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\begin{center} {\Large\bf Honors Algebra II \\ Solutions: Assignment 11} \end{center}
For the first two problems let $f(x)\in Q[x]$ be an irreducible
quartic with roots $\ah,\beta\gam,\del$ and set $K=Q(\ah,\beta,\gam,\del)$,
its splitting field. We will {\em assume} $\Gam[K:Q]$ is isomorphic to
$S_4$, the full symmetric group on $4$ elements. We will represent
$\sig\in \Gam[K:Q]$ as permutations on $\ah,\beta,\gam,\del$. Use
the following result from Group Theory with $n=4$: The {\em only}
subgroup of $S_n$ with index two is the alternating group $A_n$.
\ben
\item
Set $\kappa = \ah\beta+\gam\del$.
\ben
\item
Find a group $H\subset S_4$ of eight permutations
so that $\sig(\kappa)=\kappa$ for all $\sig\in H$.
\So You can flip $\ah,\beta$ and/or $\gam,\del$ giving
$e,(\ah\beta),(\gam\del),(\ah\beta)(\gam\del)$. Also
you can interchange $\{\ah,\beta\}$ with $\{\gam,\del\}$
giving $(\ah\gam)(\beta)(\del)$, $(\ah\del)(\beta\gam)$
and two cycles $\ah\ra\\gam\ra\beta\ra\del\ra\ah$ and
$\ah\ra\del\ra\beta\ra\gam\ra\ah$. This is a nice
nonabelian group on eight elements.
\item Assume $\sig(\ah)=\ah$, $\sig(\beta)=\gam$, $\sig(\gam)=\beta$, $\sig(\del)=\del$.
Show that $\sig(\kappa)\neq \kappa$.
\So $\sig(\kappa)=\ah\gam+\beta\del$. If $\kappa=\sig(\kappa)$ then $(\ah-\del)(\beta-\gam)=0$.
But irreducible polynomials have distinct roots, a contradiction.
\item Show that $Q(\kappa)^*=H$.
\So It contains $H$, which has eight elements but isn't all of $S_4$, which has $24$ elements. By
Lagrange's Theorem there is no group $H'$ strictly between $H$ and $S_4$.
\item Deduce that $[Q(\kappa):Q]= 3$.
\So From the counting part of the Galois Correspondence Theorem, as $H$ has index three in $S_4$, $H^{\dag}=Q(\kappa)$ has dimension three over $Q$.
\een
\item Set $L=Q(\ah)$. Set $H=L^{*}$.
\ben
\item Give six $\sig\in H$.
\So The six permutation of $\beta,\gam,\del$, keeping $\ah$ fixed.
\item Using the counting relations of the Galois Correspondence
Theorem show that $H$ consists of precisely the six $\sig$ you
described in the first part.
\So We are assumming $\Gam[K:Q]$ is $S_4$ with $24$ elements so
$[K:Q]=24$. As $[L:Q]=4$, $|L^*|=6$. As we've found six elements
of $L^*$ we have found them all.
\item Show that there is no group $H^+$ with $H\subset H^+\subset
S_4$ that has precisely $12$ elements.
\So From group theory we need only consider $H^+=A_4$. But
$H$ has the permutation $(\beta\gam)$ which is an odd
permutation so $H$ is not a subgroup of $H^+$.
\item Now using the Galois Correspondence Theorem show that
there is no field $M$, $Q\subset M \subset L$, with $[M:Q]=2$.
\So If such an $M$ existed we would have $L^*\subset M^*\subset S_4$
with $|M^*|=12$ which we showed above doesn't happen.
\een
\item Find $\Phi_{15}(x)$, the fifteenth cyclotomic polynomial,
explicitly.
\So The formula is $(x^{15}-1)/(\Phi_1(x)\Phi_3(x)\Phi_5(x))$. As
$x^5-1=\Phi_1(x)\Phi_5(x)$ we divide by that giving
$(x^{10}+x^5+1)/(x^2+x+1)$ which, after a bit of honest labor, yields
$x^8-x^7+x^5-x^4+x^3-x+1$.
{\tt Remark:} It appears from small examples that the coefficients
of the cyclotomic polynomials are either $0,1$ or $-1$. Indeed,
it was conjectured that they are bounded in absolute value, but
this was {\em disproved} by the great twentieth century mathematician
Paul Erd\H{o}s.
\item Let $p=2k+1$ be an odd prime.
Set $\eps=e^{2\pi i/p}$ and set $K=Q(\eps)$.
Let $\tau$ denote complex conjugation and set $H=\{e,\tau\}$.
Let $L$ denote the set of real numbers in $K$.
\ben
\item Show that $L=H^{\dag}$.
\So $\ah\in H^{\dag}$ if and only if $e(\ah)=\ah$ and $\tau(\ah)=\ah$.
But $e(\ah)=\ah$ always so the condition is that $\ah$ equals its
complex conjugate which occurs precisely when $\ah$ is real.
\item Find $[L:Q]$.
\So As $[K:Q]=p-1$ we again apply the Galois Correspondence Theorem
with values so that $[L:Q]=[H^{\dag}:Q]=|\Gam[K:Q]|/|H| = (p-1)/2$.
\item Set $\gam=\eps+\eps^{-1}$. Show $Q(\gam)^*=H$.
\So As $\eps^{-1}=\tau(\eps)$, $\tau(\eps+\eps^{-1})=\eps^{-1}+\eps$
so $\tau\in Q(\gam^*)$. The elements of $\Gam[K:Q]$ are given by
$\sig_j$, $1\leq j\leq p-1$, where $\sig_j(\eps)=\eps^J$. If
$\sig_j(\gam)=\gam$ we have $\eps+\eps^{-1}=\eps^j+\eps^{-j}$. We
can suppose $j\leq (p-1)/2$ as otherwise we could replace it
by $p-j$. Multiplying by $\eps^j$:
$\eps^{j+1}+\eps^{j-1}=\eps^{2j}+ 1$. With $j<(p-1)/2$ this
contradicts the independence of $1,\eps,\ldots,\eps^{p-2}$.
With $j=(p-1)/2$ you'd have
$\eps^{j+1}+\eps^{j-1}=\eps^{p-1}+ 1$. Replacing
$\eps^{p-1}$ with $-(1+\eps+\ldots+\eps^{p-2})$ one again
contradicts the independence of $1,\eps,\ldots,\eps^{p-2}$.
\item Write $\gam$ in terms of trig functions.
\So $\eps = \cos(2\pi/p) + i\sin(2\pi/p)$. $\eps^{-1}$ is
the complex conjugate of $\eps$. (For any $\beta$ on the unit
circle its inverse is its complex conjugate.) Thus
$\eps^{-1} = \cos(2\pi/p) - i\sin(2\pi/p)$. Thus
$\gam = 2\cos(2\pi/p)$.
\item Find the degree of minimal polynomial $f(x)\in Q[x]$ with
$\cos(2\pi/p)$ as a root.
\So As $|Q(\gam)*|=2$, $[Q(\gam):Q] = (p-1)/2$. But
$\cos(2\pi/p)=\gam/2$ so $[Q(\cos(2\pi/p):Q]=(p-1)/2$ and
so the minimal polynomial has degree $(p-1)/2$. It is an
interesting trigonometry problem, using formulae for
multiple angles, to find the polynomial.
\een
\een
\end{document}