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\title{MATH0037 Galois Theory 2000}
\author{GCS -- Solution Sheet 1}
\date{}
\begin{document}
\maketitle

{\it All rings are
commutative with 1 as far as this course is concerned}
\begin{enumerate}
\item Suppose that $R$ is a ring and that $I$ and $J$ are both ideals
of $R$. Let $I + J$ be the subset
$\{ a + b \mid a \in I,\ b \in J \}$ of
$R$. Prove that $I + J \unlhd R.$
\newline\noindent{\bf Solution}
Let $X = I + J$. Now $0 = 0 + 0 \in X$ since $0 \in I \cap J$.
Thus $X \not = \emptyset$. In order to show that $X$ is an additive
subgroup of $R$ it suffices to show that if $x,y \in X,$ then
$x - y \in X.$ Choose any such $x$ and $y$. Then 
$x = i_1 + j_1$ and $y = i_2 + j_2$ for $i_1,i_2 \in I$ and
$j_1, j_2 \in J.$ Therefore $x-y = [i_1 - i_2] + [j_1 - j_2]$.
Now $I$ and $J$ are additive subgroups of $R$ so 
$ i_1 - i_2\in I$ and $j_1 - j_2 \in J$ so $x - y \in I+J.$
Finally to establish that $X$ is an ideal of $R$ we must
show that if $r \in R$ and $x \in X$ then $rx \in X.$
Choose any such $r$ and $x = a + b$ for $a \in I$ and $b \in J$.
Now $rx = r(a+b) = ra + rb$ where $ra \in I$ and $rb \in J$
since $I, J \unlhd R.$ Thus $I + J \unlhd R$.
\item Suppose that $R$ is a ring and
that $I$ and $J$ are both ideals
of $R$. Let $IJ$ be the set of all sums
\[ \sum_{i=1}^m a_i b_i \]
where $m$ is any natural number, and for each $i$ we have
$a_i \in I$ and $b_i \in J$. Prove that $IJ \unlhd R.$
\newline\noindent{\bf Solution}
First observe that $0 = 0 \cdot 0 \in IJ$ so $IJ \not =
\emptyset$. Next suppose that $x,y \in IJ$,
so $x = \sum_{i=1}^u c_i d_i$ and $y = \sum_{j=1}^v e_j f_j$
where $c_i, e_j \in I$ and $d_i, f_j \in J$ for all legal
values of $i$ and $j$. Now put $c_{u+j} = -e_j \in I$ for $1 \leq j \leq v$
and $d_{u+j} = f_j \in J$ for $1 \leq j \leq v$.
Now $x - y =  \sum_{i=1}^{u+v} c_i d_i \in IJ$ so $IJ$ is
an additive subgroup of $R$.
Now choose any  $x = \sum_{i=1}^u c_i d_i$
where $c_i \in I$ and $d_i \in J$ for all currently legal 
values of $i$. Also choose $r \in R.$
Then $rx = \sum_{i=1}^u (rc_i) d_i \in IJ$ since
each $rc_i \in J$ because $J \unlhd R$.
\item Suppose that $R$ is a ring and that $I$ is an ideal
of $R$. The previous question given meaning to the ideal
$II$ and we refer to this ideal as $I^2$. How should we define
$I^3$? Give a sensible definition
of an ideal $I^n$ when $n$ is any natural number.\newline\noindent{\bf Solution}
Define $I^0$ to be $R$. For $m \geq 1$ inductively define
$I^m$ to be $I I^{m-1}.$ Informally, $I^m$ will be the 
set of all possible sums of terms each of which is
the product of $m$ elements of $I$. 
\item Recall that every subgroup of a cyclic group is cyclic.
Use this result to classify (list and understand)
the additive subgroups of $\mathbb Z$
(the additive group of integers). Now classify the
ideals of the ring $\mathbb Z$.
\newline\noindent{\bf Solution}
>From an earlier course, the subgroups of ${\mathbb Z}$
are exactly the cyclic groups $(m)$ where $m$ is any non-negative
integer. It so happens that these additive subgroups of ${\mathbb Z}$
are all ideals of $\mathbb Z$ since they are all principal ideals.
\item Suppose that $m$ is an integer. Let
$(m)$ denote the set of integer multiples of
$m$, so $(m) = \{km \mid k \in {\mathbb Z}\}$.
\begin{enumerate}
\item How many elements are there in the ring 
${\mathbb Z}/(2)$?\newline\noindent{\bf Solution}
There are 2.
\item How many elements are there in the ring 
${\mathbb Z}/(m)$?\newline\noindent{\bf Solution}
There are $m$.
\item Write out the addition and multiplication tables of
${\mathbb Z}/(3)$ (it may be helpful to write
${\bold 0} := (3) + 0,$
${\bold 1} := (3) + 1$ and
${\bold 2} := (3) + 2$.
\end{enumerate}
\noindent{\bf Solution}
All is as for $\mathbb Z$ save that 
\[{\bold 2} + {\bold 1} ={\bold 0} = {\bold 1} + {\bold 2}\] 
and
\[{\bold 2} + {\bold 2} ={\bold 1} = {\bold 2} + {\bold 2}.\] 
\item Suppose that $R$ is a ring and that $I$ and $J$ are both ideals
of $R$. Show that $I \cup J$ is an ideal of $R$ if and only if
$I \cap J \in \{ I, J\}$.
\newline\noindent{\bf Solution}
If $I \cap J \in \{ I, J\}$ then either $ I \subseteq J$
or $J \subseteq I$ so $I \cup J \in \{I,J\}$ so
$I \cup J \unlhd R.$ The converse
is more interesting. Suppose that $I \cup J \unlhd R.$ 
Now assume (for contradiction that neither $I \subseteq
J$ nor $J \subseteq I$. we may therefore choose
$i, j$ such that $i \in I$, $j \in J$, $i \not \in J$
and $j \not \in I$. Now $i + j \in I \cup J$ since
$i, j \in I \cup J \unlhd R$.
If $i + j \in I$ then $j \in I$ which is absurd.
Therefore %i + j \in J$. However, this forces
$i \in J$ which is also absurd. Therefore 
either  $I \subseteq
J$ or $J \subseteq I$. In either event,
$I \cap J \in \{I,J\}$ and we are done.

\item An ideal $I$ of a ring $R$ is called a
{\it principal ideal} if there is $r \in R$ such that
$I = (r)$ where $(r) = \{sr \mid s \in R\}$.
Find an ideal of a ring which is not a principal ideal
of that ring. {\it Hint: Look in ${\mathbb Z}[X]$, the
ring of polynomials in the variable $X$ with
integral coefficients.}\newline\noindent{\bf Solution}
The set of polynomials in  ${\mathbb Z}[X]$ with even constant term
form an ideal. However $2$ is in this ideal, and the only 
polynomials in our ring which divide $2$
are $1,2,-1,-2$ from degree considerations. Now
$X \not \in (2) = (-2)$ [this is the ideal of all polynomials 
in ${\mathbb Z}[X]$ with all coefficients being even].
Also $(1) = (-1) = {\mathbb Z}[X]$. However our ideal is not
${\mathbb Z}[X]$ since it does not contain 1. 
Therefore our ideal is not principal.
\item Let ${\mathbb C}[[X]]$ denote the set of formal power series
in $X$ with coefficients in ${\mathbb C}.$ These series are simply
expressions of the form
\[ a_0 + a_1 X + a_2 X^2 + \ldots \]
where we pay no heed to convergence questions. Show how to
define addition and multiplication on ${\mathbb C}[[X]]$
in a natural way so that it becomes a ring.

When you have done this, show that $f = f_0 + f_1 X + \ldots \in
{\mathbb C}[[X]]$ has a multiplicative inverse
in ${\mathbb C}[[X]]$ if and only
if $f_0 \not = 0$.\newline\noindent{\bf Solution}
Let $\sum_{i = 0}^\infty a_i X^i$ and $\sum_{i = 0}^\infty b_i X^i$
be elements of $ {\mathbb C}[[X]]$. We define their sum and product
to respectively be 
$\sum_{i = 0}^\infty s_i X^i$ and   
$\sum_{i = 0}^\infty p_i X^i$ where for every $n \geq 0$
we let $s_n = a_n + b_n$ and $p_n = \sum_{i=0}^n a_i b_{n-i}$.
It is a routine and somewhat tedious matter to verify that
${\mathbb C}[[X]]$ so equipped has become a ring.

If $f = f_0 + f_1 X + \ldots \in
{\mathbb C}[[X]]$ has $f_0 =0$ then $fh$ will have constant term $0$
for every $h \in {\mathbb C}[[X]]$, and so $f$ will have no multiplicative
inverse in ${\mathbb C}[[X]]$. On the other hand, if $f_0 \not = 0$
then $f - f_0$ has constant term $0$ and so has the form
$f_0Xg$ for some $g \in {\mathbb C}[[X]]$. Now $f = f_0(1 + gX)$
and $f_0^{-1}( 1 - gX + g^2 X^2 + g^3 X^3 - \cdots)$ can easily be verified
to be a multiplicative inverse of $f$.
\end{enumerate} 
\end{document}