $N$ solutions for every
quadratic nonresidue $r$. Also, let $Z$ be the number of solutions to $x^2+y^2=0$.
We apply (\ref{i16b}) to find
\beq \label{i17} \kappa^2 = \frac{1}{4}[Z+R\kappa+N\lam] = \frac{1}{4}[Z+R\kappa+L(-1-\kappa)]
= \frac{1}{4}[(Z-L)+(R-L)\kappa] \eeq
which we can solve by the quadratic formula. Actually, we won't know the choice of $\pm$ in
the quadratic formula, but in either case we get $Q(\kappa)=Q(\sqrt{d})$ for the same explicit
$d$.
{\tt Example:} Take $p=11$ and $\eps=e^{2\pi i/11}$. The residues are $1,4,9,16=5,25=3$ so the nonresidues are
$2,6,7,8,10$. Then
\beq \label{i17b} \kappa = \eps+\eps^3+\eps^4+\eps^5+\eps^9 \eeq
Now consider the terms in $\kappa^2$, always reducing the exponent modulo $11$. We get (this
is not always the case!) no terms of $\eps^0$. We get $2\eps^3\eps^9=2\eps^1$ as well as
$2\eps^3,2\eps^4,2\eps^5,2\eps^9$. For the nonresidues we get $\eps^1\eps^1+2\eps^4\eps^9=
3\eps^2$ as well as $3\eps^6,3\eps^7,3\eps^8,3\eps^{10}$. Thus
\beq\label{i18} \kappa^2 = 2\kappa+3\lam = 2\kappa + 3(-1-\kappa) = -3-\kappa \eeq
so that
\beq\label{i19} \kappa = \frac{-1 \pm \sqrt{-11}}{2} \eeq
The unique quadratic field inside $Q(\eps)$ is therefore $Q(\sqrt{-11})$.
Hmmmmm, when $p=5$ the quadratic field was $Q(\sqrt{5})$ and when $p=11$ the
quadratic field was $Q(\sqrt{-11})$. Coincidence? No! The quadratic
field will be $Q(\sqrt{p})$ when $p$ is a prime of the form $4k+1$ and
will be $Q(\sqrt{-p})$ when $p$ is a prime of the form $4k+3$. But we'll
leave this nice fact unproven.
\section{Assorted Consequences}
Suppose that an irreducible $p(x)\in Q[x]$ of degree $n$
has complex roots $\ah_1,\ldots,\ah_n$ and we set
$K=Q(\ah_1,\ldots,\ah_n)$. Each $\sig\in \Gamma[K:Q]$
permutes the roots though not every permutation of the
roots yields an automorphism $\sig$.
Suppose $\rho$ is a polynomial function of $\ah_1,\ldots,\ah_n$
which is symmetric. Then every $\sig\in \Gamma[K:Q]$ has
$\sig(\rho)=\rho$. Hence $\rho\in Q$. As an example suppose
a cubic $p(x)\in Q[x]$ has roots $\ah,\beta,\gam$ and let
\beq \label{j1} \rho=(\ah-\beta)^2(\ah-\gam)^2(\beta-\gam)^2 \eeq
Any permutation of $\ah,\beta,\gam$ fixes $\rho$ and hence $\rho$
is a rational number.
(FYI: this is called the {\em discriminant} and generalizes the
famous $b^2-4ac$ term with quadratics.)
When $\rho$ is not fully symmetric in $\ah_1,\ldots,\ah_n$ there
is still some information to be gleaned. Suppose $\kappa$ is fixed
by the alternating group, the even permutations of $\ah_1,\ldots,\ah_n$.
If $\Gamma[K:Q]$ is contained in the alternating group then $\kappa\in Q$
as before. Otherwise, $\Gamma[K:Q]$ would have more than $n!/2$ elements
and so would be the full symmetric group of $\ah_1,\ldots,\ah_n$. In
that case $\kappa$ would not be in $Q$ since it isn't fixed by all $\sig\in \Gamma[K:Q]$.
Letting $H$ be the alternating group, as $|H|=|G|/2$, $[H^{\dag}:Q]=2$. Then
$\kappa$ would be in a quadratic extension of $Q$. Continuing the cubic
example above, now set
\beq \label{j2} \kappa=(\ah-\beta)(\ah-\gam)(\beta-\gam) \eeq
Assume $\Gamma[K:Q]\cong S_3$. Of the six permutations of $\ah,\beta,\gam$,
three send $\kappa$ to itself and the other three send $\kappa$ to $-\kappa$
(which is not $\kappa$ as $\kappa\neq 0$ as $\ah,\beta,\gam$ are distinct).
(For example, if $\ah,\beta$ are flipped and $\gam$ stays where it is then
$\kappa$ goes to $-\kappa$ but if $\ah$ goes to $\beta$ which goes to $\gam$
which goes to $\ah$ then $\kappa$ goes to $\kappa$.) Then $[Q(\kappa):Q]=2$
so $\kappa$ can be expressed in terms of a square root. Since, further,
$\kappa^2=\rho\in Q$, $\kappa$ will be the square root of a rational number.
\end{document}