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\title{MATH0037 Galois Theory 2002, Sheet 2}
\author{Geoff Smith, http://www.bath.ac.uk/$\sim$masgcs}
\date{}
\begin{document}
\maketitle
\begin{enumerate}

\item {\em Let $q$ be a rational number. 
Define a map $\varepsilon_q : \mathbb Q [X ] \rightarrow \mathbb Q$ via
$f \mapsto f(q)$. Show that $\varepsilon_q$ is a ring homomorphism.
Describe $\mbox{Ker } \varepsilon_0$ explicitly.} \newline
\noindent {\bf Solution}\ It is routine to verify that this map
is a ring homomorphism. The kernel is the set of all polynomials
which have $q$ as a root. This is $(X-q)$, the (principal) ideal
consisting of all possible multiples of $X-q$ by a rational 
polynomial.
\item {\em Let $\mathbb Z_2$ denote the integers modulo 2.
In lectures we constructed \[\mathbb R[X]/(X^2+1)\] which turned out
to be $\mathbb C$ in disguise. Construct the rings
\begin{enumerate}
\item $\mathbb Z_2 [X]/(X^2+1)$ and 
\item $\mathbb Z_2 [X]/(X^2 + X + 1).$
\end{enumerate}}
\noindent {\bf Solution}\
Each of these structures is a ring with 4 elements. One of 
them is a field and the other is not. Which is which, and further
more, what properties of the polynomials $X^2+1$ and $X^2 + X + 1$
(in $\mathbb Z_2[X]$) are causing this difference? \newline
\noindent {\bf Solution}\ Let $f$ denote the polynomial in each case.
As before, we find that a complete list (without repetitions)
of the elements of  $\mathbb Z_2 [X]/(f)$  is:
cosets of the form $a + bX + (f)$ where $a,b \in \mathbb Z_2[X]$.
The additive structure is that of addition of polynomials 
of dgeree at most two, and $\mathbb Z_2 [X]/(f) \longrightarrow 
\mathbb Z_2 \oplus \mathbb Z_2$, where $a + bX + (f)  \mapsto (a,b)$ is
an isomorphism of the additive groups. 
The multiplicative structures $\mathbb Z_2 [X]/(f)$
for the two different polynomials 
$f$ are very different. The points is that $X^2 + 1 = [X+1]^2$,
so $[1+X + (X^2+1)]^2 = 0 \not = [1+X + (X^2+1)].$
Thus this ring is not an integral domain, and is therefore
certainly not a field. However, $\mathbb Z_2 [X]/(X^2 + X + 1)$
is a field (as you will find out by writing out its multiplication
table. The reason for the difference is that if $f$ is reducible
(factorizable) in $Z_2[X]$ then this will ensure that
$ \mathbb Z_2 [X]/(f)$  has non-zero elements with product $0$.
On the other hand, if $f$ cannot be factored, it is fairly easy
to show that $Z_2[X]/(f)$ is an integral domain (but it is
finite and therefore a field). This is entirely analogous
to the situation when you try to determine for which
natural numbers $n$ is it that $\mathbb Z_n$ is a field.

\item {\em Construct $\mathbb R[X]/(X^2 + X + 1).$ Can you recognize
this structure?} \newline
\noindent {\bf Solution}\ Let $omega = e^{2\pi i/3} \in \mathbb C$.
Thus $\omega^2 + \omega + 1 = 0$. Each complex number is
uniquely of the form $a + b \omega$ for $a,b \in \mathbb R$.
Consider the map $\mathbb R[X] \rightarrow  \mathbb C$
defined by evaluating polynomials at $\omega$. This is a 
ring homomorphism, is surjective, and has kernel
$(X^2 + X + 1)$ (why?). Thus by the first isomorphism 
theorem $\mathbb R[X]/(X^2 + X + 1)$ is ismorphic to the field $\mathbb C$. 
\item {\em Construct $\mathbb R[X]/(X^2 - 3X + 2).$ Can you recognize
this structure?} \newline
\noindent {\bf Solution}\ 
This is not a field. In fact its structure is $\mathbb R \oplus
\mathbb R$ under co-ordinatewise addition and multiplication.
To see this, we have ring homomorphism $\varepsilon_1, \varepsilon_2$
from $\mathbb R[X]$ to $\mathbb R$ which have evaluation at 1 and 2
respectively. They are both surjective, and have kernels
$(X-1)$ and $(X-2)$. We can use these maps to define a ring
homomorphism $\varepsilon : \mathbb R[X] \rightarrow \mathbb R \oplus
\mathbb R$ by $\varepsilon(r) = (\varepsilon_1(r),\varepsilon_2(r)).$ 
Then show that $\varepsilon$ is surjective with kernel 
$(X^2 -3X + 2)$ and apply the first isomorphism theorem.
\item {\em Construct addition and multiplication tables for a field
which has exactly 9 elements.} \newline
\noindent {\bf Solution}\ Find a quadratic
$f \in \mathbb Z_3[X]$  which is irreducible. This is 
easy since $X^2 +1$ has no root in $\mathbb Z_3$ and therefore
no linear factor. The multiplication is rather like that of the
complex numbers. The multiplicative structure of the non-zero
elements will be a cyclic group of order 8.
\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)] Suppose that $r^2 = r\ \forall r \in R$. Prove that
$R$ has commutative multiplication.
\item[(b)$^*$] Suppose that $r^3 = r\ \forall r \in R$. Prove that
$R$ has commutative multiplication.
\end{enumerate}} \noindent
{\bf Solution} This has been discussed previously.  
\end{enumerate}
\end{document}
