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\title{MATH0037 Galois Theory 2002, Solutions 4}
\author{Geoff Smith, http://www.bath.ac.uk/$\sim$masgcs}
\date{}
\begin{document}
\maketitle
\begin{enumerate}
\item {\em Factorize the following polynomials:
$X^2 + 1$,\ $2X^2 - 1$,\ $X^5-1$}
\begin{enumerate}
\item {\em in $\mathbb C[X]$,}\newline
{\bf Solution}\ $(X+i)(X-i),\ (\sqrt2 X +1)(\sqrt 2 X -1)$ and 
\[ (X-1)(X-\lambda)(X - \lambda^2)(X-\lambda^3)(X - \lambda^4)\]
where $\lambda = e^{2\pi i/5}\in \mathbb C$.
\item {\em in $\mathbb R[X]$,}\newline
{\bf Solution }\ $X^2 +1$, $2X^2 -1$ and using $\lambda$ as before,
we have a factorization
$(X-1)(X^2- (\lambda + \lambda^4)X+ 1)(X^2 - (\lambda^2 + \lambda^3)X + 1$.
Note here that $\overline \lambda = \lambda^{-1}$ and $\overline
\lambda^2 = \lambda^{-3}$
\item {\em in $\mathbb Q[X]$.}\newline
{\bf Solution }\ $X^2 +1$, $2X^2 -1$ and $(X-1)(X^4 + X^3 + X^2 + X + 1)$.
\end{enumerate}
\item \begin{enumerate} 
\item {\em Consider the ring $\mathbb G$ of Gaussian integers.
Show that $z \in G$ has the property that $z^{-1} \in \mathbb G$ if and only
if $z \overline z = 1$. Determine which Gaussian integers have this property.}
\newline
{\bf Solution}\ For the complex number $z$ we let $N(z) = z \overline z$.
Now $N(ab) = N(a)N(b)$ forll $a,b \in \mathbb C$.
Now if $z^{-1}$ is a Gaussian integer we have 
$zz^{-1} = 1$ so $N(z) N(z^{-1}) = N(1) =1$.
Now the norm of a non-zero Gaussian integer is a positive
integer so $N(z) = N(z^{-1}) = 1$. Coonversely if $z \in \mathbb G$
and $N(z) =1$, then $z \overline z = 1$ so $z^{-1} = \overline z \in \mathbb G$.
The Gaussian integers $a + bi$ ($a,b \in \mathbb Z$) is therefore a unit
of $\mathbb G$ if and only if $a^2 + b^2 =1$. Thus the units
of $\mathbb G$ are $1, -1, i, -i$.
\item {\em Let $\omega = e^{2\pi i/3} \in \mathbb C$, 
Consider the ring $R = \{ a + b \omega \mid a, b \in \mathbb Z\}$
(a subring of the complex numbers). Suppose that $r  = x + \omega y$ with
$x, \in \mathbb Z$. Show that 
that $r^{-1} \in R$ if and only if $x^2 + xy + y^2 = 1$. Determine
which elements of $R$ have this property.}\newline
{\bf Solution}\ Defining the norm $N(z)$ of a complex number $z$ 
to be $z \overline z$ as before, we obtain that 
the `norm' is a multiplicative function (again). The formula
for the norm of an element of $R$ is
\[ N(a + b \omega) = (a + b \omega)(a + b \omega^2)
= a^2 + b^2 -ab.\] Whoops. The sign of the ab term is wrong!
Well, let us press on. Arguing as before, we discover that
if $a, b \in \mathbb Z$ then $a + b \omega$ has inverse in $R$
if and only if $a^2 -ab + b^2 = 1$. The elements of $R$ correspond to
the vertices of an equilateral triangular lattice in the plane.
The only points of norm 1 are $1, \omega, \omega^2, -1, 1+ \omega,
1 + \omega^2$ (forming a regular hexagon on the unit circle).

\item[(c)$^*$] Consider the ring $S = \{ a + b \sqrt 2 \mid a, b \in \mathbb Z\}$.
Determine which elements $s$ of $S$ have the property that $s^{-1} \in S$.
For $a, b \in \mathbb Q$ let $N(a + b \sqrt 2) =
(a + b \sqrt 2)(a - b \sqrt 2) = a^2 - 2b^2.$ Arguing as before, we end up
worrying which $a, b \in \mathbb Z$ have the property that 
$a^2 - 2b^2 = \pm 1$. There is a whole (very interesting) theory
of this type of equation (so called {\em Pell's equation}). See
any book on elementary number theory, for example
Davenport's {\em Higher Arithmetic} (C.U.P.).
\end{enumerate}
\item 
\begin{enumerate}
\item {\em Suppose that $F$ is a subring of the ring $R$, and that $F$ is a field.
Show that addition and multiplication in $R$ render
$R$ into a vector space over $F$.}\newline
{\bf Solution}\ This is routine.
\item {\em Suppose that $R$ is a finite field. Let $S_0 = \{ 0 \} \subseteq R$.
For each integer $i>0$ let $S_{i+1} = S_i \cup \{ x + 1_R \mid  x \in S_i\}.$
Let \[S = \cup_{i \geq 0} S_i.\] Prove that 
$-1_R \in S$ and that $S$ is a subfield of $R$.}\newline
{\bf Solution}\ Now $S$ consists of $0$ and everything which expressible as
a sum of 1's. Thus $S$ is multiplicatively closed.
Also the finiteness of $R$ ensures that there are distinct natural numbers
$n < m$ such that 
\[ \sum_{i=1}^n 1 = \sum_{i=1}^m 1\]
so subtracting we deduce that there is a natural number $l$ such that
the sum of $l$ 1's is $0$. Thus $-1$ is the sum of $l-1$ 1's
so $-1 \in S$. It follows that $S$ is an additive group and that
$S$ is a subring of $R$. Of course $S$ (being a subring of a field)
is an integral domain. However, $S$ is finite so $S$ is a field.
\newline
{\em Show that $S$ is isomorphic to $\mathbb Z_p$ for some prime number $p$.} 
Notice that $S$ is additively a cyclic group generated by $1$. If this subgroup
has order $p$, then $p$ must be prime, else $p = ab$ for natural numbers
$a$, $b$ where $a,b <p$. In this case the product
of the sum of $a$ 1's and the sum of $b$ 1's is $0$. This is impossible
because $S$ is an integral domain, and neither the sum of $a$ 1's nor
the sum of $b$ 1's in $0$. Thus $p = |S|$ is prime.


\item {\em Deduce that if $R$ is a finite field then $|R| = p^n$ for some
integer $n$.} Hint: in the notation of part (b), let
$n = \mbox{dim}_S R$. Pick an $S$-basis $v_1, v_2, \ldots, v_n$ for $R$.
Now every element of $R$ is uniquely expressible as 
$\sum_{i=1}^n s_i v_i$ for $s_i \in S$ for all $i$. It follows that
$|R| = p^n$ (the number of possible sequences of coefficients).
\end{enumerate}
\end{enumerate}
\end{document}
