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\usetikzlibrary{lindenmayersystems} % This is for drawing Sierpinski triangles.
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% 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{\calF}{\mathcal{F}}
\newcommand{\Z}[1]{\mathbb{Z}/#1\mathbb{Z}} % integers modulo k
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\newcommand{\GL}{\operatorname{GL}} % general linear group
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\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{\Ann}{\operatorname{Ann}} % annihilator
\newcommand{\symd}{\mathbin{\bigtriangleup}} % symmetric difference of sets
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\newcommand{\mono}{\hookrightarrow} % Monomorphism/injection.
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% Arrows for xymatrix:
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%	METADATA
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\newcommand{\myname}{Darij Grinberg} % ENTER YOUR NAME HERE
\newcommand{\mymail}{darij.grinberg@drexel.edu} % ENTER YOUR EMAIL HERE
\newcommand{\psetnumber}{2} % ENTER THE NUMBER OF THIS PSET HERE

%----------------------------------------------------------------------------------------
%	HEADER AND FOOTER
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\ihead{Solutions to midterm \#\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} \\ [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 2023:
Midterm \psetnumber\\% The assignment title
\horrule{2pt} \\[0.5cm] % Thick bottom horizontal rule
Please solve \textbf{at most 3 of the 6 problems!} \\
\textbf{No collaboration} is allowed on the midterm.
% \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}

\begin{enumerate}

\item[\textbf{(a)}]
Prove that there are no ring morphisms from
$\ZZ\ive{i}$ to $\ZZ$.

\end{enumerate}

Now, let $p$ be a prime number. Prove the following:

\begin{enumerate}

\item[\textbf{(b)}]
There are no ring morphisms from $\ZZ\ive{i}$
to $\ZZ/p$ if $p \equiv 3 \mod 4$.

\item[\textbf{(c)}]
There are exactly two ring morphisms from
$\ZZ\ive{i}$ to $\ZZ/p$ if $p \equiv 1 \mod 4$.

\item[\textbf{(d)}]
There is a unique ring morphism from $\ZZ\ive{i}$ to
$\ZZ/p$ if $p = 2$.

\end{enumerate}

More generally, prove the following:

\begin{enumerate}

\item[\textbf{(e)}]
If $R$ is any ring, then the number of ring morphisms from
$\ZZ\ive{i}$ to $R$ is the number of all elements $r \in R$
satisfying $r^2 = -1$.

\end{enumerate}

\subsection{Hint}

If $f$ is a ring morphism from $\ZZ\ive{i}$ to $R$, then
what equation must $f \tup{i}$ satisfy?

\subsection{Solution}

...

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

\section{Exercise 2}

\subsection{Problem}

Let $\omega$ denote the complex number
$\dfrac{-1 + \sqrt{-3}}{2} \in \CC$.

\begin{enumerate}

\item[\textbf{(a)}]
Prove that $\omega^3 = 1$ and $\omega^2 + \omega + 1 = 0$.

\item[\textbf{(b)}]
Prove that $\abs{a + b \omega} = \sqrt{a^2 - ab + b^2}$
for any $a, b \in \RR$.

\item[\textbf{(c)}]
Define a subset $\ZZ\ive{\omega}$ of $\CC$ by
\[
\ZZ\ive{\omega} := \set{ a + b \omega \ \mid \ a, b \in \ZZ }.
\]
Prove that $\ZZ\ive{\omega}$ is a subring of $\CC$.
(It is called the ring of \emph{Eisenstein integers}.)

\item[\textbf{(d)}]
Prove that $\ZZ\ive{\sqrt{-3}}$ is a subring of
$\ZZ\ive{\omega}$.

\item[\textbf{(e)}]
Prove that the ring $\ZZ\ive{\omega}$ is Euclidean, and
that the map
\begin{align*}
N : \ZZ\ive{\omega} &\to \NN, \\
a + b \omega &\mapsto a^2 - ab + b^2
\qquad \tup{\text{for $a, b \in \ZZ$}}
\end{align*}
is a Euclidean norm for it.

\textbf{[For the sake of brevity, you are allowed to reason
from a picture here.]}

\end{enumerate}

\subsection{Hint}

Geometrically speaking, the three complex numbers
$1,\ \omega,\ \omega^2$ are the vertices of an equilateral triangle
inscribed in the unit circle. The elements of $\ZZ\ive{\omega}$
are the grid points of a triangular lattice that looks as
follows (imagine the picture extended to infinity all on sides):
\[
\pgfmathsetmacro{\rows}{5}
\begin{tikzpicture}[baseline, scale=2.5]
\clip(-2.5, -2.5) rectangle (2.5, 2.5);
\foreach\row in {-\rows, ..., \rows} {
\draw[gray, very thin] ($(-\rows,0)+\row*(0.5, {0.5*sqrt(3)})$) -- ($(\rows
,0)+\row*(-0.5, {0.5*sqrt(3)})$);
\draw[gray, very thin] ($({-\rows/2},{-\rows/2*sqrt(3)})+\row
*(0.5,{-0.5*sqrt(3)})$) -- ($(\rows/2,{\rows/2*sqrt(3)})+\row
*(0.5,{-0.5*sqrt(3)})$);
\draw[gray, very thin] ($(\rows/2,{-\rows/2*sqrt(3)})+\row
*(-0.5,{-0.5*sqrt(3)})$) -- ($({-\rows/2},{\rows/2*sqrt(3)})+\row
*(-0.5,{-0.5*sqrt(3)})$);
}
\draw[green!50!black] (0, 0) circle [radius=1];
\fill(0, 0) circle (1pt) node[above left=2pt] {$0$};
\fill(1, 0) circle (1pt) node[above left=2pt] {$1$};
\fill(2, 0) circle (1pt);
\fill(-1, 0) circle (1pt) node[below left=2pt] {$-1$};
\fill(-2, 0) circle (1pt);
\fill(0, {sqrt(3)}) circle (1pt) node[above left=2pt] {$\sqrt{-3}$};
\fill(-1/2, {sqrt(3)/2}) circle (1pt) node[above left=2pt] {$\omega$};
\fill(-1/2, {-sqrt(3)/2}) circle (1pt) node[below left=2pt] {$\omega^2$};
\fill(3/2, {sqrt(3)/2}) circle (1pt) node[above right=2pt] {$\omega+1$};
\fill(5/2, {sqrt(3)/2}) circle (1pt);
\fill(-3/2, {sqrt(3)/2}) circle (1pt);
\fill(-5/2, {sqrt(3)/2}) circle (1pt);
\fill(-5/2, {-sqrt(3)/2}) circle (1pt);
\fill(-3/2, {-sqrt(3)/2}) circle (1pt);
\fill(3/2, {-sqrt(3)/2}) circle (1pt);
\fill(5/2, {-sqrt(3)/2}) circle (1pt);
\fill(-2, {-sqrt(3)}) circle (1pt);
\fill(-1, {-sqrt(3)}) circle (1pt);
\fill(0, {-sqrt(3)}) circle (1pt);
\fill(1, {-sqrt(3)}) circle (1pt);
\fill(2, {-sqrt(3)}) circle (1pt);
\fill(-2, {sqrt(3)}) circle (1pt);
\fill(-1, {sqrt(3)}) circle (1pt);
\fill(1, {sqrt(3)}) circle (1pt);
\fill(2, {sqrt(3)}) circle (1pt);
\fill(1/2, {-sqrt(3)/2}) circle (1pt) node[below=2pt] {$-\omega$};
\fill(1/2, {sqrt(3)/2}) circle (1pt) node[below=2pt] {$-\omega^2$};
\end{tikzpicture}
\]

For part \textbf{(d)}, don't forget to show that
$\ZZ\ive{\sqrt{-3}}$ is a \textbf{subset} of
$\ZZ\ive{\omega}$ in the first place!

\subsection{Solution}

...

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

\section{Exercise 3}

\subsection{Problem}

Let $R$ be a ring. Let $M$ be a left $R$-module.

For any subset $K$ of $M$, let $\Ann K$ denote the subset
$\set{ r \in R \ \mid \ rk = 0 \text{ for all } k \in K }$
of $R$. (This is called the \emph{annihilator} of $K$.)

\begin{enumerate}

\item[\textbf{(a)}]
Prove that $\Ann M$ is an ideal of $R$.

\item[\textbf{(b)}]
Let $K$ be any subset of $M$. Prove that $\Ann K$ is a left
ideal of $R$. (Recall that a \emph{left ideal} of $R$ means
a subset $L$ of $R$ that is closed under addition and
contains $0$ and satisfies $ra \in L$ for all $r \in R$ and
$a \in L$.)

\item[\textbf{(c)}]
Find an example showing that the $\Ann K$ in part
\textbf{(b)} is not always an ideal of $R$.

\end{enumerate}

\subsection{Solution}

...

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

\section{Exercise 4}

\subsection{Problem}

Let $R$ be a ring. Let $M$ be a left $R$-module.
Prove the following:

\begin{enumerate}

\item[\textbf{(a)}]
Let $I$ be an ideal of $R$. 
An \emph{$\tup{I, M}$-product} shall mean a
product of the form $im$ with $i \in I$ and
$m \in M$. Then,
\[
IM :=
\set{ \text{finite sums of $\tup{I, M}$-products} }
\]
is an $R$-submodule of $M$.

\item[\textbf{(b)}]
Let $a$ be a central element of $R$. Prove that
\[
aM := \set{ am \ \mid \ m \in M}
\]
is an $R$-submodule of $M$.

\end{enumerate}

\subsection{Solution}

...

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

\section{Exercise 5}

\subsection{Problem}

Let $n$ and $k$ be two positive integers.
Let $V$ be the subset
\[
\set{ \tup{a_1, a_2, \ldots, a_n} \in \ZZ^n
\ \mid\ a_1 \equiv a_2 \equiv \cdots \equiv a_n
\mod k }
\]
of the $\ZZ$-module $\ZZ^n$.

It is straightforward to see that $V$ is a
$\ZZ$-submodule of $\ZZ^n$.

Show that $V$ is free, and find a basis of $V$.

\subsection{Solution}

...

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

\section{Exercise 6}

\subsection{Problem}

Let $R$ be any ring. Consider the map
\begin{align*}
S : R^{\NN}  &\to R^{\NN}, \\
\tup{a_0, a_1, a_2, \ldots}
&\mapsto \tup{a_0,\ a_0 + a_1,\ a_0 + a_1 + a_2,
              \ a_0 + a_1 + a_2 + a_3,\ \ldots}
\\
& \qquad =
\tup{b_0, b_1, b_2, \ldots}
\text{ where } b_i = a_0 + a_1 + \cdots + a_i .
\end{align*}
Consider furthermore the map
\begin{align*}
\Delta : R^{\NN}  &\to R^{\NN}, \\
\tup{a_0, a_1, a_2, \ldots}
&\mapsto \tup{a_0,\ a_1 - a_0,\ a_2 - a_1,
              \ a_3 - a_2, \ \ldots} , \\
& \qquad =
\tup{c_0, c_1, c_2, \ldots}
\text{ where }
c_0 = a_0 \text{ and }
c_i = a_i - a_{i-1} \text{ for all } i \geq 1.
\end{align*}

\begin{enumerate}

\item[\textbf{(a)}]
Prove that $S$ and $\Delta$ are $R$-linear maps
and are mutually inverse.

\item[\textbf{(b)}]
Recall the $R$-submodule
\[
R^{\tup{\NN}}
= \set{ \tup{a_0, a_1, a_2, \ldots} \in R^{\NN}
        \ \mid\ \text{only finitely many } i \in \NN
        \text{ satisfy } a_i \neq 0 }
\]
of $R^{\NN}$.
Define a further $R$-submodule $R^{\tup{\NN}+}$
of $R^{\NN}$ by
\begin{align*}
R^{\tup{\NN}+}  &:=
\left\{ \tup{a_0, a_1, a_2, \ldots} \in R^{\NN}
\ \mid\ \text{there exists a }c\in R
\text{ such that} \right. \\
& \qquad \qquad \left.
\text{ only finitely many } i \in \NN
\text{ satisfy } a_i \neq c
\right\}  .
\end{align*}
(Thus, a sequence
$\tup{a_0, a_1, a_2, \ldots} \in R^{\NN}$ belongs to
$R^{\tup{\NN}+}$ if and only if starting from
some point on, all its entries are equal.)

Clearly, $R^{\tup{\NN}}$ is a proper subset of
$R^{\tup{\NN}+}$ (unless $R$ is trivial).

Prove that $R^{\tup{\NN}}\cong R^{\tup{\NN}+}$ as left
$R$-modules, and in fact the restriction of the map $S$ to
$R^{\tup{\NN}}$ is a left $R$-module isomorphism from
$R^{\tup{\NN}}$ to $R^{\tup{\NN}+}$.

\end{enumerate}

\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},
% 12 January 2023.
% \url{https://www.cip.ifi.lmu.de/~grinberg/t/23wa/23wa.pdf}

% \end{thebibliography}

\end{document}

