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\title{MATH0037 Galois Theory 2000}
\author{GCS - Solution Sheet 4}
\date{}
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\maketitle

\begin{enumerate}
\item Let $R$ be a principal ideal domain which is not
a field. Prove that $x \in R$ is irreducible
if and only if $(x)$ is a maximal ideal {\em (apparently I omitted one
of the
two implications in lectures, and now is a good opportunity to fill this
lacuna).}
\newline{\bf Solution} Suppose that $x$ is irreducible, so $x \not = 0$
and $x \not \in U(R)$. Suppose that $(x) \subset I \unlhd R$ and $(x)
\not = I$. Now
$R$ is a PID so $I = (y)$ for some $y \in R$. Thus $x = ty$ for
some $t \in R$. Now $t \not \in U(R)$ else $(x) = (y)$, so
the irreducibility of $x$ forces $y \in U(R)$ so $I = R$. Thus
$(x)$ is a maximal ideal.

Now begin again, this time supposing that $(x)$ is a maximal ideal.
Now if $x = 0$, then $(y) = R$ for every $y \not = 0$ so all non-zero
elements of $R$ are units and thus $R$ is a field which is absurd.
Thus $x \not = 0$. Suppose that $x = ab$ for $a,b \in R$.
Thus $(x) \subseteq (a)$.
Now if $a \not \in U(R)$ we have $(a) \not = R$
and so  by maximality of $(x)$ we have $(x) = (a)$.
It follows that $b \in U(R)$. Thus $x$ is irreducible.

\item Suppose that $f \in {\mathbb Z}[X]$ and that $f$ is not the zero
polynomial. We define
the {\em content} $c(f)$ of $f$ to be the greatest common
divisor of the coefficients of $f$. We say that $f$ is {\em primitive}
if $c(f) = 1$.
\begin{enumerate}
\item Prove that if $f, g \in {\mathbb Z}[X]$ are primitive, then $fg$
is primitive. {\em Hint:
If $fg$ is not primitive, then there is a prime number
$p$ which divides each of its coefficients.}
\newline{\bf Solution}
Let $f = \sum_{i=0}^\infty a_i X^i$ where $a_n \not = 0$ but $a_i = 0$
for every $i > n$. Similarly $g = \sum_{i=0}^\infty b_i X^i$ where $b_m
\not = 0$ but $a_i = 0$
for every $i > m$. Suppose that $fg$ is not primitive, so that there
is a prime number $p$ such that $p$ divides each coefficient of $fg$.
Since $f$ and $g$ are primitive there exist minimal $N$, $M$
such that $p$ divides neither $a_N$ nor $b_M$. The coefficient of
$X^{N+M}$ in $fg$ is
\[ a_0 b_{N+M} + \cdots a_N b_M + \cdots a_{N+M} b_0\]
which is a sum of terms exactly one of which is not divisible by
$p$, and so this coefficient is not a multiple of $p$ which is
absurd. Thus $fg$ is primitive.
\item Show that if $f \in {\mathbb Z}[X]$ and $f$ is not the zero
polynomial,
then $f = c(f) \cdot \hat f$ where $\hat f \in {\mathbb Z}[X]$ is a
primitive
polynomial.
\newline{\bf Solution} Clearly $f = c(f) \cdot \hat f$ where
$\hat f \in {\mathbb Z}[X]$. If $q = c(\hat f)$, then $q c(f)$
divides $c(f)$ so $q = 1$. We conclude that $\hat f$ is primitive.
\item Show that if $f, g \in {\mathbb Z}[X]$, then $c(fg) = c(f) c(g)$.
\newline{\bf Solution}
We have \[ c(f) \hat f \cdot c(g) \hat g = fg = c(fg) \widehat
{fg}\]
where the hatted polynomials are primitive. Now $\hat f \hat g$
is primitive. Take the contents of each side of this equation
to deduce that $c(f)c(g) = c(fg)$ (and incidentally notice that
$\hat f \hat g = \widehat{fg}$).
\item Suppose that $f, g, h \in {\mathbb Q}[X]$ with $0 \not = f
\in {\mathbb Z}[X].$ Suppose also that $f = gh$. Show that there are
rational numbers $\alpha, \beta$ such that $\alpha\beta = 1$ and
$\alpha g, \beta h \in {\mathbb Z}[X]$. What does this say
for the factorization of $f$?
{\em Hint: Choose non-zero integers $m,n$ such that $g_1 = mg \in
{\mathbb Z}[X]$ and $h_1 = nh
\in {\mathbb Z}[X]$. Now $mnf = g_1 h_1.$ Get
very interested in the content of each side.}
\newline{\bf Solution}
Follow the hint and take the contents of each side of the equation so
that
$mn c(f) = c(g_1)c(h_1)$. Now
$mn f = c(g_1) \hat g_1 \cdot c(h_1) \hat h_1$ in the notation we have
established. Thus $f = (c(g_1) c(h_1)/mn) \hat g_1 \hat h_1$, where
$0 \not = z = c(g_1) c(h_1)/mn \in {\mathbb Z}.$
Let $g_2 = z g_1$ and $h_2 = h_1$ so $g_2, h_2 \in {\mathbb Z}[X]$.
Note that $g h = g_2 h_2$.
Now $g_1 = \gamma g$ for $0 \not = \gamma \in \mathbb Q$.
Also $g_2 = \delta g_1$ for $0 \not = \delta \in \mathbb Q$.
Thus $g_2 = \alpha g$ for $ 0 \not = \alpha \in \mathbb Q$.
Now $g_2 h_2 = \alpha g h_2 = g h$ so $g( \alpha h_2 - h) = 0$.
However, ${\mathbb Z}[X]$ is an integral domain so $h = \alpha h_2$
and therefore $h_2 = \beta h$ where $\alpha \beta = 1$.
\end{enumerate}
\end{enumerate}
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