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\begin{center} {\Large\bf Algebra V63.0349 \\ Assignment 11 Solutions} \end{center}
\ben
\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 $\ah,\eps$.
\So $\ah\eps^t$ for $0\leq t \leq p-1$.
\item Show that $Q(\eps):Q$ is a normal extension.
\So It is the splitting field of $x^p-1$ as all the other roots
$\eps^t\in Q(\eps)$.
\item Show that $Q(\ah):Q$ is not a normal extension.
\So $x^p-2$ is irreducible (Eisenstein's criterion) and has
$\ah$ as a root and $\ah\eps$ as root but $\ah\in Q(\ah)$ and
$\ah\eps\not\in Q(\ah)$ as all elements of $Q(\ah)$ are real.
\item\label{one} Show that $[K:Q]=p(p-1)$. (Hint: $[Q(\eps):Q]$ and
$[Q(\ah):Q]$ are relatively prime.)
\So $[Q(\eps):Q]=p-1$ and $[Q(\ah):Q]=p$ and since they are
relatively prime (see Assignment $5$), $[Q(\eps,\ah):Q]=p(p-1)$.
\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 $0\leq t \leq p$.
\So $\sig(\eps)$ must satisfy $\Phi_p(x)=(x^p-1)/(x-1)$ so must be
$\eps^s$. $\sig(\ah)$ must satisfy $x^p-2$ so must be $\ah\eps^t$.
\item Show that the values $s,t$ above would determine $\sig$.
\So As $K=Q(\ah,\eps)$ the values $\sig(\ah),\sig(\eps)$ determine $\sig$.
\item Show (use the Galois Correspondence theorem and \ref{one}) that
for each such $s,t$ there is such a $\sig$.
\So As $[K:Q]=p(p-1)$ there must be precisely $p(p-1)$ automorphisms
and there are only $p(p-1)$ possible automorphisms above so they must
all {\em be} automorophisms.
\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
$G=\{\sig_{s,t}: 1\leq s\leq p-1, 0\leq t \leq p-1\}$ from
the previous problem. 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 $s,t,u,v$.
\So We calculate
\[ (\sig_{s,t}\sig_{u,v})(\eps) = \sig_{u,v}(\eps^s) = \eps^{su} \]
where that $w=su$, calculated modulo $p$. Also
\[ (\sig_{s,t}\sig_{u,v})(\ah) = \sig_{u,v}(\ah\eps^t) =
= \sig_{u,v}(\ah)\sig_{u,v}(\eps^t) =
= \ah\eps^v(\eps^t)^u \]
where $x=tu+v$ calculated modulo $p$.
\item Is $G$ Abelian? (Give reason!)
\So No. If it were you would need (always!) $tu+v=sv+t$ which
is false. Take, for exampel, $s=t=v=1, u=2$.
\item Precisely which $\sig_{s,t}$ are in $Q(\eps)^*$?
\So $\sig\in Q(\eps)^*$ iff $\sig(\eps)=\eps$ iff $s=1$ so this
is $H$ as defined above.
\item Precisely which $\sig_{s,t}$ are in $Q(\ah)^*$?
\So $\sig\in Q(\ah)^*$ iff $\sig(\ah)=\ah$ iff $t=0$
\item Show that $H$ is a subgroup of $G$
\So Well, since it is $Q(\eps)^*$ it must be a group. But
we can show it directly (good finals practice!):
\ben
\item {\tt Identity:} $e=\sig_{1,0}\in H$
\item {\tt Product:} Let $\sig_{1,t},\sig_{1,t'}\in H$.
Then $\sig_{1,t}\sig_{1,t'}= \sig_{1,t+t'}\in H$.
\item {\tt Inverse:} As $\sig_{1,t}\sig_{1,-t}=e$ (with
$-t$ defined in $Z_p$) the inverse of $\sig_{1,t}$ is
indeed in $H$.
\item Show that $H$ is an Abelian group.
\So $\sig_{1,t+t'}=\sig_{1,t'+t}$ as $t+t'=t'+t$.
\item Show that $H$ is a Normal subgroup of $G$.
\So We can do this directly, calculating $\sig_{s,t}^{-1}\sig_{1,t'}\sig_{s,t}$.
Another way is to note $\sig\in H$ iff $\sig(\eps)=\eps$. Let $\sig\in H$
and $\tau\in G$. Then $(\tau^{-1}\sig\tau)(\eps) = \tau(\sig(\tau^{-1}(\eps)))$.
But $\tau^{-1}(\eps)$ must be some power of $\eps$ so $\sig$ doesn't move it
and $\tau(\sig(\tau^{-1}(\eps)))=\tau(\tau^{-1}(\eps))=\eps$.
\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 $\{e\}$ 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.
\So $\sig_{s,t},\sig_{s',t'}$ are in the same $H$-coset
if and only if $t=t'$. (One can calculate this directly
but its easier to note that $\sig_{s,t}\sig_{s',t'}^{-1}\in H$
iff
$\sig_{s,t}\sig_{s',t'}^{-1}(\ah)=\ah$ iff
$\sig_{s,t}(\ah)=\sig_{s',t'}(\ah)$ iff $t=t'$.)
Take $\sig_{1,t}$ as the coset representatives. As
$\sig_{1,t}\sig_{1,t'}=\sig_{1,t+t'}$, $G/H$ is
isomorphic to $Z_p$ under addition. ({\tt Note:} It
is {\em NOT} true that $G$ is isomorphic to $H\times (G/H)$.
Indeed, here is an instance where $H$and $G/H$ are both
Abelian but $G$ is not!)
\een
{\tt Remark:} Here is yet another way of looking at this interesting group.
Associate $s,t$ with the linear function $f:Z_p\ra Z_p$ given
by $f(x)=sx+t$. Then $G$ is isomorphic to the group of these
linear functions under composition.
\item In $Z[i]$ find $d=\gcd(18+23i,9+8i)$.
\So As $\frac{18+23i}{9+8i}=
\frac{346+63i}{145}$ we take the quotient $2+0i$ and write
\[ (18+23i) = 2(9+8i) + 7i \]
Now $\frac{9+8i}{7i} = \frac{8-9i}{7}$ so take quotient $1-i$ and write
\[ (9+8i) = (1-i)(7i) + (2+i) \]
Now $\frac{7i}{2+i} = \frac{7+14i}{5}$ so take quotient $1+3i$ and write
\[ 7i = (1+3i)(2+i) + 1 \]
Now $1$ divides $2+i$ so the gcd is $1$.
\item Let $F$ be a finite field with $q=p^n$ elements. Define
$\sig: F\ra F$ by $\sig(\ah)=\ah^p$. In assignment $7$ you
showed that $\sig$ (called the Frobenius automorphism) is
an automorphism of $F$ over $Z_p$.
\ben
\item Show that $\sig^n=e$.
\So $\sig^n(\ah)=\ah^q$ and $\ah^q=\ah$ for all $\ah\in F$.
\item Show that $\sig^i\neq e$, $1\leq i