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\title{MATH0037 Galois Theory 2002, Solutions 7}
\author{Geoff Smith, http://www.bath.ac.uk/$\sim$masgcs}
\date{}
\begin{document}
\maketitle
 \begin{enumerate}
\item {\em Suppose that $K$ is a field. Suppose that $f, h \in K[X] = R$
and that neither $f$ nor $h$ is the zero polynomial.
We have two natural maps $\pi_1 : R 
\mapsto K[X]/(f) = R_1$ and $\pi_2 : K[X] \mapsto
K[X]/(h) 
= R_2$. Define a map
$\pi : R \mapsto R_1 \oplus R_2$ by
$g \mapsto (\pi_1(g), \pi_2(g))$ for all $g \in R$.
Suppose that $(f) + (h) = R$.
\begin{enumerate}
\item Prove that
$\pi$ is a surjective ring homomorphism and
a linear map between $K$-vector spaces.
\item Prove that Ker $\pi = (f) \cap (h) = (fh)$.
\item By considering dimensions, prove that
$\pi$ is surjective.
\end{enumerate}} \noindent 
{\bf Solution}\ This has been dealt with in the model answers associated
with Sheet 6.
\item
\begin{enumerate}
\item {\em Suppose that $K \leq L \leq M$ are extensions 
of fields. let $l_1, \ldots, l_t$ be a $K$-basis of $L$ and
$m_1, \ldots, m_s$ be an $L$-basis of $M$. Show that
$(l_im_j)$ where $1 \leq i \leq t$ and $1 \leq j \leq s$ is
(in any order) a $K$-basis of $M$.}\newline
\noindent {\bf Solution}\ Suppose that for $1 \leq i \leq t$
and $1 \leq j \leq s$ we have elem$\kappa_{ij}$
such that $\sum_{i, j} \kappa_{ij}l_im_j = 0$.
Now
\[ \sum_j \left( \sum_i \kappa_{ij} l_i \right) m_j.\]
quantities $m_j$ are linearly independent over $L$,
so \[ \sum_i \kappa_{ij} l_i = 0 \ \ \mbox{ for all }j.\]
Now the quantities $l_i$ are linearly independent
over $K$, so $\kappa_{ij} = 0$ for all $i$ and for all $j$.


\item {\em Suppose that $\alpha = \sqrt[3]2 \in \mathbb R$. 
Notice that $\alpha$ is algebraic so $\mathbb Q[\alpha]$ is
a field.}
\begin{enumerate}
\item What is dim$_{\mathbb Q} \mathbb Q[\alpha]$? \newline
\noindent {\bf Solution }\ 3, since this field is isomorphic to
$\mathbb Q[X]/(X^3 -2).$
\item {\em Show that if $\mathbb Q \leq K$ is a field extension
and dim$_{\mathbb Q} K = 2^m$ for some integer $m$, then
$\alpha \not \in K$.} \newline
\noindent {\bf Solution} Suppose for contradiction that 
$Q(\alpha) \leq\mathbb K$, then 
\[ |K: \mathbb Q| = |K:\mathbb Q(\alpha)|| \mathbb Q(\alpha): \mathbb Q|\]
and so 3 divides $2^t$ which is absurd.
\end{enumerate}
\end{enumerate}
\item {\em Let $1 \not = \omega \in \mathbb C$ be a fifth
root of unity.}
\begin{enumerate}
\item {\em What is dim$_{\mathbb Q} \mathbb Q[\omega]$?} \newline
\noindent {\bf Solution}\ 4, since $X^4 + X^3 + X^2 + X + 1 \in 
\mathbb Q[X]$ is irreducible (and this polynomial has $\omega$ as
a root).
\item {\em What is dim$_{\mathbb R} \mathbb R[\omega]$?} \newline
\noindent {\bf Solution}\ $\omega \not \in \mathbb R$ so 
the dimension is not 2. Also $[X - \omega][X - \overline \omega] \in
\mathbb R[X]$ so $\omega$ is the root of a quadratic real polynomial
so the extension has degree at most 2.
Therefore the required dimension is 2.
\item {\em What is dim$_{\mathbb A} \mathbb A[\omega]$?}   \newline
\noindent {\bf Solution} 1, since $\omega \in \mathbb A$. 
\item {\em What is dim$_{\mathbb C} \mathbb C[\omega]$?} \newline
\noindent {\bf Solution} 2, since $\omega \in \mathbb C$.          
\end{enumerate}
\end{enumerate}


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