\documentclass{article}
\pagestyle{empty}
\usepackage{amssymb}
\usepackage{latexsym}
 
\title{Exam Solution for Galois Theory 2001}
\author{Geoff Smith, http://www.bath.ac.uk/$\sim$masgcs}
\date{}
\begin{document}
\maketitle
\begin{enumerate}
\item {\em Let $R$ be a Principal Ideal Domain.}
\begin{enumerate}
\item {\em Define the term {\rm irreducible element} of $R$.}
\newline \noindent {\bf Solution}\
$r \in R$ is irreducible if it is not $0$, not a unit, and
if $r = uv$ then either $u$ or $v$ is a unit.
\item {\em Show that if $x \in R$, then $x$ is irreducible if and only
if $(x)$ is a non-zero maximal ideal.}
\newline \noindent {\bf Solution}\
If $x$ is irreducible then it follows that $(x)$ is neither $0$ nor $R$.
Suppose that $(x) \subseteq I$ where $I$ is an ideal. Since $R$ is a PID
it follows that $I = (a)$. Now $x \in I$ so $x = ab$ for some $b \in R$.
Now if $b$ is not a unit, then $a$ must be so $(a) = R$. Conversely if
$b$ is a unit, then $a = b^{-1}x \in (x)$ si $I = (a) = (x)$. Thus
$(x)$ is a non-zero maximal ideal. 

Next suppose that $(x)$ is a non-zero maximal ideal, so $x$ is neither
$0$ nor a unit. Suppose that $x = yz$ for $y, z \in R$. we must show that
one of $y, z$ is a unit. Thus we suppose that $z$ is not a unit and try
to deduce that $y$ must be a unit.
Now if $z$ is not a unit then $(x) \subseteq (y) \not = R$ so $(x) = (y)$
by maximality of $(x)$. Thus $y = wx$ for some $w \in R$. 
Now $x = wxz$ so $x(1-wz) = 0$. Now $x$ is not $0$ so neither is
$y$ and nor is $x$. The ring $R$ is an integral domain and so
$1 - wz = 0$ so $wz = 1$ and $z$ is a unit as required.

\item {\em State carefully the unique factorization theorem for elements
of $R$.}
\newline \noindent {\bf Solution}\                                                                 Suppose that $r \in R$ and that $r$ is neither $0$ nor a unit.
Then $r$ may be written as a product of irreducible elements of $R$
via (say)
\[ r = \prod_{i=1}^m r_i.\]
Suppose that 
\[ r = \prod_{j=1}^n s_j\] is a rival factorization of $r$ as 
a product of irreducible elements of $R$, then $m = n$ and after suitable
permutation of the subscripts, $r_i$ and $s_i$ are associates for all $i$.

\item {\em Let $R$ be a principal ideal domain containing exactly one maximal ideal.
Prove that if $I$, $J$ are ideals of $R$ then either $I \subseteq J$ or $J
\subseteq I$.} 
\newline \noindent {\bf Solution}\                                           
If either $I$ or $J$ is $0$ or $R$ the result is clear so we may assume 
that this is not the case (and in particular that $R$ is not a field
so the unique maximal ideal $M = (r)$ of $R$ is not the zero ideal.
Thus $r$ is irreducible. By the unique factorization theorem
$I = (ur^s)$ and $J = (vr^t)$ where $u, v$ are units and 
$s, t$ are positive integers.
Thus $I = (r^s)$ and $J = (r^t)$ where either $s \leq t$ or $t \leq s$.
The result now follows.
\end{enumerate}

\item
\begin{enumerate}
\item {\em Prove that a finite integral domain $R$ must be a field.}
\newline \noindent {\bf Solution}\                                        
If $r \in R$ and $r \not = 0$, define $m_r : R \rightarrow R$
by $m_r(s) = rs$ for all $s \in R$.
Now the fact that $R$ is an integral domain forces $m_r$ to be injective,
but $R$ is a finite set so $m_r$ must also be surjective, and will
therefore have $1$ in its image. Thus there is $t \in R$ such that
$rt = 1$, so $r$ is invertible and we are done.

\item {\em Suppose that $R$ is an integral domain. Show that
$R[X]$ is an integral domain.}
\newline \noindent {\bf Solution}\                                               
If $f, h \in R[X]$ are not $0$, then $f = aX^m +$ terms of smaller degree
and $h = bX^n +$ terms of smaller degree with neither $a$ nor $b$ being
$0$. Now 
$fh = abX^{m+n} +$ terms of smaller degree. Since $R$ is an integral domain,
$ab \not = 0$ so $fh \not = 0$.

\item {\em Let $R$ be a ring with the property that whenever $r \in R$ and $0
\not = 0$, then there is $f_r \in R[X]$ such that $f_r(0) = 1$ and 
$f_r(r) = 1$. Prove that $R$ is a field.}
\newline \noindent {\bf Solution}\                                                 
Let $f_r = a_0 + a_1X + \cdots + a_t X^t.$ Since $f_r(0) = 1$ it
follows that $a_0 = 1$. Therefore
\[ 1 + a_1r + a_2r^2 + \cdots + a_t r^t = 0\]
and so
\[ 1 = r ( -a_1 -a_2r - \cdots -a_tr^{t-1})\]
and so $r$ has a multiplicative inverse in $R$. Thus
$R$ is a field.
\end{enumerate}

\item
\begin{enumerate}
\item {\em What is an algebraic number?}
\newline \noindent {\bf Solution}\                  
$\alpha \in \mathbb C$ is an algebraic number if it is a root
of a non-zero rational polynomial, or equally well if
$|\mathbb Q(\alpha) : \mathbb Q|$ is finite.
\item {\em Suppose that $\alpha, \beta$ are algebraic numbers.
Prove that $\alpha^2 + 3\alpha \beta - \beta^3$ is algebraic,
clearly stating any theorems to which you appeal.}
\newline \noindent {\bf Solution}\ Let $\beta$ be a root of the
non-zero polynomial $f \in \mathbb Q[X]$. Now we may regard $f$
as an element of $\mathbb Q(\alpha)(X)$, so $\beta$ is algebraic
over $\mathbb Q (\alpha),$ so $|\mathbb Q(\alpha, \beta): \mathbb Q(\alpha)|$
is finite. Now we use the theorem of the multiplicativity of the degrees
in a tower of field extensions, so
\[|\mathbb Q(\alpha, \beta): \mathbb Q|
= |\mathbb Q(\alpha, \beta): \mathbb Q(\alpha)| \cdot
 |\mathbb Q(\alpha) : \mathbb Q|\]
is finite. All elements of $\mathbb Q(\alpha, \beta)$ are therefore
algebraic numbers, including $\alpha^2 + 3\alpha \beta - \beta^3$.

\item {\em Suppose that $\theta, \psi \in \mathbb C$ have the property that
$\theta + \psi$ and $\theta^2 + \psi^2$ are both rational, show that
$\theta, \psi$ are both algebraic.}
\newline \noindent {\bf Solution}\             
\[ \theta\psi = \frac{2\theta\psi}{2} = 
\frac{(\theta + \psi)^2 - (\theta^2 + \psi^2)}{2} \in \mathbb Q.\]
Thus $X^2 - (\theta + \psi)X + \theta\psi$ is a non-zero rational
polynomial having $\theta, \psi$ as roots, and these numbers are therefore
algebraic.

\end{enumerate}

\item 
\begin{enumerate}
\item {\em Explain the term {\rm normal field extension}.}
\newline \noindent {\bf Solution}\                  
Suppose that $K, L$ are fields and that $K$ is a subfield of $L$.
We say that $L$ is a normal extension of $K$ if whenever $f \in K[X]$
is irreducible, and $f$ has a root in $L$, then $f$ factorizes
into linear polynomials in $L[X]$.
\item {\em State the {\rm Fundamental Theorem of Galois Theory} for fields
of characteristic $0$.}
\newline \noindent {\bf Solution}\        
See lecture notes.

\item {\em Suppose that $K$ is a normal extension of $\mathbb Q$ of finite degree.
Also suppose that there is a unique subfield $L$ of $K$ which is strictly
intermediate between $\mathbb Q$ and $K$ (so if $M$ is a subfield of $K$
and $\mathbb Q \not = M \not = K$, then $M = L$). Prove that
$\mbox{Gal}_{\mathbb Q}K$ is a cyclic group.}
\newline \noindent {\bf Solution}\                     
Let $G = \mbox{Gal}_{\mathbb Q}K$ so by FTGT $G$ is a finite group
and has a unique subgroup $H$ such that $1 \not = H \not = G$.
Choose $x \in G$ but $x \not \in H$. Now the set of powers of
$x$ is $\langle x \rangle,$ a subgroup of $G$ which by design
is neither $1$ nor $H$. That only leaves the possibility that
$\langle x \rangle = G$ which must therefore be a cyclic group.
\end{enumerate}
\end{enumerate}
\end{document}
