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\begin{document}
\begin{center} {\Large\bf Algebra V63.0344 \\ Assignment 1}
\end{center}
\ben
\item An element $u\in R$ ($R$ a ring) is called a {\em unit} if it
has a multiplicative inverse, that is, there exists an element $v\in R$
such that $uv=1$. (When this occurs we can write $v=u^{-1}$ or $v=\frac{1}{u}$.)
Let $U$ be the set of units of $R$. Show that $U$ forms a group under multiplication.
The template for this is: \ben \item {\tt Identity:} $1\in U$ \item {\tt Inverse:}
If $u\in U$ then $u^{-1}\in U$. \item {\tt Product:} If $u,v\in U$ then $uv\in U$. \een
\noi {\tt Solution:}
\ben
\item {\tt Identity:} As $1\cdot 1=1$, $1\in U$.
\item {\tt Inverse:} If $u\in U$ then there is an element $v=u^{-1}\in R$. As $vu=1$,
$v\in U$.
\item {\tt Product:} If $u,v\in U$ there exist $w,x\in R$ with $uw=vx=1$ but then
$(uv)(wx)=1$ so $uv\in U$.
\een
\item Define (this will be standard)
\[ Z[\sqrt{2}] = \{a+b\sqrt{2}: a,b\in Z \} \]
Show that $Z[\sqrt{2}]$ is a Ring.
The template for showing $R$ is a ring when $R$ is a subset of the complex numbers is:
\ben
\item {\tt Identity:} $0\in R$
\item {\tt Inverse:} If $u\in R$ then $-u\in R$.
\item {\tt Sum:} If $u,v\in R$ then $u+v\in R$.
\item {\tt Product:} If $u,v\in R$ then $uv\in R$.
\een
\noi {\tt Solution:}
\ben
\item {\tt Identity:} $0=0+0\sqrt{2}\in Z[\sqrt{2}]$
\item {\tt Inverse:} If $u\in Z[\sqrt{2}]$ then $u=a+b\sqrt{2}$ for some $a,b\in Z$
so $-u=(-a)+(-b)\sqrt{2}\in Z[\sqrt{2}]$.
\item {\tt Sum:}
If $u,v\in Z[\sqrt{2}]$ then $u=a+b\sqrt{2}$, $v=c+d\sqrt{2}$ so
$u+v= (a+c)+(b+d)\sqrt{2}\in Z[\sqrt{2}]$.
\item {\tt Product:}
If $u,v\in Z[\sqrt{2}]$ then $u=a+b\sqrt{2}$, $v=c+d\sqrt{2}$ so
\[ uv = (ac+2bd)+ (ad+bc)\sqrt{2} \]
\een
\item Let $\ah= a+b\sqrt{2}\in Z[\sqrt{2}]$. Show
\ben
\item If $a^2-2b^2=\pm 1$ then $\ah$ is a unit.
\So ({\tt Caution:} Yes, the inverse of a nonzero element will exist
as a real number, the problem here is to show that that inverse is in
the ring.) We ``rationalize the denominator":
\[
\frac{1}{a+b\sqrt{2}}
= \frac{1}{a+b\sqrt{2}} \frac{a-b\sqrt{2}}{a-b\sqrt{2}}= \frac{a-b\sqrt{2}}{a^2-2b^2} \]
With the denominator $\pm 1$, both $a/(a^2-2b^2)$ and $-b/(a^2-2b^2)$ are integers.
\item (*) [The asterisk indicates a more difficult, but still required,
problem.] If $\ah$ is a unit then $a^2-2b^2=\pm 1$.
\So If $\ah$ is a unit then we must have the coefficients of $\ah^{-1}$ integral so that
$a^2-2b^2|a$
and $a^2-2b^2|b$.
Squaring,
$(a^2-2b^2)^2|a^2$
$(a^2-2b^2)^2|b^2$
so
$(a^2-2b^2)^2|(a^2-2b^2)$
so
$(a^2-2b^2)|1$ and so $a^2-2b^2=\pm 1$.
\\ {\tt Another Approach:} (This will tie in to work we will do later.) For
$\ah=a+b\sqrt{2}$ set $\ol{\ah}=a-b\sqrt{2}$ (this is {\em not} complex
conjugation!) and $f(\ah)=\ah\ol{\ah} = a^2-2b^2$. Then $f$ is multiplicative,
that is,
\[ (\ah\beta)= \ah\ol{\ah}\beta\ol{\beta} = (\ah\beta)\ol{\ah\beta} = f(\ah)f(\beta) \]
If $\ah\beta=1$ then $1=f(\ah)f(\beta)$. But for all nonzero $\ah\in Z[\sqrt{2}]$,
$f(\ah)=a^2-2b^2$ is a nonzero integer. For the product of two nonzero integers to
be one each must be $\pm 1$, so $f(\ah)=\pm 1$ as desired.
\een
\item Define (this will be standard)
\[ Q(\sqrt{2}) = \{a+b\sqrt{2}: a,b\in Q \} \]
Prove that $Q(\sqrt{2})$ is a field. The template for showing $R$ is a field when $R$
is a subset of the complex numbers is
that given for a ring above plus: If $\ah\in F$ and $\ah\neq 0$ then $\ah^{-1}\in F$.
\So The ring properties are basically the same as for $Z[\sqrt{2}]$ above. For
multiplicative inverse we again write
\[ \frac{1}{a+b\sqrt{2}} = \frac{a}{a^2-2b^2} + \frac{-b}{a^2-2b^2} \]
Now the coefficients $\frac{a}{a^2-2b^2}, \frac{-b}{a^2-2b^2}$ are ratios
of elements of $Q$ and so are in $Q$. The only problem would be if the
denominator $a^2-2b^2=0$. If $b\neq 0$ that would imply $(a/b)^2=2$,
contradicting the irrationality of $\sqrt{2}$. If $b=0$ then $a=0$ and
$0+0\sqrt{2}=0$, which is not supposed to have a multiplicative inverse.
\item Set
\[ \omega = e^{2\pi i/3} = \frac{-1+i\sqrt{3}}{2} \]
and set
\[ Z[\omega] = \{a+b\omega: a,b\in Z\} \]
[ Notations: $e^{i\theta} = \cos\theta + i\sin\theta$. Any nonzero complex
$\ah$ may be uniquely written $\ah=re^{i\theta}$ with $r>0$ real and $0\leq \theta< 2\pi$
and has polar coordinates $(r,\theta)$ when placed on the complex plane. For any complex
$\gam=c+di$ we write $\ol{\gam}=c-di$, the complex conjugate of $\gam$. We define the
absolute value $|\gam|=\sqrt{c^2+d^2}$, the distance from $\gam$ to the origin in the
complex plane. Note $\gam\ol{\gam} = c^2+d^2= |\gam|^2$.]
\ben
\item Draw a {\em careful} picture marking the points of $Z[\omega]$ on the complex plane.
You should get a pleasing pattern.
\So You should get a nice hexagonal picture.
\item Show $Z[\omega]$ is a ring. ({\tt Product} is the hard part!)
\So Same as $Z[\sqrt{2}]$ above except for {\tt Product}. As $\omega^2=-1-\omega$ we have
\[ (a+b\omega)(c+d\omega)=ac+(bc+ad)\omega + bd(-1-\omega)= (ac-bd)+(bc+ad-bd)\omega \]
\item For $\kappa=a+b\omega\in Z[\omega]$ write $|\kappa|^2$ as a function of $a,b$.
Show that $|\kappa|^2$ is an integer. Show that $|\kappa|^2$ is a positive integer
when $\kappa\neq 0$.
\So $|\kap|^2 = \kap\ol{\kap}= (a+b\omega)(a+b\ol{\omega})= a^2 + ab(\omega+\ol{\omega})
+ b^2\omega\ol{\omega}$. But $\omega+\ol{\omega} = -1$ and $\omega\ol{\omega}=1$ (indeed,
$\ol{\omega}=\omega^2$) so $|\kap|^2=a^2-ab+b^2$. This is an integer. As it is the square
of a distance it cannot be negative. It is zero only when $\kap=0$, so otherwise it
is a positive integer.
\item (*) Find all units (with proof that you have all!) of $Z[\omega]$.
\So The units are $1,1+\omega,\omega,-1,-1-\omega,-\omega$. These are the points of
a regular hexagon on the unit circle and can also be written (changing the order)
as
$e^{2\pi ij/6}$,
$0\leq j\leq 5$. Seen this way they are all units as
$e^{2\pi ij/6}e^{2\pi i(6-j)/6}=e^{2\pi i} =1$.
Showing these are the only units is harder. From above every nonzero element
$\ah\in Z[\omega]$ has $|\ah|^2$ a positive integer, and so $|\ah|\geq 1$.
\par That is, no nonzero elements of $Z[\omega]$ lie inside the unit circle. Let
$\ah$ be a unit. Then $|\ah|\geq 1$ {\em and} $|\ah^{-1}|\geq 1$. As $|\ah^{-1}|=|\ah|^{-1}$
this can only occur when $|\ah|=1$. So the units of $Z[\omega]$ must lie {\em on} the unit
circle. If you drew a good picture you can ``see" that there are precisely six such points
but how to prove that. We have the equation
\[ |\ah|^2 = a^2+b^2-ab=1 \]
Hmmmm. Well, if $a=0$ we have $b^2=1$, $b=\pm 1$, giving the units $\pm\omega$. If $b=0$
then $a^2=1$, $a=\pm 1$, giving the units $\pm 1$. Now suppose $a,b\neq 0$. If $a,b$ have
opposite signs we rewrite
$1 = (a+b)^2-3ab$ but $(a+b)^2\geq 0$ and $-3ab\geq 3$ so no solutions there. Finally,
suppose $a,b$ have the same sign. We rewrite $1=(a-b)^2+ab$. Both terms are nonnegative
and the second is strictly positive so we must have $a-b=0$ and $ab=1$ and there are
two solutions $a=b=1$ (the unit $1+\omega$) and $a=b=-1$ (the unit $-1-\omega$).
\par Whew! That was pretty complicated. There are other more geometric approaches.
In general, the determining the units of a ring {\em is} a complicated and interesting
undertaking.
\\ {\tt Another approach:} To check if $\ah=a+b\omega$ is a unit we calculate $\ah^{-1}$
by rationalizing the denominator, by multiplying by the complex conjugate. Here
$\omega$ has complex conjugate $\ol{\omega}=-1-\omega$. We have
\[ \frac{1}{a+b\omega}
=\frac{1}{a+b\omega}\frac{a+b(-1-\omega)}{a+b(-1-\omega)}
=\frac{a+b(-1-\omega)}{a^2+b^2-ab}
\]
where the denominator is $\ah\ol{\ah}= |\ah|^2$. If $a^2+b^2-ab=1$ (it can't be $-1$ as it
is the square of $|\ah|$) then $\ah$ is a unit. Conversely, to be a unit we would have
to have
$(a^2+b^2-ab)$ dividing both $a-b$ and $-b$, so both $a,b$, so then
$(a^2+b^2-ab)^2$ would divide all of $a^2,b^2,ab$ and thus would divide
$(a^2+b^2-ab)2$. We conclude that for $\ah$ to be a unit we must have $a^2+b^2-ab=\pm 1$.
That is, $a^2+b^2-ab=1$ is a necessary and sufficient condition for $\ah=a+b\omega$ to be
a unit.
\\ {\tt Yet another approach:} Which points of $Z[\omega]$ lie on or inside the unit circle?
Well, we are looking at $i+j\omega$ which are vectors (thinking of $C$ as the plane) of the form
\[ j(-\frac{1}{2},+\frac{\sqrt{3}}{2}) + i(1,0) \]
So the $y$-value is a multiple of $\sqrt{3}/2$. If it is on or inside the unit circle this
must be in $[-1,+1]$ and so we need only consider $j=0,1,-1$. For each fixed $y$ value of this
form the points on the horozontal line are spread out exactly one apart. For $j=0$ they are
the integers and $i=-1,0,1$ gives the complex (and real) numbers $-1,0,1$. For $j=1$ the $x$-values
are $-\frac{1}{2}+i$ and to be in or on the unit circle we must have this in $[-1,+1]$ and the
only possibilities are $i=0,1$ giving the complex numbers $\omega$ and $\omega+1$. Similarly for
$j=-1$ the $x$-values
are $\frac{1}{2}+i$ and to be in or on the unit circle we must have this in $[-1,+1]$ and the
only possibilities are $i=0,-1$ giving the complex numbers $-\omega$ and $-\omega-1$. Other
than $0$, there are no points in the interior of the unit circle and these six points on the
unit circle. Any nonzero $\ah\in Z[\omega]$ not on the unit circle would then be outside
the unit circle (i.e., $|\ah|>1$) so its inverse (as a complex number) would be inside the
unit circle and hence not in $Z[\omega]$. Thus the only possibilities for units are the six
points {\em on} the unit circle and they all work.
\een
\een
\end{document}
\item For $\kappa=a+b\omega\in Z[\omega]$ write $|\kappa|^2$ as a function of $a,b$.
Show that $|\kappa|^2$ is an integer. Show that $|\kappa|^2$ is a positive integer
when $\kappa\neq 0$.
\item (*) Find all units (with proof that you have all!) of $Z[\omega]$.
\een
\een
\beq
What a chimera then is man! What a novelty, what a monster, what a
chaos, what a contradiction, what a prodigy! Judge of all things, feeble
earthworm, repository of truth, sewer of uncertainty and error, the
glory and the scum of the universe.
\\ -- Blaise Pascal
\enq
\end{document}