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\author{Jason Wojciechowski}
\title{Notebook Problem Fraleigh 9.9.8}

\newtheorem*{thm}{Theorem}

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\maketitle

We will construct a polynomial in $\Q[x]$ of degree $5$ which is
insoluble by radicals.

Suppose we have $H$, a subgroup of $S_5$, which contains a cycle of
length $5$ and a transposition.  We will show that $H$ is $S_5$.

Suppose, without loss of generality, that $H$ contains $(12345)$ and
$(12)$.  Then $H$ must contain products of these, as well as inverses,
so it must contain $(12345)(12)(12345)^{-1} = (12345)(12)(15432) =
(1)(23)(4)(5) = (23)$, so it must contain $(12345)(23)(15432) =
(1)(2)(34)(5) = (34)$ and $(12345)(34)(15432) = (45)$.

Now we see that $H$ must contain $(12)(23)(12) = (13)$, $(13)(34)(13)
= (14)$, and $(14)(45)(14) = (15)$.  Thus $H$ has all transpositions
of the form $(1m)$, so it has $(1m)(1n)(1m) = (mn)$, so $H$ has all
transpositions, but since every permutation is a product of
transpositions, $H$ contains every permutation of $5$ elements, so $H
= S_5$.

Now, given $f(x)$, an irreducible polynomial of degree $5$ in $\Q[x]$
with three real (and thus two complex) roots, we will show that the
Galois group of the splitting field $\Sigma$ of $f$ over $\Q$ is
$S_5$.

Let $G$ be the Galois group of $\Sigma : \Q$.  Since $G$ is the group
of automorphisms that leave $\Q$ fixed, we can consider $G$ to be the
permutations of the five zeros of $f$, so that $G$ is a subgroup of
$S_5$.  Since $f$ has two complex zeros, there is an automorphism that
takes one to the other, which can be seen as an element of order $2$,
or a transposition.  At some point in constructing $\Sigma$, we adjoin
an element of order $5$ (since the polynomial is of degree $5$), so
the order of the extension $\Sigma : \Q$ is divisible by $5$, so the
order of $G$ must also be divisible by $5$.  Since $5$ is a prime, we
can apply Theorem $13.14$ from Stewart to get that there is an element
of order $5$ in $G$.  The only elements of order $5$ is $S_5$ are
$5-\text{cycles}$, so we know that $G$ contains a five cycle and a
transposition, so $G$ must be $S_5$.

Consider $f(x) = 2x^5-5x^4 + 5$.  Since $5$ does not divide $2$,
$5|5$, $5|0$, and $25$ does not divide $5$, this polynomial is
irreducible by Eisenstein's criterion.

\begin{figure}
\centering \includegraphics{graph}
\end{figure}

The graph of $f$ shows that it has three real roots.  We can verify
this using calculus.  The derivative of $2x^5 - 5x^4 + 5$ is $10x^4 -
20x^3$.  Setting this equal to zero and solving gives that there are
critical points at $x=0$ (this is a solution of multiplicity $3$) and
$x=2$. $f(0) = 5$ and $f(2) = -9$, so $x=0$ is a local maximum and
$x=2$ is a minimum, and since there are only two local minima or
maxima, there are three real roots.

We will combine the second result above (that the Galois group of the
splitting field of a polynomial with three real and two complex roots
is $S_5$) with the contrapositive of Stewart's Theorem $14.6$, namely:

\begin{thm}
  Let $f$ be a polynomial over a field $K$ of characteristic $0$.  If
  the Galois group of $f$ over $K$ is not soluble, then $f$ is not
  soluble by radicals.
\end{thm}

First note that our field $\Q$ has characteristic $0$.  Next, since
the Galois group of $f$ over $\Q$ is $S_5$ and $S_5$ is not a soluble
group, we've met both conditions of the theorem, so we have as a
result that $f$ is not soluble by radicals.

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