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\title{MATH0037 Galois Theory 2000}
\author{GCS -- Sheet 3 - Solutions}
\date{}
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\maketitle

{\it All rings are
commutative with 1 as far as this course is concerned}
\begin{enumerate}
\item Suppose that $R$ is a ring and that $S$ is a subring
of $R$. Suppose that $S$ happens to be a field. Show that
$R$ equipped with $+$ and $\times$ can be regarded as
an vector space over $S$. \newline
\noindent{\bf Solution} Addition of vectors is addition in $R$.
Multiplication of scalars by vectors is multiplication in $R$.
The vector space axioms follow from the axioms of the field
$S$ and the ring $R$.
\item Suppose that $K$ is a subfield of $L$ which in turn
is a subfield of $M$.  Let the finite sequence $(a_i)$
($1 \leq i \leq m$) be a $K$-basis of $L$ (i.e. an ordered
basis of $L$ regarded as a vector space over $K$).
Let the finite sequence $(b_j)$
($1 \leq j \leq n$) be an $L$-basis of $M$.
Show that the $mn$ quantities $a_ib_j$
form a $K$-basis of $M.$
Deduce that
\[ \hbox{dim}_K M = \hbox{dim}_K  L \ \cdot \ \hbox{dim}_L M.\]
\noindent{\bf Solution} First we address spanning: suppose that
$x \in M$. Thus there are scalars $\mu_j \in L$ such that
$x = \sum_j \mu_j b_j$. Now each $\mu_j$ is in $L$ so there
are scalars $\lambda_{ij} \in K$ such that $\mu_j = \sum_{i}
\lambda_{ij}a_i$.
Putting these facts together we obtain that
$x = \sum_{i,j} \lambda_{ij}a_ib_j$ and spanning is established.

Next we address linear independence. Suppose that there are\noindent{\bf
Solution}
$mn$ scalars $\theta_{ij} \in K$ such that $\sum_{ij}\theta_{ij}a_ib_j =
0$.
Now this forces \[ \sum_j \left( \sum_i \theta_{ij}a_i\right) b_j.\]
The quantities $b_j$ are linearly independent over $L$ and so
$\sum_i \theta_{ij}a_i = 0$ for each $j$. However, the quantities
$a_i$ are linearly independent over $K$ and so for each $i$ and for each

$j$ we have $\theta_{ij} = 0$. Linear independence
is established.
\item What are the following?
\begin{enumerate}
\item $\hbox{dim}_{\mathbb R} {\mathbb R}?$\newline
\noindent{\bf Solution} 1.
\item $\hbox{dim}_{\mathbb C} {\mathbb C}?$\newline
\noindent{\bf Solution} 1.
\item $\hbox{dim}_{\mathbb R} {\mathbb C}?$\newline
\noindent{\bf Solution} 2.
\item $\hbox{dim}_{\mathbb Q} {\mathbb Q}?$\newline
\noindent{\bf Solution} 1.
\item $\hbox{dim}_{\mathbb Q} {\mathbb R}?$\newline
\noindent{\bf Solution} infinity. More specifically, the cardinality of
$\mathbb R$ (often called $c$, the ``cardinality of the continuum'').
\item $\hbox{dim}_{\mathbb Q} {\mathbb C}?$\newline
\noindent{\bf Solution} $c$.
\end{enumerate}

\item Let $K$ be the subset of $\mathbb R$ consisting
of numbers of the form $a + b \sqrt 2$ where $a, b \in \mathbb Q$.
\begin{enumerate}
\item Prove that $K$ is a subfield of $\mathbb R$.
{\it Feel free to skip the obvious parts; the interesting
aspect of this problem is to show that a non-zero element
of $K$ has a multiplicative inverse in $K$.}\newline
\noindent{\bf Solution} Suppose that $a, b \in \mathbb Q$
and $z = a + b \sqrt 2$. We define $\bar z = a - b \sqrt 2 \in K$.
Now $z \bar z = a^2 - 2b^2$. Notice that $a^2 - 2b^2 = 0$ if and
only if both $a$ and $b$ are $0$ since $\sqrt 2$ is an irrational
number. Thus if $z \not = 0$ then $z \bar z$ is a non-zero rational.
Let $z' = \bar z /z \bar z \in K$, then $zz' = 1$ and so any
non-zero
element of $K$ is invertible.
\item Calculate $\hbox{dim}_{\mathbb Q} K$.\newline
\noindent{\bf Solution} $1$ does not span $K$ as a $\mathbb Q$
so $K$ has $\mathbb Q$-dimension bigger than $1$.
Also $1, \sqrt 2$ spans $K$ as a $\mathbb Q$-space so
$K$ has $\mathbb Q$-dimension no more than $2$.
Thus $
\hbox{dim}_{\mathbb Q}K = 2$.
\item Demonstrate that $\sqrt[3]{2} \not \in K.$\newline
\noindent{\bf Solution} Let $\alpha = \sqrt[3]{2}$.
Notice that $\alpha$ is a root of $X^3 - 2$, a
rational polynomial which is irreducible by Eisenstein's
criterion. If (for contradiction) $\alpha \in K$, then
the sequence $1, \alpha, \alpha^2$ would be linearly dependent
over $\mathbb Q$. Thus there would be a non-zero
rational polynomial $f$ of degree at most 2 having $\alpha$
as a root. Divide $X^3 -2$ by $f$ to obtain
\[ X^3 - 2 = q \cdot f + r\] where $q,r$ are rational
polynomials and $r$ has degree at most 1. Evaluate at $\alpha$
to obtain that $0 = 0 + r(\alpha)$. We cannot have $r = 0$
since $X^3 - 2$ is irreducible as a rational polynomial.
Now divide $X^3 - 2$ by $r$ so that
\[ X^3 - 2 = h \cdot r + c\] where $h,c$ are rational
polynomials and $c$ is a constant. Evaluating at $\alpha$
we discover that $c = 0$ and this violates the irreducability
of $X^3 -2$ as a rational polynomial. {\em There are many other ways
to do this question}.
\end{enumerate}
\item Given a rectangle in the Euclidean plane, show how
to to construct a square of the same area using only an
infinitely fine pencil and an arbitrarily long
unmarked straight edge.\newline
\noindent{\bf Solution} Let the sides of the rectangle be
of length $a$ and $b$. Copy these lengths onto a straight line
through $A, B$ and $C$ so that $|AB| = a$ and $|BC| = b$.
This can be done using the compasses and the straight edge.
Find the midpoint $D$ of $AB$ (a standard rules and compasses
construction). Draw the circle with centre at $D$ and radius $|AD|$.
Draw the perpendicular to $AC$ through $B$. Let this perpendicular
meet the circle at $E$ and $F$. Now verify that $|EB|^2 = ab$.
\end{enumerate}
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