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\title{Galois Theory Handout 1}
\author{Geoff Smith, http://www.bath.ac.uk/$\sim$masgcs}
\date{}
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\maketitle
\noindent {\bf Theorem} {\em If $K$ is a if characteristic $0$ (i.e.
it contains a copy of $\mathbb Q$) and $|K(a,b):K|$ is finite,
then there is $c \in K(a,b)$ such that $K(a,b) = K(c)$.}

Observe that this means that all field extensions $K \leq L$
where $|L:K| < \infty$ are {\em simple} extensions in that
$L = K(d)$ for some $d \in L$. This is a straightforward indeuction
on the degree of the extension. We now address the proof of the main result.
\newline \noindent{\bf Proof}
Let $p, q \in K[X]$ be 
irreducible polynomials having $a$ and $b$ as roots (respectively).
Construct a splitting field $\Sigma$ for $pq$. Let the roots of 
$p$ and roots of $q$ in $\Sigma$ be $a = a_1, a_2, \ldots, a_m$
and $b = b_1, b_2, \ldots, b_n$ (respectively). Among the 
infinitely many elements of $K$, choose $\lambda$ which is not
one of the finitely many quantities
\[ \frac{a_i - a}{b - b_j}\]
where $i$ is arbitrary but $j > 1$.
Thus $a_i \not = a + \lambda (b - b_j)$ for all $j >1$.

We will show that $c = a + \lambda b \in K(a,b)$ will so the trick.
Clearly $K(c) \subseteq K(a,b).$ It will suffice to show that
$b \in K(c)$, so then $a = c - \lambda b \in K(c)$ so 
$a, b \leq K(c)$ and therefore $K(a,b) \leq K(c)$ and so we
will be done.

Let $r = r(X) = q(c - \lambda X) \in K(c)[X] \ni q$. Now
$b$ is a root of both these polynomials.  Let $s \in K(c)[X]$
be monic irreducible with $b$ as a root. It follows that $(s)$
is the ideal consisting of all polynomials which have $b$ as a root,
so $s$ must divide them all in $K(c)[X].$
It will suffice to show that $s = X -b$ for then $b \in K(c)$
so that $K(a,b) \leq K(c)$ and we will be done.

We do this by showing that $b$ is the only root of $s$. This
is because of the choice of $\lambda$. The only possible 
roots of $s$ are common roots of $q$ and of $r$.
The roots of $q$ are the various $b_j$, but
$r(b_j) = p(c - \lambda b_j) 
= p(a +\lambda(b - b_j))$ and we have arranged that this is not 0
for $j > 1$. Now $s \in K(c)[X]$ is irreducible and 
$K(c)$ has characteristic $0$, so $s$ does not have repeated roots
in any extension field.
Thus $s = X -b$ and we are done. 
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