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\title{MATH0037 Galois Theory 2002, Solutions 6}
\author{Geoff Smith, http://www.bath.ac.uk/$\sim$masgcs}
\date{}
\begin{document}
\maketitle
 \begin{enumerate}
\item {\em Suppose that $K$ is a field. Define a map
$D_K : K[X] \rightarrow K[X]$ by
\[\sum_{i=0}^n a_i X^i \mapsto \sum_{i=0}^n ia_iX^{i-1}.\]
This map is familiar in the context that
$K = \mathbb R$, but that is not our assumption.
We may write $f'$ for $D_K(f)$.}
\begin{enumerate}
\item  {\em The ring $K[X]$ is vector space over $K$. Show
that the map $D_K$ is a linear map.} \newline
{\bf Solution}\ This is routine.
{\em \item  Suppose that $f, h \in K[X]$. Prove that
$D_K(fh) = D_K(f)h + fD_K(h).$} \newline
{\bf Solution}\ This is routine. The proof of Leibniz's rule
for differentiation given in first year courses is valid
for formal differentation using coefficients in an arbitrary field.
The details involve writing $f$ and $h$ as sums of monomials,
and calculating.
\item {\em Suppose that $K$ is a subfield of $L$. Prove that
if $f \in K[X],$ then $D_K(f) = D_L(f)$.} \newline
{\bf Solution}\  This is true because of the recipe for 
`differentiation' in each case. This is an apparently trivial
observation$^\dagger$ which has powerful consequences. 
\item {\em Suppose that $f \in \mathbb Q[X]$ is irreducible.
Show that there is no field $L$ with subfield $\mathbb Q$
such that there is $\alpha \in L$ with $f = [X-\alpha]^2 h$
for some polynomial $h \in L[X]$.} \newline
\noindent {\bf Solution}\ Suppose that such 
$h$ and $\alpha$ exist (for contradiction). Now \[D_L(f)
= D_K(f) = 2[X-\alpha]h + [X-\alpha]^2D_K(h)\] has
$\alpha$ as a root (we just used that observation $\dagger$. 
Therefore $\alpha$ is a root of
both $f$ and $D_K(f)$, and so their g.c.d. 
(which is of the form $af + bD_K(f)$ for rational polynomials
$a,b$) must have $\alpha$ as a root. Therefore their g.c.d. 
is not 1. Now the irreducibily of $f$ and the fact that
$D_K(f)$ has smaller degree than $f$ ensures that 
$D_K(f) = 0$. However, since $K$ contains $\mathbb Q$ then
only way that a derivative of a polynomial can vanish
is if that polynomial is constant (this argument fails when
the characeteristic is not 0). However, $f$ is irreducible 
as an element of $K[X]$, so is not a constant polynomial.
We are done.

\item $Harder$ {\em Let $F=\mathbb Z_p$ be the field which
consists of the integers modulo
$p$. The polynomial ring $F[Y]$ is an integral domain so
we may form its field of fractions $K$ by a construction
found on problem sheet 3 (Question 1). Now consider
$X^p - Y \in K[X]$. Show that in $K[X]$ this polynomial
is irreducible, but that given any field $L$ with 
$K < L$ such that $L$ contains a root $\alpha$ of $X^p-Y$,
then $X^p-Y = [X-\alpha]^2 h$ for some polynomial $h \in L[X]$. }\newline
\noindent {\bf Solution}\ Let $f \in K[X]$ be a monic irreducible factor
of $X^p - Y \in K[X]$. Note that $f$ is not linear since 
$Y$ is not the $p$-th power of an element of $K$. This is
because if $u, v \in K[Y]$ and $v \not = 0$ are such
that $Y = (u/v)^p$, then $v^pY = u^p$. The degree
(in the variable $Y$) of $v^pY$ is one more than a 
multiple of $p$, whereas the degree of $u^p$ is a multiple
of $p$ or is $-\infty$, so this is absurd.

Having established that $f$ is not linear, choose a root
$\alpha$ of $f$ in some extension field. Now $[X - \alpha]^p
= X^p - \alpha^p = X^p - Y$. It follows that $\{ \alpha\}$ is 
a complete set of roots of $f$. Thus $f = [X - \alpha]^t$ for
some $t$ in the range $1 \leq t \leq p$. Thus $\alpha^t \in K[Y]$.
Now suppose that $t < p$ (for contradiction). Then 
the g.c.d. of $t$ and $p$ is 1, so there are integers
$\lambda, \mu$ such that $\lambda t + \mu p = 1$.
Now \[
\alpha = \alpha^1 = (\alpha^p)^\mu + (\alpha^t)\lambda \in K[Y].\]
However, $Y$ has no $p$-th root in $K[Y]$ by degree considerations,
but $\alpha^p = Y$. This is absurd so $t = p$ and 
$X^p - Y = f$ is irreducible in $K[X]$, yet has a single root
(repeated $p$ times) in the extension field $K(\alpha)$. 

\end{enumerate}
\item {\em Suppose that $K$ is a field and that
$0 \not = f \in K[X]$. Now $R = K[X]/(f)$ has a natural
vector space structure. Prove that dim$_K R = \mbox{deg }f$.
What happens if $f = 0$?} \newline
\noindent {\bf Solution}\ This is bookwork, but we repeat is
here. First suppose that $f \not = 0$ and that
$f = \sum_{i=0}^n a_i X^i$. Recall that if $h \in K[X]$
then $h = q f + r$ where $q, r \in K[X]$ and deg $r$ < deg $f$.
Thus the cosets $h + (f)$ and $r + (f)$ are the same.
We deduce that \[1 + (f), X + (f), \ldots, X^{n-1} + (f)\]
spans $R$. Suppose that $\lambda_0, \lambda_1, \ldots, \lambda_{n-1}$
are elements of $K$ and that  
\[ \sum_{i=1}^{n-1} [\lambda_i X^i + (f)] = 0,\]
then 
\[ \sum_{i=1}^{n-1} \lambda_i X^i \in (f).\]
From degree considerations it follows that $\sum_{i=1}^{n-1} \lambda_i X^i$
is the zero polynomial, so $\lambda_i = 0$ for each $i$ in the range
$0 \leq i \leq n-1$. Now the sequence of elements
\[ 1 + (f), X + (f), \ldots , X^{n-1} + (f)\]
forms a basis of the $K$-vector space $R$.

Finally, if $f = 0$, then a similar argument yields that
the infinite sequence of cosets $X^i + (0)$ is a basis of $R$,
which is therefore an infinite dimensional vector space over $K$.
 
\item {\em Suppose that $K$ is a field. Suppose that $f, h \in K[X] = R$
and that neither $f$ nor $h$ is the zero polynomial.
We have two natural maps $\pi_1 : R 
\mapsto K[X]/(f) = R_1$ and $\pi_2 : K[X] \mapsto
K[X]/(h) 
= R_2$. Define a map
$\pi : R \mapsto R_1 \oplus R_2$ by
$g \mapsto (\pi_1(g), \pi_2(g))$ for all $g \in R$.
Suppose that $(f) + (h) = R$.}
\begin{enumerate}
\item {\em Prove that
$\pi$ is a surjective ring homomorphism and
a linear map between $K$-vector spaces.} \newline
\noindent {\bf Solution}\ The surjectivity is best postponed. 
Linearity of this map is routine.
\item {\em Prove that Ker $\pi = (f) \cap (h) = (fh)$.} \newline
\noindent {\bf Solution} This is the most interesting part.
Clearly $(fh) \subseteq (f)$ and $(fh) \subseteq (fh)$ so
$(fh) \subseteq (f) \cap (h)$. Converselt suppose that
$t = ah = bf \in (f) \cap (h)$, with $a,b \in R$.
Since $(f) + (h) = R$ there are $c, d \in R$ such that
$1 = cf + dh$. Therefore \[t = tcf + tdh = acfh + bdfh \in (fh).\] 
Thus $(f) \cap (h) \subseteq (fh)$ and so $(f) \cap (h) = (fh).$ 
\item {\em By considering dimensions, prove that
$\pi$ is surjective.} Notice that Ker $\pi = (f) \cap (h)
= (fh)$. The first isomorphism theorem tells us that
$R/(fh) \simeq \mbox{Im }\pi$, so dim Im $\pi = $ deg$fh
= \mbox{deg }f + \mbox{deg }h$. However dim $R_1 = $ deg $f$ and
dim $R_2 = $ deg $h$. Moreover dim $R = $ dim $R_1 + $ dim $R_2
= $ dim Im $\pi$. Therefore Im $\pi = R$.
\end{enumerate}
\end{enumerate}
\end{document}
