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\begin{center} {\Large\bf Algebra V63.0344 \\ Assignment 8} \\ Due,
Friday, April 4 in Recitation \end{center}
\beq
We are surrounded by ideas and objects infinitely more ancient than
we imagine and yet at the same time everything is in motion.
-- Tielhard
\enq
\ben
\item Let $\eps=e^{2\pi i/18}$. (You might check the
posted Old
Midterm, with solutions.)
Now assume (and this is true) that $[Q(\eps):Q]=6$.
Let $p(x)\in Q[x]$ be the minimal polynomial for $\eps$.
\ben
\item What is the degree of $p(x)$?
\item Give all the roots of $p(x)$.
\item Describe all $\sig\in \Gamma[Q(\eps):Q]$ in a nice way. Give a
table of all products in $\Gamma[Q(\eps):Q]$
\item What well known group is $\Gamma[Q(\eps):Q]$ isomorphic to?
Give the isomorphism explicitly.
\een
\item Let $K=Q(\sqrt{a_1},\ldots,\sqrt{a_s})$ with $a_1,\ldots,a_s\in Q$.
\ben
\item Prove $[K:Q]\leq 2^s$.
\item Let $\sig\in \Gamma[K:Q]$. Prove $\sig^2=e$.
\item Let $\sig,\tau\in \Gamma[K:Q]$. Prove $\sig\tau=\tau\sig$.
\een
\item Let $p(x)\in Q[x]$ be an irreducible cubic with roots $\ah,\beta,\gam\in C$.
Let $K=Q(\ah,\beta,\gam)$. Suppose $\sqrt{a}\in K$ where $a\in Q$ and $\sqrt{a}\not\in Q$.
Set $L=Q(\sqrt{a})$. Prove $p(x)$ is still irreducible when considered in $L[x]$.
(Hint: When cubics reduce they have a root.)
\item Set $\ah=2^{1/4}$, $\beta=i\ah$, $\gam=-\ah$, $\del=-i\ah$. Set
$p(x)=x^4-2$. Set $K=Q(\ah,\beta,\gam,\del)$. Set $L=Q(i)$. Set $\Gam=\Gam[K:Q]$,
the Galois Group of $K$ over $Q$.
\ben
\item Show that $p(x)$ is irreducible in $Q[x]$.
\item \label{one} Give the factorization of $p(x)$ into irreducible factors in $K[x]$.
\item Show that $p(x)$ is irreducible in $L[x]$. (As $p(x)$ is quartic it
is not enough to look for factors as, a priori, it could be the product of
two quadratics. But given your factorization for problem \ref{one} any factorization
over the smaller field $L$ must come from joining together factors from \ref{one}.
Show that none of them work.)
\item Show $K=Q(\ah,i)$.
\item Show that $[K:Q]=8$ and give a basis for $K$ over $Q$.
\item Show that $|\Gam|\leq 24$. (This follows from general principles.)
\item Show that actually $|\Gam|\leq 8$. (Idea: $\sig(\ah)$ determines $\sig(\gam)$.)
\item List the eight possible permutations of $\ah,\beta,\gam,\del$ that could
come from a $\sig\in \Gam$.
\item Actually $\Gam$ is given by the eight permutations that you just found. Given
this, show that $\Gam$ is not Abelian.
\een
\een
\beq
There is a theory which states that if ever anybody discovers exactly what
the Universe is for and why it is here, it will instantly disappear and be
replaced by something even more bizarre and inexplicable. There is another
theory which states that this has already happened.
\\ Douglas Adams
\enq
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