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\begin{center} {\Large\bf Algebra V63.0349 \\ Assignment 10} \\ Due Friday, April 24 in recitation \end{center}
\beq
Simplicity is the highest goal, achievable when you have overcome
all difficulties. After one has played a vast quantity of notes and
more notes, it is simplicity that emerges as the crowning reward of
art.
\\ Fryderyk Chopin
\enq
For the first three 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 $\ah,\beta,\gam,\del$.
For problem \ref{one} below 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\label{two}
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 $\sig\in H$.
\item Deduce that $[Q(\kappa):Q]\leq 3$.
\item Show that $Q(\kappa)^{*}$ must be either all of $S_4$
or the group $H$ you found above.
\item Deduce that either $\kappa\in Q$ or $[Q(\kappa):Q]=3$.
\een
\item\label{three} Continuing, let $\sig(\ah)=\beta$,
$\sig(\beta)=\gam$, $\sig(\gam)=\ah$, $\sig(\del)=\del$.
\ben
\item Show that if $\sig(\kappa)=\kappa$ then some two of
$\ah,\beta,\gam,\del$ must be equal.
\item Why can't that happen?
\item Deduce the value of $[Q(\kappa):Q]$.
\item Prove that $\ah$ is square root constructible from
$\kappa$. By this we mean that there is a sequence
$u_1,\ldots,u_n=\ah$ with the usual conditions except
that we now allow $u_j=\kappa$.
\een
\item Set $L=Q(\ah)$. Set $H=L^{*}$.
\ben
\item Give six $\sig\in H$.
\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.
\item\label{one} Show that there is no group $H^+$ with $H\subset H^+\subset
S_4$ that has precisely $12$ elements.
\item Now using the Galois Correspondence Theorem show that
there is no field $M$, $Q\subset M \subset L$, with $[M:Q]=2$.
\een
\item Let $f(x)=x^6+ax^4+bx^2+c$ with $a,b,c\in Q$ be an irreducible
polynomial in $Q[x]$. ({\em Note:} All exponents even!) Let $K$ be the splitting field of $f(x)$
over $Q$ and let $G=\Gamma[K:Q]$. Prove $|G|\leq 48$.
\item Find $\Phi_{15}(x)$, the fifteenth cyclotomic polynomial,
explicitly. (Some grunt work here.)
\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 $L=H^{\dag}$.
\item Find $[L:Q]$.
\item Set $\gam=\eps+\eps^{-1}$. Show $Q(\gam)^*=H$.
\item Write $\gam$ in terms of trig functions.
\item Find the degree of minimal polynomial $f(x)\in Q[x]$ with
$\cos(2\pi/p)$ as a root.
\een
\een
\beq
She had been born with a map of time in her mind. She
pictured other abstractions as well, numbers and the
letters of the alphabet, both in English and in Bengali.
Numbers and letters were like links on a chain. Months
were arrayed as if along an orbit in space.
\\ Jhumpa Lahiri, The Lowland
\enq
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