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\begin{center} {\Large\bf Algebra V63.0349 \\ Assignment 11} \\ Due Friday, May 1 in recitation \end{center}
\beq
Every universe, our own included, begins in conversation.
\\ -- Michael Chabon
\enq
\ben
\item Let $f(x)=x^3+ax^2+bx+c\in Q[x]$ be an irreducible cubic with
roots $\ah,\beta,\gam\in C$. Using the
ideas of Cardano's Formula find a sequence
\[ Q = K_0 \subset K_1 \subset \cdots \subset K_r = K \]
such that $\ah,\beta,\gam\in K$ and so that
for each $K_i=K_{i-1}(\beta_i)$ with either $\beta_i^2 \in K_{i-1}$
or $\beta_i^3\in K_{i-1}$. Try to make the sequence as short as you
can. [It helps to add $\omega = e^{2\pi i/3}$ first. Then when
you want all cube roots of some $\zeta$ you only need one extension.]
\item Let $f(x)=x^4+ax^3+bx^2+cx+d \in Q[x]$ be an irreducible quartic with
roots $\ah,\beta,\gam,\delta\in C$. Using the
ideas for solving quartics find a sequence
\[ Q = K_0 \subset K_1 \subset \cdots \subset K_r = K \]
such that $\ah,\beta,\gam,\del\in K$ and so that for
each $K_i=K_{i-1}(\beta_i)$ with either $\beta_i^2 \in K_i$
or $\beta_i^3\in K_i$. Try to make the sequence as short as you
can. (Note that you don't actually need fourth roots!)
\item Set $\eps=e^{2\pi i/23}$. Find an explicit nonsquare $m$ so
that $Q(\sqrt{m})\subset Q(\eps)$.
\item Let $p$ be an odd prime. Set $\ah=2^{1/p},\eps=e^{2\pi i/p}$,
$f(x)=x^p-2$. Let $K$ denote the splitting field of $f(x)$ over $Q$.
\ben
\item Express {\em all} the roots of $f(x)$ in terms of $\ah,\eps$.
\item Show that $Q(\eps):Q$ is a normal extension.
\item Show that $Q(\ah):Q$ is not a normal extension.
\item\label{one} Show that $[K:Q]=p(p-1)$. (Hint: $[Q(\eps):Q]$ and
$[Q(\ah):Q]$ are relatively prime.)
\item Let $\sig\in \Gam(K:Q)$. Show that $\sig(\eps)=\eps^s$ and
$\sig(\ah) = \ah\eps^t$ for some $1\leq s\leq p-1$ and $0\leq t \leq p-1$.
\item Show that the values $s,t$ above would determine $\sig$.
\item Show (use the Galois Correspondence theorem and (\ref{one}) that
for each such $s,t$ there is such a $\sig$.
\een
\item Continuing the previous problem, let $\sig_{s,t}$ denote that
$\sig$ with $\sig(\eps)=\eps^s$ and $\sig(\ah)=\ah\eps^t$. Let $G$
denote the Galois Group $\Gamma(K:Q)$ so that
$G=\{\sig_{s,t}: 1\leq s\leq p-1, 0\leq t \leq p-1\}$.
Further, let $H=\{\sig_{1,t}:0\leq t\leq p-1\}$.
\ben
\item Suppose
$\sig_{s,t}\sig_{u,v} = \sig_{w,x}$.
Express $w,x$ in terms of $s,t,u,v$.
\item Is $G$ Abelian? (Give reason!)
\item Precisely which $\sig_{s,t}$ are in $Q(\eps)^*$?
\item Precisely which $\sig_{s,t}$ are in $Q(\ah)^*$?
\item Show that $H$ is a subgroup of $G$
\item Show that $H$ is an Abelian group.
\item Show that $H$ is a Normal subgroup of $G$.
\item Show that $G/H$ is an Abelian group. [{\tt Remark:}
Our big theorem was that if $\ah$ is expressible there
is a tower from $\{e\}$ to the Galois Group.
Our $\ah$ is immediately expressible (it {\em is} a $p$-th
root) and this exercise gives the explicit tower
$\{e\}\subset H \subset G$ with the desired properties.
\een
\een
\beq
If you know a thing only qualitatively you know it no more
than vaguely. If you know it quantitatively - grasping some
numerical measure that distinguishes it from
an infinite number of other possibilities - you are beginning
to know it deeply. You comprehend some of its beauty and you
gain access to its power and the understanding it provides. Being
afraid of quantification is tantamount to disenfranchising yourself,
giving up on one of the most potent prospects for understanding and
changing the world.
\\ -- Carl Sagan
\enq
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