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\newcommand{\myname}{Darij Grinberg} % ENTER YOUR NAME HERE
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\newcommand{\psetnumber}{4} % ENTER THE NUMBER OF THIS PSET HERE

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\ihead{Solutions to homework set \#\psetnumber} % Page header left
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\title{	
\normalfont \normalsize 
\textsc{Drexel University, Department of Mathematics} \\ [15pt] % Your university, school and/or department name(s)
\horrule{0.5pt} \\[0.4cm] % Thin top horizontal rule
\huge Math 332: Undergraduate Abstract Algebra II, \\
Winter 2025:
Homework \psetnumber\\% The assignment title
\horrule{2pt} \\[0.5cm] % Thick bottom horizontal rule
Please solve \textbf{at most 3 of the 6 problems!}
% \footnote{%
% I recommend solving as many problems as you can and wish,
% but I will only grade and score 3 solutions per submission
% (and if you submit more, I get to pick which ones I grade).
% \\
% Results stated in class, and claims of previous problems
% (even if you did not solve these previous problems), can
% be used without proof. For example, in solving Problem 5,
% you can use the result of Problem 2 without proof.
% \\
% I expect approximately the level of detail that I give in class.
% Purely straightforward arguments (like checking the ring
% axioms for a direct product of rings) need not be spelled out;
% I only expect a note of their necessity.
% }
\horrule{2pt} \\[0.5cm] % Thick bottom horizontal rule
}
\author{\myname}

\begin{document}

\maketitle % This command causes the title part to be printed.

%----------------------------------------------------------------------------------------
%	EXERCISE 1
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 1}

\subsection{Problem}

In this exercise, we shall see how idempotent central
elements are responsible for rings decomposing as direct
products.

Let $R$ be a ring, and let $e$ be an idempotent
central element of $R$.
(Recall: ``Central'' means that $eb = be$ for all
$b \in R$.)

Exercise 4 \textbf{(a)} on homework set \#1 shows that
$1 - e \in R$ is again idempotent.

\begin{enumerate}

\item[\textbf{(a)}]
Show that $1 - e$ is furthermore central.

\item[\textbf{(b)}]
Show that the principal ideal
$eR$ is itself a ring, with addition and multiplication
inherited from $R$ and with zero $0_R$ and with unity $e$.
(This almost makes $eR$ a subring of $R$, but not quite,
since a subring would have to have unity $1_R$.)

\item[\textbf{(c)}]
Show that the same holds for the
principal ideal $\tup{1-e}R$ (except that its unity will
be $1-e$ instead of $e$).

\item[\textbf{(d)}]
Consider the map
\begin{align*}
f : \tup{eR} \times \tup{\tup{1-e}R} & \to R , \\
\tup{a, b} & \mapsto a+b .
\end{align*}
Prove that this map $f$ is a ring isomorphism.

\end{enumerate}

\subsection{Remark}

Part \textbf{(d)} of this exercise shows that if a
ring $R$ has an idempotent central element $e$, then
$R$ can be decomposed (up to isomorphism)
as a direct product $A \times B$
of two rings $A$ and $B$ (namely, $A = eR$ and
$B = \tup{1-e}R$). If $e$ is not one of the two trivial
idempotents $0$ and $1$, then these two rings $A$ and $B$
will be nontrivial, so the decomposition really deserves
its name.\footnote{As an example, take $R = \ZZ / 6 \ZZ$,
and let $e$ be the idempotent element $\overline{3}
= 3 + 6 \ZZ$ of $R$ (this is idempotent since
$3^2 = 9 \equiv 3 \mod 6$ and thus $\overline{3}^2
= \overline{3^2} = \overline{3}$). Then,
$eR = \set{\overline{0}, \overline{3}}
\cong \ZZ / 2 \ZZ$ and
$\tup{1-e}R = \set{\overline{0}, \overline{2}, \overline{4}}
\cong \ZZ / 3 \ZZ$.
Hence, the ring isomorphism
$R \cong \tup{eR} \times \tup{\tup{1-e}R}$ becomes the
ring isomorphism
$\ZZ / 6 \ZZ \cong \tup{\ZZ / 2 \ZZ} \times \tup{\ZZ / 3 \ZZ}$
that we have seen in \cite[\S 2.12.1]{notes} (as an
instance of the Chinese Remainder Theorem).}

Conversely, any direct product of two nontrivial rings
has nontrivial central idempotents: If $R$ and $S$ are two
rings, then $\tup{1_R, 0_S}$ and $\tup{0_R, 1_S}$ are two
idempotent central elements of the direct product $R \times S$.

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 2
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 2}

\subsection{Problem}

Let $R$ be any nontrivial ring.
Consider the ideals
\begin{align*}
I & := \set{ \begin{pmatrix}
          0 & b \\
          0 & d
        \end{pmatrix}
        \ \mid\ b, d \in R }, \\
J & := \set{ \begin{pmatrix}
          a & b \\
          0 & 0
        \end{pmatrix}
        \ \mid\ a, b \in R }, \\
K & := \set{ \begin{pmatrix}
          0 & b \\
          0 & 0
        \end{pmatrix}
        \ \mid\ b \in R }
\end{align*}
of the upper-triangular matrix ring $R^{2 \leq 2}$
defined in Exercise 2 of homework set \#1.
Note that $K = I \cap J$.

\begin{enumerate}

\item[\textbf{(a)}]
Prove that the ideals $I$ and $J$ are comaximal
(i.e., we have $I + J = R^{2 \leq 2}$).

\item[\textbf{(b)}]
Prove that $IJ = \set{0}$.

\item[\textbf{(c)}]
Prove that $JI = K$.

\end{enumerate}

\subsection{Remark}

This exercise illustrates that $I \cap J = IJ$
does not always hold in the noncommutative case.

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 3
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 3}

\subsection{Problem}

Consider the quotient ring $\QQ^{3 \leq 3} / \QQ^{3 << 3}$
that was constructed in \cite[\S 2.9.3]{notes}.
Recall that we write its elements (which are residue
classes of upper-triangular $3 \times 3$-matrices modulo
the ideal $\QQ^{3 << 3}$) as ``partly determined''
upper-triangular $3 \times 3$-matrices with a question mark
in the top-right corner.

\begin{enumerate}

\item[\textbf{(a)}]
Prove that the map
\begin{align*}
f : \QQ^{3 \leq 3}
&\to \QQ^{2 \times 2},\\
\begin{pmatrix}
a & b & c\\
0 & d & e\\
0 & 0 & g
\end{pmatrix}
&  \mapsto
\begin{pmatrix}
a & b\\
0 & d
\end{pmatrix}
\end{align*}
is a ring morphism.

\item[\textbf{(b)}]
Prove that this morphism $f$ satisfies
$f \tup{\QQ^{3 << 3}} = 0$.

\item[\textbf{(c)}]
Use the universal property of quotient rings
to conclude that there is a ring morphism
\begin{align*}
f' : \QQ^{3 \leq 3} / \QQ^{3 << 3}
&\to \QQ^{2 \times 2}, \\
\begin{pmatrix}
a & b & ?\\
0 & d & e\\
0 & 0 & g
\end{pmatrix}
& \mapsto
\begin{pmatrix}
a & b\\
0 & d
\end{pmatrix} .
\end{align*}

\item[\textbf{(d)}]
Use a similar reasoning to prove the existence of
a ring morphism
\begin{align*}
F' : \QQ^{3 \leq 3} / \QQ^{3 << 3}
&\to \QQ^{4 \times 4}, \\
\begin{pmatrix}
a & b & ?\\
0 & d & e\\
0 & 0 & g
\end{pmatrix}
&  \mapsto
\begin{pmatrix}
a & b & 0 & 0\\
0 & d & 0 & 0\\
0 & 0 & d & e\\
0 & 0 & 0 & g
\end{pmatrix} ,
\end{align*}
which is furthermore injective.

\item[\textbf{(e)}]
Conclude that the ring $\QQ^{3 \leq 3} / \QQ^{3 << 3}$ is
isomorphic to a subring of $\QQ^{4 \times 4}$.

\end{enumerate}

\textbf{[When proving that a map is a ring morphism,
feel free to only check the ``respects multiplication''
axiom, as the other axioms are essentially obvious in
the present context.
Some parts of this exercise can be solved in 1 sentence.]}

\subsection{Remark}

Part \textbf{(e)} shows that if you want to represent the
elements of $\QQ^{3 \leq 3} / \QQ^{3 << 3}$ as actual honest
matrices (i.e., not matrices with question marks or other
kinds of residue classes), then you can do it using
$4 \times 4$-matrices (by duplicating the second entry
on the diagonal).
Doing it by $3 \times 3$-matrices alone does not work
(see \url{https://mathoverflow.net/questions/439996} for
a proof).

More generally, ``partly determined'' upper-triangular
$n \times n$-matrices with a question mark in their
top-right corner can be represented as honest
$\tup{2n-1} \times \tup{2n-1}$-matrices (but not as matrices
of any smaller size).

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 4
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 4}

\subsection{Problem}

Let $R$ be a ring.
Prove the following:

\begin{enumerate}

\item[\textbf{(a)}]
If $I$ and $J$ are two ideals of $R$,
then $IJ$ is an ideal of $R$ as well.

\item[\textbf{(b)}]
The set of all ideals of $R$ is a monoid with respect to the
binary operation $\cdot$, with neutral element $R = 1R$.
That is, we have
\begin{align*}
\tup{IJ} K &= I \tup{JK}
\qquad \text{for any three ideals $I, J, K$ of $R$;}\\
IR &= RI = I \qquad \text{for any ideal $I$ of $R$.}
\end{align*}

\end{enumerate}

\subsection{Remark}

This is part of \cite[Proposition 2.11.2]{notes}
(which you cannot use without proof, of course).

I recommend convincing yourself that the rest of said
proposition is true as well (but don't bother writing it
down unless you want to).
Most of the work required here is bookkeeping.
Note that it is easier to talk abstractly
about sums of $\tup{I, J}$-products than to write them
out as $i_1 j_1 + i_2 j_2 + \cdots + i_k j_k$.
For the proof of $\tup{IJ} K = I \tup{JK}$,
a good approach is to first show that any
$\tup{IJ, K}$-product belongs to $I \tup{JK}$.

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 5
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 5}

\subsection{Problem}

Let $R$ be a ring.
Let $I$ and $J$ be two comaximal ideals of $R$.
Let $n \in \NN$ and $m \in \NN$.
Prove that $I^n$ and $J^m$, too, are comaximal.

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 6
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 6}

\subsection{Problem}



Let $R$ be a ring. Let $n \in \NN$.

For each subset $I$ of $R$, let $I^{n \times n}$ be the
subset
\[
\set{ A \in R^{n \times n} \ \mid \ \text{all entries of }
                A \text{ belong to } I }
\]
of the matrix ring $R^{n \times n}$.

Prove the following:

\begin{enumerate}

\item[\textbf{(a)}]
If $I$ is an ideal of $R$, then $I^{n \times n}$
is an ideal of the matrix ring $R^{n \times n}$.

\item[\textbf{(b)}]
Any ideal of $R^{n \times n}$ has the form $I^{n \times n}$
for some ideal $I$ of $R$.

\end{enumerate}

\subsection{Remark}

In particular, if $F$ is a field, then the matrix ring
$F^{n \times n}$ has only two ideals, namely $\set{0}$ and
the whole $F^{n \times n}$ (where $0$ stands for the zero
matrix). This is because the field $F$ has only two ideals
($\set{0}$ and $F$).

(In contrast, the matrix ring $F^{n \times n}$ has many
more left ideals and right ideals -- i.e., almost-ideals
that satisfy only one half of the second ideal axiom.%
\footnote{For example, a \emph{left ideal} $L$ of a ring
$R$ must satisfy $a\ell \in L$ for all $a \in R$ and
$\ell \in L$, but not necessarily $\ell a \in L$.})

\subsection{Hint}

For each $i, j \in \set{1, 2, \ldots, n}$, let
$E_{i,j} \in R^{n \times n}$ be the
\emph{$\tup{i, j}$-th elementary matrix} -- i.e., the
$n \times n$-matrix whose $\tup{i, j}$-th entry
is $1$ and whose all remaining entries are $0$.
What happens when you multiply a given matrix
$A \in R^{n\times n}$ by $E_{i,j}$ from the left or from
the right? I.e., how can you describe the matrices
$E_{i,j} A$ and $A E_{i,j}$ ?
This exercise will make you truly appreciate elementary
matrices for all the operations that they can perform on
a matrix.

\subsection{Solution}

...


\horrule{0.3pt} \\[0.4cm]

% If you want to cite some sources, uncomment the following
% lines (i.e., remove the "%" symbols at the start of each
% line) and adapt it accordingly.

\begin{thebibliography}{99999999}

% \bibitem[DF]{DF}
% David S. Dummit, Richard M. Foote,
% \textit{Abstract Algebra},
% 3rd edition,
% Wiley 2004.

\bibitem[23wa]{notes}
Darij Grinberg,
\textit{Math 332 Winter 2023 notes: An introduction
to the algebra of rings and fields},
17 January 2025.
\url{https://www.cip.ifi.lmu.de/~grinberg/t/23wa/23wa.pdf}

\end{thebibliography}

\end{document}

