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\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.
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