% LaTeX solution template for Math 332
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% Like most advanced LaTeX files, this one begins with a lot of
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% and the number of the pset. (Search for "METADATA" to find the place
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\usepackage[T1]{fontenc} % Use 8-bit encoding that has 256 glyphs
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\usepackage{amsmath,amsfonts,amsthm,amssymb} % Math packages
\usepackage{mathrsfs}    % More math packages
\usepackage{sectsty}  % Allows customizing section commands
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\usepackage{hyperref} % Turns cross-references into hyperlinks,
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\usepackage{graphicx} % For embedding graphics files.
\usepackage{framed}   % For the "leftbar" environment used below.
\usepackage{ifthen}   % Used for the \powset command below.
\usepackage{lastpage} % for counting the number of pages
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\usepackage[height=10in,a4paper,hmargin={1in,0.8in}]{geometry}
\usepackage[usenames,dvipsnames]{xcolor}
\usepackage[all]{xy}  % Package for commutative diagrams.
\usepackage{tikz}     % This is a powerful tool to draw vector
                      % graphics inside LaTeX. In particular, you can
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\usepackage{verbatim} % For the "verbatim" environment, in which
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                      % Useful, e.g., for quoting code (or ASCII art).

\usetikzlibrary{lindenmayersystems} % This is for drawing Sierpinski triangles.

%\numberwithin{table}{section} % Number tables within sections (i.e. 1.1, 1.2, 2.1, 2.2 instead of 1, 2, 3, 4)

\setlength\parindent{20pt} % Makes indentation for paragraphs longer.
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%	VARIOUS USEFUL COMMANDS
%----------------------------------------------------------------------------------------
% The commands below might be convenient. For example, you probably
% prefer to write $\powset[2]{V}$ for the set of $2$-element subsets
% of $V$, rather than writing $\mathcal{P}_2(V)$.
% Notice that you can easily define your own commands like this.
% Caveat: Some of these commands need to be properly "guarded" when
% they occur in subscripts or superscripts. So you should not write
% $K_\CC$, but rather $K_{\CC}$.
\newcommand{\CC}{\mathbb{C}} % complex numbers
\newcommand{\RR}{\mathbb{R}} % real numbers
\newcommand{\QQ}{\mathbb{Q}} % rational numbers
\newcommand{\NN}{\mathbb{N}} % nonnegative integers
\newcommand{\Z}[1]{\mathbb{Z}/#1\mathbb{Z}} % integers modulo k
                                            % (syntax: "\Z{k}")
\newcommand{\ZZ}{\mathbb{Z}} % integers
\newcommand{\id}{\operatorname{id}} % identity map
\newcommand{\lcm}{\operatorname{lcm}}
% Lowest common multiple. For historical reasons, LaTeX has a \gcd
% command built in, but not an \lcm command. The preceding line
% rectifies that.
\newcommand{\diag}{\operatorname{diag}} % diagonal matrix
\newcommand{\ord}{\operatorname{ord}} % order of a group element
\newcommand{\rank}{\operatorname{rank}} % rank of a matrix
\newcommand{\Hom}{\operatorname{Hom}} % group of homomorphisms
\newcommand{\End}{\operatorname{End}} % ring of endomorphisms
\newcommand{\Iso}{\operatorname{Iso}} % set of isomorphisms
\newcommand{\GL}{\operatorname{GL}} % general linear group
\newcommand{\SL}{\operatorname{SL}} % special linear group
\newcommand{\Nil}{\operatorname{Nil}} % nilradical
\newcommand{\Jac}{\operatorname{Jac}} % Jacobson radical
\newcommand{\Tor}{\operatorname{Tor}} % torsion submodule or Tor functor
\newcommand{\Ext}{\operatorname{Ext}} % Ext functor
\newcommand{\Ker}{\operatorname{Ker}} % kernel
\newcommand{\Coker}{\operatorname{Coker}} % cokernel
\newcommand{\powset}[2][]{\ifthenelse{\equal{#2}{}}{\mathcal{P}\left(#1\right)}{\mathcal{P}_{#1}\left(#2\right)}}
% $\powset[k]{S}$ stands for the set of all $k$-element subsets of
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\newcommand{\ive}[1]{\left[ #1 \right]}
% $\ive{...}$ compiles to [...] (Iverson bracket, aka truth value).
\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}
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\newcommand{\underbrack}[2]{\underbrace{#1}_{\substack{#2}}}
% $\underbrack{...1}{...2}$ yields
% $\underbrace{...1}_{\substack{...2}}$. This is useful for doing
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% $\explain{...}$ interprets the "..." as text (you can
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% would in any other text block) and puts it in parentheses.

% Arrows for normal use:
\newcommand{\mono}{\hookrightarrow} % Monomorphism/injection.
\newcommand{\epi}{\twoheadrightarrow} % Epimorphism/surjection.
\newcommand{\iso}{\overset{\cong}{\to}} % Epimorphism/surjection.

% Arrows for xymatrix:
\newcommand{\arinj}{\ar@{_{(}->}} % Monomorphism.
\newcommand{\arinjrev}{\ar@{^{(}->}} % Monomorphism.
\newcommand{\arsurj}{\ar@{->>}} % Epimorphism.
\newcommand{\arelem}{\ar@{|->}} % Arrow to be used in describing a map on elements.
\newcommand{\arback}{\ar@{<-}} % Backward arrow.

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%	MAKING SUMMATION SIGNS ALWAYS PUT THEIR BOUNDS ABOVE AND BELOW
%	THE SIGN
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% The following are hacks to ensure that sums (such as
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% $n$) underneath and above the sign, as opposed to on its right.
% Same for products (\prod), set unions (\bigcup) and set
% intersections (\bigcap). Remove the 8 lines below if you do not want
% this behavior.
\let\sumnonlimits\sum
\let\prodnonlimits\prod
\let\cupnonlimits\bigcup
\let\capnonlimits\bigcap
\renewcommand{\sum}{\sumnonlimits\limits}
\renewcommand{\prod}{\prodnonlimits\limits}
\renewcommand{\bigcup}{\cupnonlimits\limits}
\renewcommand{\bigcap}{\capnonlimits\limits}

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  {8pt plus 2pt minus 4pt}% <Space above>
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  \newtheorem{conjecture}[theorem]{Conjecture}
% Environments that don't:
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%----------------------------------------------------------------------------------------
%	METADATA
%----------------------------------------------------------------------------------------
\newcommand{\myname}{Darij Grinberg} % ENTER YOUR NAME HERE
\newcommand{\mymail}{darij.grinberg@drexel.edu} % ENTER YOUR EMAIL HERE
\newcommand{\psetnumber}{0} % ENTER THE NUMBER OF THIS PSET HERE

%----------------------------------------------------------------------------------------
%	HEADER AND FOOTER
%----------------------------------------------------------------------------------------
\ihead{Solutions to homework set \#\psetnumber} % Page header left
\ohead{page \thepage\ of \pageref{LastPage}} % Page header right
\ifoot{\myname} % left footer
\ofoot{\mymail} % right footer

%----------------------------------------------------------------------------------------
%	TITLE SECTION
%----------------------------------------------------------------------------------------
\title{	
\normalfont \normalsize 
\textsc{Drexel University, Department of Mathematics} \\ [25pt] % 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 2023:
Homework \psetnumber\\% The assignment title
\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}

Open the Gradescope shell of this course and make sure you
know how to upload your homework submission.
Try it out.

(Don't worry about uploading an unfinished version; you
can update it as often as you wish until the deadline.)

\subsection{Solution}

...

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

\section{Exercise 2}

\subsection{Problem}

How familiar are you with the notions of
\begin{enumerate}
\item normal subgroup of a group;
\item determinant;
\item ring;
\item Cayley--Hamilton theorem;
\item quotient vector space $V/W$;
\item exact sequence;
\item complex number;
\item Gaussian integer;
\item primitive $n$-th root of unity;
\item discrete Fourier transform?
\item greatest common divisor of two (univariate) polynomials;
\item tensor product;
\item cyclotomic polynomial;
\item Reed--Muller code;
\item Elkies--Stanley code;
\item Bose--Chaudhuri--Hocquenghem code?
\end{enumerate}

(Write in a number between $0$ (for ``never seen it'') and $5$ (for ``could teach a lecture about it with no preparation'') for each one.)
\subsection{Solution}

...

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

\section{Exercise 3}

\subsection{Problem}

\begin{itemize}

\item[\textbf{(a)}]
Factor the polynomial $a^3 + b^3 + c^3 - 3abc$.

\item[\textbf{(b)}]
Factor the polynomial $bc\tup{b-c} + ca\tup{c-a} + ab\tup{a-b}$.

\item[\textbf{(c)}]
How general have your methods been?
Did you use tricks specific to the given polynomials, or do you have an algorithm for factoring any polynomial (say, with integer coefficients)?

\end{itemize}

\subsection{Solution}

...

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

\section{Exercise 4}

\subsection{Problem}

Simplify $\sqrt[3]{2 + \sqrt 5} + \sqrt[3]{2 - \sqrt 5}$.

\subsection{Solution}

...

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

\section{Exercise 5}

\subsection{Problem}

Let $n \in \NN$.
Let $a_1, a_2, \ldots, a_n$ be $n$ integers, and
let $b_1, b_2, \ldots, b_n$ be $n$ further integers.
The Gaussian elimination algorithm
tells you how to solve the system
\begin{align*}
a_1 x_1 + a_2 x_2 + \cdots + a_n x_n &= 0; \\
b_1 x_1 + b_2 x_2 + \cdots + b_n x_n &= 0
\end{align*}
for $n$ unknowns $x_1, x_2, \ldots, x_n \in \QQ$.
The answer, in general, will have the form
``all $\QQ$-linear combinations (i.e., linear
combinations with rational coefficients) of
a certain bunch of vectors''.
(More precisely, ``a certain bunch of vectors''
are $n-2$ or $n-1$ or $n$ vectors with rational
coordinates, depending on the rank of the
$2 \times n$-matrix
$\begin{pmatrix}
a_1 & a_2 & \cdots & a_n \\
b_1 & b_2 & \cdots & b_n
\end{pmatrix}$.)

Now, how can you solve the above system for $n$
unknowns $x_1, x_2, \ldots, x_n \in \ZZ$ ?
Will the answer still be
``all $\ZZ$-linear combinations (i.e., linear
combinations with integer coefficients) of
a certain bunch of vectors''?

What about more general systems of linear
equations to be solved for integer unknowns?

\subsection{Solution}

...

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

\section{Exercise 6}

\subsection{Problem}

You are given a $5 \times 5$-grid of lamps,
each of which is either on or off.
For example, writing $1$ for ``on'' and $0$ for ``off'',
it may look as follows:
\[
\begin{tabular}{|c|c|c|c|c|}
\hline
1 & 0 & 0 & 1 & 1 \\
\hline
1 & 1 & 0 & 0 & 1 \\
\hline
1 & 0 & 0 & 1 & 0 \\
\hline
0 & 1 & 1 & 1 & 1 \\
\hline
0 & 1 & 0 & 0 & 0 \\
\hline
\end{tabular}
\]
In a single move, you can toggle any lamp (i.e.,
turn it on if it was off, or turn it off if it was
on); however, this will also toggle every lamp
adjacent to it.
(``Adjacent to it'' means ``having a grid edge in
common with it'';
thus, a lamp will have $2$ or $3$ or $4$ adjacent
lamps.)
For example, if we toggle the second lamp (from
the left) in the topmost row in the above example grid,
then we obtain
\[
\begin{tabular}{|c|c|c|c|c|}
\hline
\textbf{1} & \textbf{0} & \textbf{0} & 1 & 1 \\
\hline
1 & \textbf{1} & 0 & 0 & 1 \\
\hline
1 & 0 & 0 & 1 & 0 \\
\hline
0 & 1 & 1 & 1 & 1 \\
\hline
0 & 1 & 0 & 0 & 0 \\
\hline
\end{tabular}
\]
(where the boldfaced numbers correspond to the lamps
that have been affected by the move).

Assume that all lamps are initially off.
Can you (by a strategically chosen sequence
of moves) achieve a state in which all lamps are
on?

[\textit{Remark:} You can play this game on
\url{https://codepen.io/wintlu/pen/ZJJLGz} .]

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 7
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 7}

\subsection{Problem}

\begin{itemize}

\item[\textbf{(a)}]
How many of the numbers $0, 1, \ldots, 6$
appear as remainders of a perfect square
divided by $7$ ?

\item[\textbf{(b)}]
How many of the numbers $0, 1, \ldots, 13$
appear as remainders of a perfect square
divided by $14$ ?

\end{itemize}

What about replacing $7$ or $14$ by $n$?
Can you do better than just squaring them all?

[For example, $3$ of the numbers $0, 1, \ldots, 4$
appear as remainders of a perfect square divided
by $5$ -- namely, the three numbers $0, 1, 4$.]

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 8
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 8}

\subsection{Problem}

Solve the following system of equations:
\begin{align*}
a^2+b+c &= 1; \\
b^2+c+a &= 1; \\
c^2+a+b &= 1
\end{align*}
for three complex numbers $a,b,c$.

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 9
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 9}

\subsection{Problem}

The following triangular table shows the binomial
coefficients $\dbinom{n}{m}$ for
$n \in \set{0, 1, \ldots, 7}$ and $m \in \set{0, 1, \ldots, n}$:
\[
\hspace{-0.5pc}
\begingroup
\setlength{\tabcolsep}{4pt} % Default value: 6pt
\renewcommand{\arraystretch}{0.5} % Default value: 1
\begin{tabular}[c]{ll|ccccccccccccccccc}
\vphantom{$\overset{k=0}{\swarrow}$} &  & $\phantom{20}$ & $\phantom{20}$ &
$\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ &  &  &  &
$\overset{k=0}{\swarrow}$ & $\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ &
$\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$\\
\vphantom{$\overset{k=0}{\swarrow}$}$n=0$ & $\rightarrow$ &  &  &  &  &  &  &
&  & $1$ &  & $\overset{k=1}{\swarrow}$ &  &  &  &  &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=1$ & $\rightarrow$ &  &  &  &  &  &  &
& $1$ &  & $1$ &  & $\overset{k=2}{\swarrow}$ &  &  &  &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=2$ & $\rightarrow$ &  &  &  &  &  &  &
$1$ &  & $2$ &  & $1$ &  & $\overset{k=3}{\swarrow}$ &  &  &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=3$ & $\rightarrow$ &  &  &  &  &  & $1$
&  & $3$ &  & $3$ &  & $1$ &  & $\overset{k=4}{\swarrow}$ &  &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=4$ & $\rightarrow$ &  &  &  &  & $1$ &
& $4$ &  & $6$ &  & $4$ &  & $1$ &  & $\overset{k=5}{\swarrow}$ &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=5$ & $\rightarrow$ &  &  &  & $1$ &  &
$5$ &  & $10$ &  & $10$ &  & $5$ &  & $1$ &  & $\overset{k=6}{\swarrow}$ & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=6$ & $\rightarrow$ &  &  & $1$ &  & $6$
&  & $15$ &  & $20$ &  & $15$ &  & $6$ &  & $1$ &  & $\overset{k=7}{\swarrow}%
$\\
\vphantom{$\overset{k=0}{\swarrow}$}$n=7$ & $\rightarrow$ &  & $1$ &  & $7$ &
& $21$ &  & $35$ &  & $35$ &  & $21$ &  & $7$ &  & $1$ &
\end{tabular}
\endgroup
\]
(This is part of what is known as \emph{Pascal's triangle}.)

Now, in this table, let us replace each even number by a $0$ and
each odd number by a $1$. We obtain
\[
\hspace{-0.5pc}
\begingroup
\setlength{\tabcolsep}{4pt} % Default value: 6pt
\renewcommand{\arraystretch}{0.5} % Default value: 1
\begin{tabular}[c]{ll|ccccccccccccccccc}
\vphantom{$\overset{k=0}{\swarrow}$} &  & $\phantom{20}$ & $\phantom{20}$ &
$\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ &
$\overset{k=0}{\swarrow}$ & $\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ &
$\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$\\
\vphantom{$\overset{k=0}{\swarrow}$}$n=0$ & $\rightarrow$ &  &  &  &  &  &  &
&  & $1$ &  & $\overset{k=1}{\swarrow}$ &  &  &  &  &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=1$ & $\rightarrow$ &  &  &  &  &  &  &
& $1$ &  & $1$ &  & $\overset{k=2}{\swarrow}$ &  &  &  &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=2$ & $\rightarrow$ &  &  &  &  &  &  &
$1$ &  & $0$ &  & $1$ &  & $\overset{k=3}{\swarrow}$ &  &  &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=3$ & $\rightarrow$ &  &  &  &  &  & $1$
&  & $1$ &  & $1$ &  & $1$ &  & $\overset{k=4}{\swarrow}$ &  &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=4$ & $\rightarrow$ &  &  &  &  & $1$ &
& $0$ &  & $0$ &  & $0$ &  & $1$ &  & $\overset{k=5}{\swarrow}$ &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=5$ & $\rightarrow$ &  &  &  & $1$ &  &
$1$ &  & $0$ &  & $0$ &  & $1$ &  & $1$ &  & $\overset{k=6}{\swarrow}$ & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=6$ & $\rightarrow$ &  &  & $1$ &  & $0$
&  & $1$ &  & $0$ &  & $1$ &  & $0$ &  & $1$ &  & $\overset{k=7}{\swarrow}%
$\\
\vphantom{$\overset{k=0}{\swarrow}$}$n=7$ & $\rightarrow$ &  & $1$ &  & $1$ &
& $1$ &  & $1$ &  & $1$ &  & $1$ &  & $1$ &  & $1$ &
\end{tabular}
\endgroup
\]
This looks rather similar to the third evolutionary stage
of \href{https://en.wikipedia.org/wiki/Sierpinski triangle}{Sierpinski's triangle}:
\[
\pgfdeclarelindenmayersystem{Sierpinski triangle}{
    \symbol{X}{\pgflsystemdrawforward}
    \symbol{Y}{\pgflsystemdrawforward}
    \rule{X -> X-Y+X+Y-X}
    \rule{Y -> YY}
}
\tikzset{
    l-system={step=2cm/(2^3), order=3, angle=-120}
}
\begin{tikzpicture}
    \fill [black] (0,0) -- ++(0:2) -- ++(120:2) -- cycle;
    \draw [draw=none] (0,0) l-system [l-system={Sierpinski triangle, axiom=X},fill=white];
\end{tikzpicture}
\]
% This graphic is from https://tex.stackexchange.com/a/119604/ .
(Each $0$ in the above table corresponds to a white $\triangle$ triangle, and
each $1$ corresponds to a black $\blacktriangle$ triangle.)

Where does this similarity come from?

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 10
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 10}

\subsection{Problem}

The number $\sqrt{2}$ is a root of the polynomial $x^2 - 2$.
The number $\sqrt{3}$ is a root of the polynomial $x^2 - 3$.
Can you find a polynomial with integer coefficients
that has $\sqrt{2} + \sqrt{3}$ as a root?

Yes, I know about the zero polynomial. Find a better one!

\subsection{Solution}

...


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

\begin{thebibliography}{99999999}                                                                                         %

\end{thebibliography}

\end{document}

