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\title{MATH0037 Galois Theory 2000}
\author{GCS -- Sheet 5}
\date{}
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\maketitle

\begin{enumerate}
\item Let $K = {\mathbb Z}/3{\mathbb Z}$, a field of size 3.
Find a cubic polynomial $f \in K[Y]$ which is irreducible.
Let $L =  K[Y]/(f)$, so $L$ is a field of size $3^3 = 27$.
Prove that $U(L)$ is a cyclic group.\newline
\noindent{\bf Solition} $f = Y^3 + 2Y + 1$ will do
since $f(0) = f(1) = f(2) = 1 \not = 0$ and so $f$
can have no linear factor in $K[Y]$.
Let $\alpha = (f) + Y$. The order of $\alpha$ must be $2,13$
or $26$. An explicit calculation yields that $\alpha^2 \not = 1$
and $\alpha^{13} = 2 = -1 \not = 1.$ Therefore 
$\langle \alpha \rangle = U(L)$. Here $U(L)$ denotes the set
of multiplicatively invertible elements of $L$.
\item Let $K = {\mathbb Z}/5{\mathbb Z}$, a field of size 5. 
Find a quadratic polynomial $f \in K[Y]$ which is irreducible.
Let $L =  K[Y]/(f)$, so $L$ is a field of size $5^2 = 25$.
Prove that $U(L)$ is a cyclic group.\newline
\noindent{\bf Solition} $f = Y^2 + 2$ is irreducible
in $K[Y]$ since $3$ is not a square modulo $5$.
A direct calculation yields that the multiplicative
order of $\alpha = (f) +Y$ is 8. 
Experiment yields that the order of $1 + 4\alpha$ is 24,
so $U(L)$ is a cyclic group.
\item  
Suppose that $f \in {\mathbb C}[X]$ is irreducible. Show
that $f$ is linear (and refer explicitly to any result
to which you appeal). \newline
\noindent{\bf Solition} The {\bf Fundamental Theorem of Algebra}
states that if $0 \not = f \in \mathbb C[X]$, then $f$
factorizes into linear polynomials in $\mathbb C[X]$.
It follows immediately that if $f \in \mathbb C[X]$ is irreducible, then
$f$ has degree 1.
\item 
Suppose that $f \in {\mathbb R}[X]$ is irreducible. Show
that $f$ is either linear or quadratic. \newline
\noindent{\bf Solition} Temporarily think of $f \in \mathbb C[X]$.
Suppose that $f$ is not linear. The irreduciblility of
$f \in \mathbb R[X]$ and the Fundamental Theorem
of Algebra ensure that 
$f$ has a root $\alpha
\in \mathbb C$ with $\alpha \not \in \mathbb R$.
Notice that $\overline{f(\alpha)} = f(\overline \alpha)$
so $\overline \alpha$ is a root of $f$. Now 
$f = [X - \alpha] g$ where $g \in \mathbb C[X]$
and $\overline \alpha$ is a root of $g$.
Thus $f = [X - \alpha] [X - \overline \alpha] h$
where $h \in \mathbb C[X]$. Now 
$t =  [X - \alpha] [X - \overline \alpha]
= X^2 - (\alpha + \overline \alpha)X + \alpha \overline \alpha
\in \mathbb R[X]$. Thus $f = t h$ and $\overline f
= \overline t \overline h$ so $f = t \overline h$ where
the bar denotes complex conjugation of all coefficients.
Now $0 = f - f = t(h - \overline h)$ but $t \not = 0$
and $\mathbb C[X]$ is an integral domain so $h - \overline h
= 0$
and therefore $h \in \mathbb R[X]$. The irreducibility
of $f$ forces $h$ to be a unit in $\mathbb R[X]$.
Therefore $h = c \in \mathbb R - \{0\}$.
Thus $f = ct \in \mathbb R[X]$ is quadratic. 
\item Suppose that $K$ is a field and that $\mathbb R \leq K$.
Show that if $|K : \mathbb R| = 2$, then $K$ and $\mathbb C$ 
are isomorphic fields (with the isomorphism fixing all elements of
$\mathbb R$).  \newline
\noindent{\bf Solition}
Choose $\alpha \in K$ with $\alpha \not \in \mathbb R$.
Let $m_\alpha \in \mathbb R[X]$ be the minimum polynomial
of $\alpha$. Thus
\[ K = \mathbb R(\alpha) = \mathbb R[\alpha] \simeq
\mathbb R[X]/(m_\alpha).\]
Now $m_\alpha \in \mathbb R[X]$ is an irreducible
quadratic. Thus it has no real root, but by the
Fundamental Theorem of Algebra it has a 
complex root $\beta$. The ring homomorphism
``evaluation at $\beta$'' is a map $\varepsilon_\beta:
\mathbb R[X] \rightarrow \mathbb C$. The kernel
is $(m_\alpha)$ and the map is clearly surjective
(the image is a 2-dimensional $\mathbb R$-subspace of 
$\mathbb C$). Now the first isomorphism theorem for
rings ensures that $\mathbb R[X]/(m_\alpha)
\simeq \mathbb C$.
We are done.
\end{enumerate}
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