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\begin{center} {\Large\bf Algebra V63.0349 \\ Assignment 11 Solutions} \end{center}
\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
\[ Q = K_0 \subset K_1 \subset \cdots \subset K_r = K \]
with $\ah,\beta,\gam\in K$ and such 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. [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.]
(This will be really messy in terms of $a,b,c$. Instead, use auxilliary
varibles.)
\So First by a translation we make it $y^3+dy+e$ and $d,e\in Q$ so
no extension. Then we need $\ah$ with $d\ah^{-2}=-3$ so $K_1=Q(\ah)$
and the equation is $z^3-3z+f=0$ with $f\in K_1$. Setting $z=v+v^{-1}$
gives $v^6+fv^3+1=0$. Setting $w=v^3$ we find $w\in K_2=K_1(\beta)$
where $\beta = (f^2-4)^{1/2}$. Following the hint we make
$K_3=K_2(\omega)$ with $\omega=e^{2\pi i/3}$. Then we make
$K_4 = K_3(v)$, so an extension by a cube root. Now we have
all three solutions to $v^3=w$ as the others are $v\omega$ and $v\omega^2$.
As $v\in K_4$, working backwards we find all three $z\in K_4$, then all
three $y\in K_4$ and finally all three $x\in K_4$. Note that the final
$K_4$ has $[K_4:Q]=2\cdot 2\cdot 2 \cdot 3 = 24$, so its actually quite
a bit bigger than the splitting field of $f(x)$.
\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
\[ Q = K_0 \subset K_1 \subset \cdots \subset K_r = K \]
with $\ah,\beta,\gam,\del\in K$ and such that
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!)
\So Let $\kappa=\ah\beta+\gam\del$, $\lam=\ah\del+\beta\gam$,
$\mu = \ah\del + \beta\gam$. Then $p(y)=(y-\kappa)(y-\lam)(y-\mu)\in Q[x]$
so by the previous problem we first find
$Q = K_0 \subset K_1 \subset \cdots \subset K_4 $ with $\kappa,\lam,\mu\in K_4$.
Then $\eps=\ah\beta-\gam\del$ has $\eps^2 = \kappa^2 - 4\ah\beta\gam\del$. As
the product is $d$, we set $K_5=K_4(\eps)$. Similarly we set $K_6=K_5(\theta)$
where $\theta=\ah\gam-\beta\del$ has $\theta^2 = \lam^2 - 4d$. With
$\eta = \ah\del - \beta\gam$ we have $\eps\theta\eta$ is a symmetric function
and so is in $Q$. Thus $\eta\in K_5$. But then (see the quartic notes)
all six products of the four roots are in $K_5$ and then $\ah,\beta\gam,\del\in K_6$
\item Set $\eps=e^{2\pi i/23}$. Find an explicit nonsquare $m$ so
that $Q(\sqrt{m})\subset Q(\eps)$.
\So The residues in $Z_{23}$ are $1,4,9,16,2,13,3,18,12.8.6$ and
the nonresidues are the other nonzero terms. So we set
\[ \kappa = \eps + \eps^2 + \eps^3 + \eps^4 + \eps^6 + \eps^8 + \eps^9 + \eps^{12}+\eps^{13} + \eps^{16} + \eps^{18} \]
and $\lam = -1-\kappa$ giving the $\eps^j$ over the nonresidues. Now we compute $\kappa^2$. We never get
an $1=\eps^{23}$ term. We get $\eps^1$ five times , $\eps^8\eps^{16}$ and $\eps^{16}\eps^8$
and $\eps^6\eps^{18}$ and $\eps^{18}\eps^6$
and $\eps^{12}\eps^{12}$.
But this means we get each residue $r$ appears five times as $\eps^{8r}\eps^{16r}$, etc. So these terms give $5\kappa$.
Take a nonresidue, we'll take $22$.
We get $\eps^{22}$ six times $\eps^3\eps^{19}$, $\eps^6\eps^{16},\eps^9\eps^{13}$ and their reverses. Any nonresidue
can be written $22r$ where $r$ is a residue and so will also appear six times. This gives $6\lam$.
Thus
\[ \kappa^2 = 5\kappa + 6\lam = 5\kappa + 6(-1-\kappa) = -6-\kappa \]
and
\[ \kappa= \frac{-1 \pm \sqrt{-23}}{2} \]
so that $\sqrt{-23}\in Q(\eps)$ and we take $m=-23$
\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.
\een
\een
\beq
Every Sunday my wife and I took a romantic little walk to
Grandchester, a lovely, lovely little town near Cambridge,
and we ate lunch at a pub there. We would stroll along the
road reciting pi to each other; she would do twenty places,
then I would do twenty and so forth.
\\ -- John Conway
\enq
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\een
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