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\title{MATH0037 Galois Theory 2002, Solutions 5}
\author{Geoff Smith, http://www.bath.ac.uk/$\sim$masgcs}
\date{}
\begin{document}
\maketitle
\begin{enumerate}
\item {\em Suppose that 
$f = \sum_{i=0}^N a_i X^i \in \mathbb Z[X]$ (with $N >0$) and 
that $p$ is a prime number with the following properties.
\begin{enumerate} 
\item $p \mid a_i$ for all $i > 0$,
\item $p \not\ \mid a_0$ and
\item $p^2 \not\ \mid a_N$,
\end{enumerate}
By considering the ring epimorphism $\overline \pi : \mathbb Z[X]
\rightarrow \mathbb Z_p[X]$ defined by taking the coefficients
modulo $p$, show that $f$ is not factorizable as a product 
of two integral polynomials of smaller degree. Deduce that
$f$ is irreducible as a rational polynomial by appeal to
a result of Gauss.} \newline
\noindent{\bf Solution}\  Suppose that $f = h \cdot k$ is
the product of two elements of $\mathbb Z[X]$. Now 
$\pi(f) = a$ for some non-zero $a \in \mathbb Z_p$.
Now the images of $h, k$ must also be non-zero
constants by degree considerations. Thus $h = h_0 + pXh_1$ 
and $k = k_0 + pXk_1$ where $h_0, k_0$ are integers coprime
to $p$ and $h_1, k_1 \in \mathbb Z[X]$. We deduce that
$h_1 = 0$ or $k_1 = 0$ else the third condition is
violated. Therefore either $h$ or $k$ is a constant polynomial.
Now Gauss's lemma tells us that if $f$ factors into polynomials
of smaller degree in $\mathbb Q[X]$, then it does so in
$\mathbb Z[X]$. It does not, so $f \in \mathbb Q[X]$
is irreducible.

\item {\em Let $F$ be a field and suppose that
$c \in F$. Consider maps $\alpha_c, \beta : F[X]
\rightarrow F[X]$ defined as follows. In each case
$0 \not = f = \sum_{i=0}^N a_i X^i$ has degree $N \geq 0$
and is otherwise arbitrary.
\begin{enumerate}
\item $\alpha_c : f \mapsto \sum_{i=0}^N a_i (X+c)^i$  and $0 \mapsto 0$.
\item $\beta : f \mapsto X^N f(1/X)$ and $0 \mapsto 0$.
\end{enumerate}
Show that each of these maps is an {\em automorphism} of the 
ring $F[X]$ (i.e. an isomorphism from $F[X]$ with itself) and
that each map preserves degree. Deduce that these automorphisms,
and their inverses, send irreducible elements of $F[X]$ to irreducible
elements of $F[X]$.}
\newline
\noindent{\bf Solution}\ 
\begin{enumerate}
\item It is easy to verify that each $\alpha_c$
is a ring homomorphism, and preserves degree. Note that
$\alpha_c \circ \alpha_{-c}$ is the identity map for al $c$,
so each $\alpha_c$ is bijective and is therefore a degree preserving
isomorphism.
\item This question is wrong as it stands. However, things
can be rescued. The point is that if $h = h_0 + \cdots + h_N X^N$
where $h_0, h_N$  are non-zero, then $\beta(f) =
h_0X^N + \cdot + h_N$. If we restrict attention to polynomials
of degree at least one which have non-zero constant terms, the
map $\beta$ is a bijection (since $\beta \circ \beta$ is the identity)
and it preserves degree and respects products. Since any non-trivial
factorization of a polynomial in thsi set must be into a product
of polynomials in this set, it follows that $h$ is irreducible
if and only if $\beta(h)$ is irreducible.
\end{enumerate}
\item {\em Which of the following polynomials are irreducible when viewed as
rational polynomials?}
\begin{enumerate}
\item $14X^7 + 7X + 2$.\newline
\noindent{\bf Solution}\ This is irreducible by Eisenstein's criterion. Use $p =7$.
\item $(\sum_{i=0}^N (i+1)! x^i) + 42X^{N+1}$ where $N \geq 0$. \newline
\noindent{\bf Solution} This is irreducible by Eisenstein's criterion. Use $p =2$.
\item $2X^7 + 7X^6 + 14$. {\em Hint. Use the previous question. The map
$\beta$ may be useful}. \newline
\noindent{\bf Solution}\ This is irreducible by Eisenstein's criterion
(after application of $\beta$ and use of $p = 7$. Typically people 
don't worry about about $\beta$ too much and
use Eisenstein's criterion with reversed coefficients and 
still call it Eisenstein's criterion. 
\item $X^2 + X + 1$. {\em Hint. Use the previous question. The map
$\alpha_1$ may be useful}. 
 \newline
\noindent{\bf Solution}\ This polynomial is $(X^3-1)/(X-1)$. Replacing $X$ by $X+1$
we obtain $X^2 + 3X + 3$ which is irreducible in $\mathbb Q[X]$ by
Eisenstein's criterion. Thus $X^2 + X + 1 \in \mathbb Q[X]$ is irreducible.
\item $X^3 + X^2 + X + 1$. \newline
\noindent{\bf Solution}\ Observe that 
\[(X-1)(X^3 + X^2 + X + 1) = X^4 -1 \] \[= (X^2-1)(X^2+1) = (X-1)(X+1)(X^2+1).\]
Thus $X^3 + X^2 + X + 1 = (X+1)(X^2+1)$ is not irreducible in $\mathbb Q[X]$.
\item $X^4 + X^3 + X^2 + X + 1$.  \newline
\noindent{\bf Solution}\ This polynomial is $(X^3-1)/(X-1)$. Replacing $X$ by $X+1$
we obtain $X^5 + 10 X^4 + 10X^3 + 5X + 1$ which is irreducible in $\mathbb Q[X]$
by Eisenstein's criterion. The point is that 5, like 3, is a prime number.
\item $X^5 + X^4 + X^3 + X^2 + X + 1$.  \newline
\noindent{\bf Solution}\ 6 is not prime so by now we should expect this 
polynomial to be reducible. We have
\[ X^6 - 1 = (X-1)(X^5 + X^4 + X^3 + X^2 + X + 1) =
(X^3-1)(X^3 + 1)\]
so
\[ X^5 + X^4 + X^3 + X^2 + X + 1 = (X^2 + X + 1)(X^3 +1)\] \[
= (X^2 + X + 1)(X+1)(X^2 - X + 1)\]
is not irreducible in $\mathbb Q[X]$.
\item $X^6 + X^5 + X^4 + X^3 + X^2 + X + 1$. \newline
\noindent{\bf Solution}\ 6 is not prime. See the next solution.
\item $X^{p-1} + X^{p-2} + \cdots + X + 1$ {\em where $p$ is a prime number.}  \newline
\noindent{\bf Solution}\ USe $\alpha_1$ as above. Replacing $X$ by $X+1$ in
$(X^p-1)/(X-1)$ we obtain 
\[ X^{p-1} + \left(^p_1 \right)X^{p-2} +  \left(^p_2 \right)X^{p-3} + \cdots + p\] 
which is irreducible by Eisensteins criterion using the prime
$p$. Note that these binomial coefficients $(^p_i)$ for $1 \leq i \leq p-1$
are divisible by $p$ by a cancelling argument.
\end{enumerate} 
\item {\em Suppose that 
$f = \sum_{i=0}^N a_i X^i \in \mathbb Z[X]$ (with $N >0$) and
that $p$ is a prime number with the following properties.
\begin{enumerate}
\item $p \mid a_i$ for all $i < N$,
\item $p \not \ \mid a_N$ and
\item $p^2 \not \ \mid a_0$.
\end{enumerate}
Prove that $f$ is irreducible when viewed as a rational polynomial.}\newline
\noindent{\bf Solution}\ Apply $\beta$ as above, then use Eisenstein's criterion.
\end{enumerate}
\end{document}
