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\title{MATH0037 Galois Theory 2002, Sheet 1 Solutions}
\author{Geoff Smith, http://www.bath.ac.uk/$\sim$masgcs}
\date{}
\begin{document}
\maketitle

\begin{enumerate}

\item {\em Algebra and geometry are intimately connected. For example:
{\rm Are there three points in the plane $\mathbb R^2$
with integral co-ordinates
which form the vertices of an equilateral triangle?} You can
address this question using one of the rings we have been 
discussing. What about squares and regular pentagons?}\newline
\noindent{\bf Solution} The question amounts to asking if
three distinct Gaussian integers can form the vertices of an
equilateral triangle in the complex plane. The ratio
of the complex numbers representing two adjacent 
sides of such a triangle would be $e^{i\pi/3}$ and also
have rational real and imaginary 
parts. However $\sin \pi/3 = \sqrt 3/2 \not \in \mathbb Q$.
We have a contradiction. af course squares can arise via $1,-1,i,-i$
and in many other ways. If (for contradiction) a
regular pentagon were to arise, then $e^{2\pi i/5}$ would have
rational real and imaginary parts. This does not happen.
You can study the cosine and sine of this $2\pi i/5$
by investigating the lengths of the sides and diagonals
of a regular pentagon (the cosine rule plays an important role).
This is an enjoyable exercise.

\item {\em We have proved a theorem that tells us that a finite 
integral domain is a field. This knowledge makes the following
result fairly easy to prove: {\rm Suppose that 
$n \in \mathbb N$, and that $\mathbb Z_n$ denotes the integers
modulo $n$. It follows that $\mathbb Z_n$ is a field if and only if 
$n$ is a prime number.} Do it.}\newline
\noindent{\bf Solution}\ If $n$ is prime then when $a, b \in \mathbb Z$
and $n \mid ab,$ it follows that $n \mid a$ or $n \mid b$. 
It follows that $\mathbb Z_n$ is an integral domain, and
since it is finite, is a field. If $n = 1$ then $\mathbb Z_1$
has only one element, so $0 = 1$ in this structure, and so it is not a 
field. Finally of $n = ab$ is a composite natural number, where $a,b$
are natural numbers both smaller than $n$, then $[a] \cdot [b] 
= [n] = 0$ in $\mathbb Z_n$, but $[a] \not = 0 \not = [b]$.
Thus $\mathbb Z_n$ is not an integral domain. However every field
is an integral domain, so $\mathbb Z_n$ is not a field.

\item {\em Suppose that $R$ is a ring and that $R[X]$ is an integral
domain. Prove that $R$ is an integral domain.}\ \newline
\noindent{\bf Solution} This follows by considering the degree function.
The details are routine.
 

\item 
\begin{enumerate}
\item[(a)] {\em Which elements of $\mathbb Q[X]$ have a multiplicative
inverse in $\mathbb Q[X]$?
.}\ \newline
\noindent{\bf Solution} Polynomials of positive degree do not have 
multiplicative inverses by degree considerations. The zero polynomial
of course has no multiplicative inverse. A non-zero constant polynomial
$q \in \mathbb Q$ has multiplicative inverse $q^{-1}$.
\item[(b)$^*$] {\em Which elements of $\mathbb Q[[X]]$ have a multiplicative
inverse in $\mathbb Q[[X]]$} \ \newline
\noindent{\bf Solution} If $f \in \mathbb Q[[X]]$ is a multiple
of $X$ (i.e. it has constant term 0), then of course it is not
invertible. However, if $f = c + Xh$ where $c$ is a non-zero
rational number and $h \in Q[[X]]$, then $c^{-1}f = 1 + t$
where $t \in Q[[X]]$ has constant term $0$.
Now \[ [1+t][1 - t + t^2 - \cdots ] = 1.\]
Let $s = 1 - t + t^2 + \cdots.$ Thus $c^{-1}s$ is a multiplicative inverse
for $f$.
\end{enumerate}

\item {\em Suppose that $\alpha \in \mathbb C$. Show that there
is a polynomial $f \in \mathbb R[X]$ with $\mbox{deg} f = 2$
such that $f(\alpha) = 0$ (this notation means that $\alpha$
is substituted for $X$ in $f$).}
\ \newline
\noindent{\bf Solution} $[X - \alpha][X - \overline \alpha]
= X - 2 \mathfrak{Re} (\alpha) + |\alpha|^2$ has $\alpha$ as a root
and is a polynomial with real coefficients.

\item {\em In this question only, $R$ not guaranteed to have 
a commutative law of multiplication, but all the other
ring axioms are satisfied.}
\begin{enumerate}
\item[(a)] {\em Suppose that $r^2 = r\ \forall r \in R$. Prove that
$R$ has commutative multiplication.} 
\ \newline
\noindent{\bf Solution} For all $a \in R$ we have
$(1+a)^2 = (1 + a)$ so $1 + 2a + a^2 = 1 + a$ so $2a = 0$.
Now suppose that $b, c \in R$, then
Now $(b + c)^2 = (b+c)$ so $b^2 + c^2 + bc + cb = b + c$
and therefore $bc + cb = 0$. Thus $bc + 2cb = cb$ and therefore
$bc = cb$ so $R$ is commutative. 
\item[(b)$^*$] Suppose that $r^3 = r\ \forall r \in R$. Prove that
$R$ has commutative multiplication.
 \ \newline
\noindent{\bf Solution} For every $r$ we also have $(2r)^3 = 2r$ so
$8r^3 = 2r$ so $8r = 2r$ and therefore $6r = 0$ for every $r \in R$. 
Now observe that for all $a, b \in R$ we have
$(a + b)^3 = a+b $ so \[a^2b + aba + ba^2 +b^2a + bab + ab^2 = 0\].
Replace $a$ by $-a$ and add to obtain the law
$2(a^2b + aba + ba^2) = 0.$ Replacing $a$ by $(ab -ba)$ in this
law we deduce that $2(ab - ba) = 0$ for all $a, b \in R$.
This is close to proving $ab = ba$ for all $a, b \in R$.
Can anyone finish this? 
\end{enumerate}
\end{enumerate}
\end{document}
