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\title{MATH0037 Galois Theory 2000}
\author{GCS - Solution Sheet 2}
\date{}
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\maketitle

{\it All rings are
commutative with 1 and $0$ is not a natural number as 
far as this course is concerned}
\begin{enumerate}
\item Consider $f, g \in \mathbb Q[X]$ where $f = x^4 + 4x +1$
and $g = x^2 +x -2$. Find $q, r \in \mathbb Q[X]$
such that $f = qg + r$ and $\hbox{deg }r < \hbox{deg }g$.
\newline\noindent{\bf Solution} Long division yields
\[ x^4 + 4x + 1 = [x^2 -x + 3][x^2 + x -2] + [-x+7].\]
\item Find $f \in \mathbb Q[X]$ of smallest possible degree
with the property that when divided by $X-j$ (in $\mathbb Q[X]$)
it leaves remainder $j$ for each $j \in \{1,2,3,4\}$.
\newline\noindent{\bf Solution} If $c$ is  constant polynomial
then $c$ leaves remainder $c$ when divided by each
of the four polynomials in question, and if $1 = c = 2$,
then $1-1 = 2-1$ so $0 \not = 1$ which violates a field axiom.
Actually, in $\mathbb Q$ we know that $1 \not = 2,$ but we 
can now see how the argument goes if $\mathbb Q$ is replaced
by an arbitrary field. Anyway, we may deduce that the required
polynomial has degree at least 1. Now observe that
$X = [X -j] + j$ for each $j \in {\mathbb Z}$, so $X$ 
is a polynomial of degree 1 which enjoys the property.
The enthusiastic student should try to prove that $X$ is the
{\it only } polynomial of degree 1 which leaves the specified
remainders when divided by the given polynomials.
\item Suppose that $f \in \mathbb C[Y]$ is a polynomial
with the property that $\mathbb C[Y]/(f)$ and  
$\mathbb C[Y]/(f^2)$ are isomorphic rings. What can you
deduce about $f$?
\newline\noindent{\bf Solution} We may assume 
that $f \not = 0$ since in that case the rings
are equal and therefore isomorphic.
If $g \in \mathbb C[Y]$, then
we showed in the problem class 
(and will show again in lectures) that 
considered as a vector space over the subfield consisting
of the constant polynomials, the dimension of  $\mathbb C[Y]/(g)$
is $\hbox{deg }g$. Since a ring isomorphism $\phi$ sends 1 to 1, 
$\phi$ must send $(f) + c$ to $(f^2) + c$ for all constant
polynomials $c$. One can check that $\phi$
is then a $\mathbb Q$-linear
map which is non-singular. The dimensions of the domain and
codomain must therefore coincide. This forces $f$ to be
a constant polynomial, and both domain and codomain
to be the trivial ring containing a single element,
and indeed these rings are isomorphic.
\item Suppose that $R$ is a ring and that $I_1, I_2$ are ideals
of $R$. For $i=1,2$ we let $\pi_i$ denote the
natural epimorphism $\pi_i : R \rightarrow R/I_i$ where
$(r)\pi_i = I_i + r$ for every $r \in R$.
Define $\pi : R \rightarrow R/I_1 \oplus R/I_2$ by
$(r)\pi = ( (r)\pi_1, (r)\pi_2)$ for every $r \in R.$
\begin{enumerate}
\item Prove that $\pi$ is a ring homomorphism.
\newline\noindent{\bf Solution} This is a formality.
\item Prove that $\pi$ is injective if and only if
$I_1 \cap I_2 = \{ 0\}$.
\newline\noindent{\bf Solution} $\hbox{Ker }\pi_j
= I_j$ for $j = 1,2$. Now \[\hbox{Ker }\pi = 
\hbox{Ker }\pi_1 \cap 
\hbox{Ker }\pi_2 
= I_1 \cap I_2\]
and we are done.
\item Prove that $\pi$ is surjective if and only if
$I_1 + I_2 = R$.
\newline\noindent{\bf Solution} Abusively write
$0$ and $1$ for the additive and multiplicative
identity elements of each of the rings mentioned in 
the question. Suppose that $I_1 + I_2 = R$.
Thus $1 = a + b$ for some $a \in I_1 = \hbox{Ker }\pi_1$
and $b \in I_2 = \hbox{Ker }\pi_2$. Now $a = 1-b$
and $b = 1-a$ so $\pi: a \mapsto (0,1)$ and $\pi: b 
\mapsto (1,0).$ Therefore if we choose any $(I_1 +c,I_2 +d)
\in R/I_1 \oplus R/I_2$, then \[\pi : c b + d a
\mapsto  (I_1 +c,I_2 +c)(1,0) + (I_1 +d,I_2 +d)(0,1)
= (I_1 +c,I_2 +d).\] Thus $\pi$ is surjective.
\item Prove that the rings $\mathbb Z/(6)$
and $\mathbb Z/(2) \oplus \mathbb Z/(3)$ are isomorphic.
\newline\noindent{\bf Solution} Use the preceding discussion
with $R = \mathbb Z$, $I = (2)$ and $J = (3)$.
Observe that $(2) + (3) = (1) = \mathbb Z$. 
Also $(2) \cap (3) = (6).$ Now apply the first
isomorphism theorem for rings.
\end{enumerate} 
\item Suppose that $R$ is a ring and that $I, J$ are ideals
of $R$ with the property that $I + J = R$. Suppose that
$n, m$ are natural numbers.
Prove that $I^m + J^n = R$. {\it Hint $1^{m+n} = 1$}.
\newline\noindent{\bf Solution} Clearly
$I^m + J^n \subseteq R$ so only the reverse inclusion
is of interest. Since $I^m + J^n$ is an ideal, it will suffice
to show that $1 \in I^m + J^n$ for then for each $r \in R$
we will have $r = r 1 \in I^m + J^n$.
Now $1 = i + j$ where $i \in I$ and $j \in J.$
\[ 1 = (i+j)^{m+n} = \sum_{k=0}^{m+n} 
= \left(\begin{array}{c}m+n\\k\end{array}\right)i^k j^{m+n -i}.\]
Notice three things: the coefficients arising in the binomial theorem
are integers, if $z \in \mathbb N$ and $R \in R$, then $zr$
is simply notation for $r + r + \cdots + r$ where there are $z$ copies
of $r$ in the sum, and finally that each $ i^k j^{m+n -k}$
is either an element of $I^m$ (when $k \leq n$) or an element of
$J^m$ (when $k \geq m$). Thus $1 \in I^m + J^n$ and we are done.
\end{enumerate}
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