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\title{MATH0037 Galois Theory 2000}
\author{GCS -- Sheet 6}
\date{}
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\maketitle

\begin{enumerate}
\item Suppose that the quadratic $f \in \mathbb Q[X]$ has roots $\alpha, \beta
\in \mathbb C$. Suppose furthermore that $f \in \mathbb Q[X]$
is irreducible. Let $K = \mathbb Q(\alpha, \beta)$.
\begin{enumerate}
\item Notice that $\mathbb Q(\alpha) \simeq \mathbb Q[X]/(f)$.
Deduce that the fields $\mathbb Q(\alpha), \mathbb Q(\beta)$ are isomorphic
via an isomorphism which fixes rational numbers and sends
$\alpha$ to $\beta$. \newline
\noindent{\bf Solution}\ The isomorphisms 
$\mathbb Q(\alpha) = \mathbb Q[\alpha] \simeq Q[X]/(f)$
and 
$\mathbb Q(\beta) = \mathbb Q[\beta] \simeq Q[X]/(f)$ 
both fix the rational subfield. The result now follows.
\item Prove that $K = \mathbb Q(\alpha) = \mathbb Q(\beta)$.\newline
\noindent{\bf Solution}\ Suppose that $f = aX^2 + bX + c$
where $a,b,c \in \mathbb R$ and $a \not = 0$. Now 
$\alpha + \beta = -b/a \in \mathbb R$ so $\beta = -b/a - \alpha$
and $\alpha = -b/a - \beta$. Thus $\beta \in \mathbb Q(\alpha)$
and $\alpha \in \mathbb Q(\beta).$ Therefore $\mathbb Q(\beta)
\subseteq \mathbb Q(\alpha)$ and 
$\mathbb Q(\alpha)
\subseteq \mathbb Q(\beta)$. Thus $\mathbb Q(\alpha) = 
\mathbb Q(\beta)$. 
\item Prove that $\hbox{Gal}_{\mathbb Q}K$ is a cyclic
group of order 2. \newline
\noindent{\bf Solution}\ Let $K = \mathbb Q(\alpha)$. 
Any automorphism of $K$ must leave the roots of $f$ invariant.
The map which fixes the roots is the identity. We have
shown above that the map which fixes the rationals and 
exchanges
the roots is an isomorphism.
Thus the Galois
group is cyclic of order 2.
\end{enumerate}
\item Suppose that $\mathbb Q \leq F$ is a field extension
and that $f: F \rightarrow F$ is an isomorphism of fields.
Prove that $f$ fixes each element of $\mathbb Q$. \newline
\noindent{\bf Solution}\ Any field 
homomorphism must send 1 to 1, and therefore
must fix each natural number and 0, and therefore each
integer. Thus it must fix ratios of integers and we are done.
\item Let $\lambda = e^{2\pi i/5} \in \mathbb C$.
\begin{enumerate}
\item Let $f(X) = X^4 + X^3 + X^2 + X + 1$. Show that 
$\lambda$ is a root of $f$. \newline
\noindent{\bf Solution}\ $X^5 - 1 = [X-1] f$
so $0 = \lambda^5 -1 = [\lambda - 1] f(\lambda).$
Now $\lambda - 1 \not = 0$ so $f(\lambda) = 0$.
\item Notice that $f(X) = [X^5 -1]/[X-1]$. Let $h(X) = f(X+1)$.
Use Eisenstein's criterion to deduce that $h(X) \in \mathbb Q[X]$ is
irreducible. \newline
\noindent{\bf Solution}\ 
$h(X) = [[1 + X]^5 - 1]/X = X^4 + 5X^3 + 10X^2 + 10X + 5$
and this is irreducible in $\mathbb Q[X]$ by Eisenstein's
criterion. 
\item Deduce that $f(X) \in \mathbb Q[X]$ is
irreducible. \newline
\noindent{\bf Solution}\ 
Any non-trivial factorization of $f = ab$ with
$a,b \in \mathbb Q[X]$ both non-units would give
rise
to a non-trivial factorization of $h$ via
$h = a(X+1) \cdot b(X+1)$ which is absurd.
\item Suppose that $\theta, \psi \in \mathbb C$
are both fifth roots of 1 and neither of them is 1.
Prove that there is an isomorphism between the
fields $\mathbb Q(\theta)$ and $\mathbb Q(\psi)$
which sends $\theta$ to $\psi$ and 
fixes each rational number.\newline
\noindent{\bf Solution}\ 
The isomorphism $\mathbb Q(\theta) = \mathbb Q[\theta]
\simeq \mathbb Q[X]/(f)$ sends $\theta$ to $(f) + X$.
Similarly there is an isomorphism 
$Q[X]/(f) \simeq \mathbb Q(\psi)$ sending
$(f) + X$ to $\psi$. Composing we are done.
\item Let $K$ be the field $\mathbb Q(\lambda)$. 
What is the Galois group $G$ of the field extension
$\mathbb Q \leq K$? (i.e. how many elements does $G$ have and 
what are they?) \newline
\noindent{\bf Solution}\
Note that the four complex
roots of $f$ are $\lambda, \lambda^2, \lambda^3$ and $\lambda^4$
and that $|K: \mathbb Q| = 4$ so $|G| \leq 4$ 
(and in fact $K$ is the splitting
field of $f$ so $|G| = 4$ but we do not need this).
The previous part shows that there are at least 4 elements
of the $G$ so we deduce (without the parenthetic
theory) that $|G| = 4$. The four elements of $G$
are specified by $\lambda \mapsto \lambda^i$ for
$i = 1, \ldots, 4$.
\end{enumerate}
\end{enumerate}
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