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\begin{document}
\begin{center} {\Large\bf Honors Algebra II \\ Assignment 8 Solutions}
\end{center}
\ben
\item Let $\eps=e^{2\pi i/18}$.
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 $p(x)$?
\So We always have that the degree of $K(\ah)$ is the same as the degree of
the minimal polynomial for $\ah$ in $K[x]$. Thus it is of degree $6$.
\item Give all the roots of $p(x)$.
\So From the midterm only $\eps,\eps^5,\eps^7,\eps^{11},\eps^{13},\eps^{17}$
can be roots but as $p(x)$ has degree $6$ it has six roots (recall that
we showed that an irreducible polynomial (provided the characteristic was $0$)
must have distinct roots) and so these are the roots, all the roots, nothing
but the roots.
\item Describe all $\sig\in \Gamma[Q(\eps):Q]$ in a nice way. Give a
table of all products in $\Gamma[Q(\eps):Q]$
\So Set $S=\{1,5,7,11,13,17\}$. For each $j\in S$ we have $\sig_j\in \Gamma[Q(\eps):Q]$
determined by $\sig_j(\eps)=\eps^j$. Then $\sig_j\sig_k(\eps) = \sig_k(\eps^j) = \eps^{jk}$
so $\sig_j\sig_k = \sig_{jk}$ where the index is determined modulo $18$. The table is
(writing $j$ for $\sig_j$):
\begin{center}
\begin{tabular}{l|rrrrrr}
- & 1 & 5 & 7 & 11 & 13 & 17 \\ \hline
1 & 1 & 5 & 7 & 11 & 13 & 17 \\
5 & 5 & 7 & 17 & 1 & 11 & 13 \\
7 & 7 & 17 & 13 & 5 & 1 & 11 \\
11 & 11 & 1 & 5 & 13 & 17 & 7 \\
13 & 13 & 11 & 1 & 17 & 7 & 5 \\
17 & 17 & 13 & 11 & 7 & 5 & 1 \\
\end{tabular}
\end{center}
\item What well known group is $\Gamma[Q(\eps):Q]$ isomorphic to?
Give the isomorphism explicitly.
\So This is a cyclic group on six elements. We can take $5$ as a
generator. Then
$\Gamma[Q(\eps):Q]$ is isomorphic to $(Z_6,+)$ with
$\phi: (Z_6,+) \ra \Gamma[Q(\eps):Q]$ is an isomorphism
with $\phi(0)=e$, $\phi(1)=\sig_5$, $\phi(2)=\sig_5^2=\sig_7$,
$\phi(3)=\sig_5^3=\sig_{17}$,
$\phi(4)=\sig_5^4=\sig_{13}$,
$\phi(5)=\sig_5^5=\sig_{11}$.
One way to see this is to note that $\Gamma[Q(\eps):Q]$ is Abelian and
to know that $(Z_6,+)$ is the only Abelian grouup on $6$ elements.
\een
\item Let $K=Q(\sqrt{a_1},\ldots,\sqrt{a_s})$ with $a_1,\ldots,a_s\in Q$.
\ben
\item Give an {\em upper bound} on $[K:Q]$.
\So $2^s$ from the Tower Theorem. But it might be less, for example $Q(\sqrt{2},\sqrt{3},\sqrt{6})$.
\item Let $\sig\in \Gamma[K:Q]$. Prove $\sig^2=e$.
\So For each $1\leq i\leq s$, $\sig(\sqrt{a_s})=\eps_s\sqrt{a_s}$ with $\eps_s\in \{-1,+1\}$.
Then \[ \sig^2(\sqrt{a_s}) = \sig(\eps_s\sqrt{a_s}) = \eps_s^2\sqrt{a_s} = \sqrt{a_s} \]
As $\sig^2$ preserves a set of elements that generate $K$ it preserves all of $K$ so it
is the identitly.
\item Let $\sig,\tau\in \Gamma[K:Q]$. Prove $\sig\tau=\tau\sig$.
\So With
$\sig(\sqrt{a_s})=\eps_s\sqrt{a_s}$
and
$\tau(\sqrt{a_s})=\gam_s\sqrt{a_s}$
we have
$\sig(\tau(\sqrt{a_s})) = \sig(\gam_s\sqrt{a_s}) = \eps_s\gam_s\sqrt{a_s}$
and
$\tau(\sig(\sqrt{a_s})) = \tau(\eps_s\sqrt{a_s}) = \gam_s\eps_s\sqrt{a_s}$
so $\sig\tau$ and $\tau\sig$ agree on all $\sqrt{a_s}$ and so they are equal.
\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 $L=Q(\sqrt{a})$. \ben \item Prove $p(x)$ is still irreducible when considered in $L[x]$.
(Hint: When cubics reduce they have a root.)
\So If $p(x)$ reduced it would have a root, say $\ah$. That is $\ah\in L$. But
we already have $[Q(\ah):Q]=3$ and all $\beta\in L$ have $Q(\beta):Q]\leq 2$,
a contradiction.
\item Prove $[K:Q]=6$.
\So We showed in general that when $p(x)\in Q[x]$ has roots $\ah_1,\ldots,\ah_n$
then $[Q(\ah_1,\ldots,\ah_n):Q]\leq n!$, in this case $3!=6$. For the lower
bound $L\subset K$ with $[L:Q]=2$. In $L[x]$, $p(x)$ is irreducible so
$[L(\ah):L]=3$ and hence $[L(\ah):Q]=6$ and $[K:Q]\geq [L(\ah):Q]=6$. Indeed,
this shows that $K=L(\ah)$, so that $\beta$ can be expressed in terms of $\ah$
and $\sqrt{a}$.
\een
\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 $Q[x]$.
\So Eisenstein's Criterion! Two divides coefficients 0,0,0,-2 and four doesn't divide
-2 and 2 doesn't divide the lead coefficient one.
\item \label{one} Give the factorization of $p(x)$ into irreducible factors in $K[x]$.
\So As $\ah,\beta,\gam,\del$ are roots and all are in $K$,
\[ p(x) = (x-\ah)(x-\beta)(x-\gam)(x-\del) \]
\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.)
\So As none of $\ah,\beta,\gam,\del$ is in $L$ if $p(x)$ reduced it would into
two quadratics and one would be either $(x-\ah)(x-\beta)$ or $(x-\ah)(x-\gam)$
or $(x-\ah)(x-\del)$ but then the constant coefficient would be either $\ah\beta=i\sqrt{2}$
or $\ah\gam= -\sqrt{2}$ or $\ah\del=-i\sqrt{2}$, none of which is in $L$.
\item Show $K=Q(\ah,i)$.
\So As $i=\beta/\ah$, $i\in K$. Clearly $\ah\in K$. so $Q(\ah,i)\subset K$.
But $\ah,\beta=i\ah,\gam=-\ah,\del=-i\ah$ are all in $Q(\ah,i)$ so $K\subset Q(\ah,i)$.
\item Show that $[K:Q]=8$ and give a basis for $K$ over $Q$.
\So We get a tower $Q\subset L \subset K$. $[L:Q]=2$ with basis $1,i$. As
$K=L(\ah)$ and $\ah$ satisfies an irreducible quartic in $L[x]$, $[K:L]=4$
with basis $1,\ah,\ah^2,\ah^3$. Thus $[K:Q]=8$ with basic all products,
namely $1,\ah,\ah^2,\ah^3,i,i\ah,i\ah^2,i\ah^3$.
\item Show that $|\Gam|\leq 24$. (This follows from general principles.)
\So Each $\sig\in \Gam$ is determined by a permutation of $\ah,\beta,\gam, \del$
and there are only $4!=24$ such permutations.
\item Show that actually $|\Gam|\leq 8$. (Idea: $\sig(\ah)$ determines $\sig(\gam)$.)
\So There are four choice for $\sig(\ah)$. But then $\sig(\gam)=-\sig(\ah)$ is
determined. Now there are two choices for $\sig(\beta)$ and then $\sig(\del)$
is determined -- giving $4$ times $2$ or $8$ choices for $\sig$.
\item List the eight possible permutations of $\ah,\beta,\gam,\del$ that could
come from a $\sig\in \Gam$.
\So We make a table:
\begin{center}
\begin{tabular}{rrrr}
$\sig(\ah)$ & $\sig(\beta)$ & $\sig(\gam)$ & $\sig(\del)$ \\ \hline
$\ah$ & $\beta$ & $\gam$ & $\del$ \\
$\ah$ & $\del$ & $\gam$ & $\beta$ \\
$\beta$ & $\ah$ & $\del$ & $\gam$ \\
$\beta$ & $\gam$ & $\del$ & $\ah$ \\
$\gam$ & $\beta$ & $\ah$ & $\del$ \\
$\gam$ & $\del$ & $\ah$ & $\beta$ \\
$\del$ & $\ah$ & $\beta$ & $\gam$ \\
$\del$ & $\gam$ & $\beta$ & $\ah$ \\
\end{tabular}
\end{center}
\item Actually $\Gam$ is given by the eight permutations that you just found. Given
this, show that $\Gam$ is not Abelian.
\So Many of the pairs don't commute, to show it isn't Abelian we need only find one.
For example, let $\sig$ be given by row two and $\tau$ by row three. Then
$(\sig\tau)(\ah)= \tau(\sig(\ah))=\tau(\ah)=\beta$ while
$(\tau\sig)(\ah)= \sig(\tau(\ah))=\sig(\beta)=\del$ so $\sig\tau\neq \tau\sig$.
\een
\een
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\een
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