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% The commands below might be convenient. For example, you probably
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\newcommand{\myname}{Darij Grinberg} % ENTER YOUR NAME HERE
\newcommand{\mymail}{darij.grinberg@drexel.edu} % ENTER YOUR EMAIL HERE
\newcommand{\psetnumber}{1} % ENTER THE NUMBER OF THIS PSET HERE

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%	HEADER AND FOOTER
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\ihead{Solutions to homework set \#\psetnumber} % Page header left
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\ifoot{\myname} % left footer
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%	TITLE SECTION
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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 533: Abstract Algebra I, \\
Winter 2021:
Homework \psetnumber\\% The assignment title
\horrule{2pt} \\[0.5cm] % Thick bottom horizontal rule
Please solve \textbf{at most 5 of the 10 problems!}\footnote{%
I recommend solving as many problems as you can and wish,
but I will only grade and score 5 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}

Fix an integer $m$.
An \emph{$m$-integer} shall mean a rational number
$r$ such that there exists a $k \in \NN$ satisfying
$m^k r \in \ZZ$.

For example:
\begin{itemize}
 \item Each integer $r$ is an $m$-integer (since $m^k r \in \ZZ$
       for $k = 0$).
 \item The rational number $\dfrac{5}{12}$
       is a $6$-integer (since $6^k \cdot \dfrac{5}{12} \in \ZZ$
       for $k = 2$), but neither a $2$-integer nor a $3$-integer
       (since multiplying it by a power of $2$ will not ``get
       rid of'' the prime factor $3$ in the denominator, and
       vice versa\footnote{To make this more rigorous:
       If we had $2^k \cdot \dfrac{5}{12} \in \ZZ$ for some
       $k \in \NN$, then we would have
       $12 \mid 2^k \cdot 5$, which would entail that
       $3 \mid 12 \mid 2^k \cdot 5$, and thus $3$ would appear
       as a factor in the prime factorization of $2^k \cdot 5$.
       But this is absurd. Hence,
       $2^k \cdot \dfrac{5}{12} \in \ZZ$ cannot hold.
       Similarly, $3^k \cdot \dfrac{5}{12} \in \ZZ$ cannot
       hold.}).
 \item The $1$-integers are the integers (since $1^k r = r$
       for all $r$).
 \item Every rational number $r$ is a $0$-integer (since
       $0^k r \in \ZZ$ for $k = 1$).
\end{itemize}

Let $R_m$ denote the set of all $m$-integers.
Prove that $R_m$ is a subring of $\QQ$.

\subsection{Remark}

The ring $R_m$ is an example of a ring ``between $\ZZ$ and $\QQ$''
(in the sense that $\ZZ$ is a subring of $R_m$, while $R_m$
is a subring of $\QQ$).
Note that $R_1 = \ZZ$ and $R_0 = \QQ$, whereas $R_2 = R_4 = R_8 = \cdots$
is the ring of all rational numbers that can be written in the form
$a / 2^k$ with $a \in \ZZ$ and $k \in \NN$.

\subsection{Solution}

...

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

\section{Exercise 2}

\subsection{Problem}

Let $R$ be a ring.

An element $a$ of $R$ is said to be \emph{idempotent}
if it satisfies $a^2 = a$.

An element $a$ of $R$ is said to be \emph{involutive}
if it satisfies $a^2 = 1$.

\begin{enumerate}
 \item[\textbf{(a)}] Let $a \in R$.
 Prove that if $a$ is idempotent, then $1 - 2a$ is
 involutive.
 
 \item[\textbf{(b)}] Now, assume that $2$ is
 \emph{cancellable} in $R$; this means that if
 $u$ and $v$ are two elements of $R$ satisfying
 $2u = 2v$, then $u = v$.
 Prove that the converse of the claim of part
 \textbf{(a)} holds:
 If $a \in R$ is such that $1 - 2a$ is involutive,
 then $a$ is idempotent.
 
 \item[\textbf{(c)}] Now, let $R = \ZZ / 4 \ZZ$.
 Find an element $a \in R$ such that $1 - 2a$
 is involutive, but $a$ is not idempotent.
\end{enumerate}

\subsection{Remark}

The idempotent elements of $\RR$ are $0$ and $1$.
The involutive elements of $\RR$ are $1$ and $-1$.
A matrix ring like $\RR^{n \times n}$ usually has
infinitely many idempotent elements (viz., all
projection matrices on subspaces of $\RR^n$) and
infinitely many involutive elements (viz., all
matrices $A$ satisfying $A^2 = I_n$; for instance,
all reflections across hyperplanes are represented
by such matrices).

Part \textbf{(a)} of this exercise assigns an involutive
element to each idempotent element of $R$.
If $2$ is invertible in $R$ (that is, if the element
$2 \cdot 1_{R}$ has a multiplicative inverse), then this
assignment is a bijection (as can be easily derived from
part \textbf{(b)}).
Note that this assignment, when applied to a matrix ring
$\RR^{n \times n}$, is exactly the assignment you would
expect from the geometric point of view: To the orthogonal
projection on a hyperplane $H$, it assigns the reflection
in the hyperplane $H$.
Part \textbf{(c)} shows that we cannot drop the
``$2$ is cancellable'' condition in part \textbf{(b)}.

\subsection{Solution}

...

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

\section{Exercise 3}

\subsection{Problem}

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

Let $R$ be a ring, and let $e$ be
an idempotent element of $R$.

\begin{itemize}

\item[\textbf{(a)}] Show that $1 - e \in R$
is again idempotent.

\end{itemize}

Now, assume that $R$ is commutative.

\begin{itemize}

\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 makes $eR$ a subring of $R$ in the sense of
\cite{DF}, but not in our sense, since its unity is not
generally the unity of $R$.)

\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{itemize}

\subsection{Remark}

Part \textbf{(d)} of this exercise shows that if a
commutative ring $R$ has an idempotent 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 class (as an instance of the Chinese
Remainder Theorem).}

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

Parts \textbf{(b)}, \textbf{(c)} and \textbf{(d)} of the
exercise can be generalized somewhat:
Instead of requiring $R$ to be commutative, it suffices
to require that $er = re$ for all $r \in R$.
We cannot, however, drop this requirement altogether
(otherwise, matrix rings would decompose as direct products
-- as they have lots of idempotent elements --, but
typically they don't).

\subsection{Solution}

...

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

\section{Exercise 4}

\subsection{Problem}

In set theory, the \emph{symmetric difference} of two sets $A$ and $B$
is defined to be the set
$\tup{A \cup B} \setminus \tup{A \cap B}
= \tup{A \setminus B} \cup \tup{B \setminus A}$.
This symmetric difference is denoted by $A \symd B$.

Now, let $S$ be any set. Let $\powset{S}$ denote the power set of $S$ (that
is, the set of all subsets of $S$).
It is easy to check that the following ten properties hold:
\begin{align*}
A \symd B  &= B \symd A
\qquad \text{for any sets $A$ and $B$;} \\
A \cap B &= B \cap A
\qquad \text{for any sets $A$ and $B$;} \\
\tup{A \symd B} \symd C
&= A \symd \tup{B \symd C}
\qquad \text{for any sets $A$, $B$ and $C$;}\\
\tup{A \cap B} \cap C &= A \cap \tup{B \cap C}
\qquad \text{for any sets $A$, $B$ and $C$;}\\
A \symd \varnothing &= \varnothing \symd A
= A \qquad \text{for any set $A$;}\\
A \symd A &= \varnothing
\qquad \text{for any set $A$;}\\
A \cap S &= S \cap A = A
\qquad \text{for any subset $A$ of $S$;}\\
\varnothing \cap A &= A \cap \varnothing
= \varnothing \qquad \text{for any set $A$;}\\
A \cap \tup{B \symd C} &=
\tup{A \cap B} \symd \tup{A \cap C}
\qquad \text{for any sets $A$, $B$ and $C$;}\\
\tup{A \symd B} \cap C &=
\tup{A \cap C} \symd \tup{B \cap C}
\qquad \text{for any sets $A$, $B$ and $C$.}
\end{align*}
Therefore, $\powset{S}$ becomes a commutative ring, where
the addition is defined to be the operation $\symd$, the
multiplication is defined to be the operation $\cap$, the zero is defined to
be the set $\varnothing$, and the unity is defined to be the set
$S$. (The ten properties listed above show that the axioms of
a commutative ring are satisfied for $\tup{\powset{S},
\symd, \cap, \varnothing, S}$. In particular, the sixth
property shows that every subset $A$ of $S$ has an additive inverse -- namely,
itself. Of course, it is unusual for an element of a commutative ring to be
its own additive inverse, but in this example it happens all the time!)

The commutative ring $\powset{S}$ has the property that
$a \cdot a = a$ for every $a \in \powset{S}$. (This simply means
that $A \cap A = A$ for every $A \subseteq S$.) Rings that have this
property are called
\href{https://en.wikipedia.org/wiki/Boolean_ring}{\emph{Boolean rings}}. (Of
course, $\powset{S}$ is the eponymic example for a Boolean
ring; but there are also others.)

\begin{itemize}

% \item[\textbf{(o)}] Check the ninth of the above ten
% properties. That is, check that
% $A \cap \tup{B \symd C} =
% \tup{A \cap B} \symd \tup{A \cap C}$ for any three
% sets $A$, $B$ and $C$.

\item[\textbf{(a)}]
Prove that the ring $\powset{S}$ is isomorphic to
the direct product
$\tup{\ZZ / 2 \ZZ}^S = \prod_{s \in S} \tup{\ZZ / 2 \ZZ}$.

\item[\textbf{(b)}]
Let $F$ be the set of all \textbf{finite} subsets of
$S$.
Prove that $F$ is an ideal of $\powset{S}$.

\item[\textbf{(c)}]
Assume that $S$ is infinite.
Prove that the ideal $F$ is not principal.

\item[\textbf{(d)}]
Instead, assume that $S$ is finite.
Prove that every ideal of $\powset{S}$ is principal.

\end{itemize}

[\textbf{Hint:} For part \textbf{(d)},
let $I$ be an ideal of $\powset{S}$, and pick a
subset $T \in I$ of largest size.
Argue that each subset of $T$ must also belong to $I$.
Conclude that every set in $I$ must be a subset of $T$.]

\subsection{Solution}

...

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

\section{Exercise 5}

\subsection{Problem}

Now we shall study Boolean rings in general.

A \emph{Boolean ring} means a ring $R$ such that
every $a \in R$ satisfies $a^2 = a$
(that is, every $a \in R$ is idempotent).
(Keep in mind that rings must have a $1$ according to
our definition.)

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

\begin{itemize}

\item[\textbf{(a)}]
We have $2a = 0$ for each $a \in R$.

\item[\textbf{(b)}]
We have $-a = a$ for each $a \in R$.

\item[\textbf{(c)}]
The ring $R$ is commutative.

\item[\textbf{(d)}]
If $R$ is finite, then $R \cong \tup{\ZZ / 2 \ZZ}^n$
for some $n \in \NN$.

\end{itemize}

[\textbf{Hint:} In part \textbf{(a)}, use
$a^2 = a$ and $\tup{a+1}^2 = a+1$.
In part \textbf{(c)}, expand $\tup{a+b}^2$
(but don't use the binomial formula, since you
don't know yet that $ab = ba$).
Finally, for part \textbf{(d)}, use strong
induction on $\abs{R}$ as follows:
Pick some $e \in R$ that is distinct from $0$ and $1$
(if no such $e$ exists, the claim is obvious).
Then, $e$ is idempotent, so Exercise 3 \textbf{(d)}
decomposes the ring $R$ as a direct product of
two smaller rings.
You can use without proof that direct products are
associative up to isomorphism (so that
$A_1 \times A_2 \times \cdots \times A_m
\cong \tup{A_1 \times A_2 \times \cdots \times A_k}
\times \tup{A_{k+1} \times A_{k+2} \times \cdots \times A_m}$
for any rings $A_1, A_2, \ldots, A_m$).]

\subsection{Solution}

...

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

\section{Exercise 6}

\subsection{Problem}

Let $A$ be the $2 \times 2$-matrix
$\begin{pmatrix}
  0 & 1 \\
  1 & 1
 \end{pmatrix}$
over $\ZZ$.
Consider also the identity matrix $I_2 \in \ZZ^{2 \times 2}$.

Let $\calF$ % I have defined \calF to mean \mathcal{F} for this problem.
be the subset
\[
 \set{ aA + bI_2 \mid a, b \in \ZZ }
 =
 \set{ \begin{pmatrix} b & a \\ a & a+b \end{pmatrix} \mid a, b \in \ZZ }
\]
of the matrix ring $\ZZ^{2 \times 2}$.

\begin{enumerate}

\item[\textbf{(a)}]
Prove that $A^2 = A + I_2$.

\item[\textbf{(b)}]
Prove that the set $\calF$ is a \textbf{commutative}
subring of $\ZZ^{2 \times 2}$.

\end{enumerate}

Let $\tup{f_0, f_1, f_2, \ldots}$ be the Fibonacci sequence.
This is the sequence of integers defined recursively by
\[
f_0 = 0, \qquad
f_1 = 1, \qquad \text{and} \qquad
f_n = f_{n-1} + f_{n-2} \text{ for all } n \geq 2 .
\]
The first entries of this sequence are
\[
\begin{tabular}[c]{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline
$n$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$ & $12$ \\\hline
$f_n$ & $0$ & $1$ & $1$ & $2$ & $3$ & $5$ & $8$ & $13$ & $21$ & $34$ & $55$ & $89$ & $144$ \\\hline
\end{tabular}
\ \ \ .
\]

\begin{enumerate}

\item[\textbf{(c)}]
Prove that $A^n = f_n A + f_{n-1} I_2$ for all positive
integers $n$.

\item[\textbf{(d)}]
Prove that
$f_{n+m} = f_n f_{m+1} + f_{n-1} f_m$ for all positive
integers $n$ and all $m \in \NN$.

\end{enumerate}

Now, define a further matrix $B \in \calF$ by
$B = \tup{-1}A + 1I_2 = I_2 - A$.

\begin{enumerate}

\item[\textbf{(e)}]
Prove that $B^2 = B + I_2$ and
$B^n = f_n B + f_{n-1} I_2$ for all positive
integers $n$.

\item[\textbf{(f)}]
Prove that
$A^n - B^n = f_n \tup{A - B}$ for all $n \in \NN$.

\item[\textbf{(g)}]
Prove that $f_d \mid f_{dn}$ for any nonnegative
integers $d$ and $n$.

% Prove that
% \[
%  A^n \begin{pmatrix} 0 \\ 1 \end{pmatrix}
%  = \begin{pmatrix} f_n \\ f_{n+1} \end{pmatrix}
%  \qquad \text{ for all } n \in \NN .
% \]
\end{enumerate}

[\textbf{Hint:} In part \textbf{(b)}, don't forget to
check commutativity! It is not inherited from
$\ZZ^{2 \times 2}$, since $\ZZ^{2 \times 2}$ is not
commutative.

One way to prove \textbf{(d)} is by
comparing the $\tup{1, 1}$-th entries of the two
(equal) matrices $A^n A^{m+1}$ and $A^{n+m+1}$, after
first using part \textbf{(c)} to expand these matrices.

For part \textbf{(g)}, compare the $\tup{1, 1}$-th
entries of the matrices $A^d - B^d$ and $A^{dn} - B^{dn}$,
after first proving that $A^d - B^d \mid A^{dn} - B^{dn}$ in
the commutative ring $\calF$.
Note that divisibility is a tricky concept in general
rings, but $\calF$ is a commutative ring, which lets
many arguments from the integer setting go
through unchanged.]

\subsection{Solution}

...

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

\section{Exercise 7}

\subsection{Problem}

Let $R$ be a ring.
An element $a \in R$ is said to be \emph{nilpotent}
if there exists an $n \in \NN$ such that $a^n = 0$.
(For example, the residue class $\overline{6}$ in
$\ZZ / 8 \ZZ$ is nilpotent, since its $3$-rd power
is $\overline{0}$.)

\begin{itemize}

\item[\textbf{(a)}]
If $a$ and $b$ are two nilpotent elements of $R$
satisfying $ab = ba$, then prove that $a+b$ is
nilpotent as well.

\item[\textbf{(b)}]
Find a counterexample to part \textbf{(a)} if
we don't assume $ab = ba$.

\item[\textbf{(c)}]
Assume that the ring $R$ is commutative.
Let $N$ be the set of all nilpotent elements of $R$.
Prove that $N$ is an ideal of $R$.

\end{itemize}

\subsection{Remark}

The ideal $N$ in part \textbf{(c)} of this exercise
is known as the \emph{nilradical} of $R$.

\subsection{Solution}

...

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

\section{Exercise 8}

\subsection{Problem}

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

\begin{itemize}

\item[\textbf{(a)}]
Let $I$ and $J$ be two ideals of a ring $R$.
Then, $I+J$ and $I \cap J$ and $IJ$ are ideals of
$R$ as well.

\item[\textbf{(b)}]
Let $I$ and $J$ be two ideals of a ring $R$.
Then,
$IJ \subseteq I \cap J \subseteq I \subseteq I+J$
and
$IJ \subseteq I \cap J \subseteq J \subseteq I+J$.

\item[\textbf{(c)}]
The set of all ideals of $R$ is a monoid with respect
to the binary operation $+$, with neutral element
$\set{0_R} = 0R$.
That is,
\begin{align*}
\tup{I + J} + K &= I + \tup{J + K}
\qquad \text{for any three ideals $I, J, K$ of $R$;}\\
I + \set{0_R} &= \set{0_R} + I = I
\qquad \text{for any ideal $I$ of $R$.}
\end{align*}

\item[\textbf{(d)}]
The set of all ideals of $R$ is a monoid with respect to the
binary operation $\cdot$, with neutral element $R = 1R$.
That is,
\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{itemize}

[\textbf{Hint:} You can be terse here, as there is a lot
to show, much of it straightforward.
I recommend using the notion of ``$\tup{I, J}$-products''
from lecture 4; it is often 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 $\tup{IJ} K = I \tup{JK}$,
I recommend first showing that any $\tup{IJ, K}$-product
belongs to $I \tup{JK}$.]

\subsection{Solution}

...

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

\section{Exercise 9}

\subsection{Problem}

Let $R$ be a ring.
Let $a$ and $b$ be two elements of $R$. Prove the following:
If $c$ is an inverse of $1 - ab$, then $1 + bca$ is an
inverse of $1 - ba$.

\subsection{Remark}

This yields a well-known result in functional analysis;
see \url{https://math.stackexchange.com/questions/79217} .

\subsection{Solution}

...

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

\section{Exercise 10}

\subsection{Problem}

Let $F$ be a field.

\begin{enumerate}

\item[\textbf{(a)}] Prove that if $a, b \in F$ satisfy
$a^2 = b^2$, then $a = b$ or $a = -b$.

\end{enumerate}

An element $\eta \in F$ is called a
\emph{square} if there exists some $\alpha \in F$
such that $\eta = \alpha^2$.
For example, the squares in $\ZZ / 7 \ZZ$ are the four elements
$\overline{0}, \overline{1}, \overline{2}, \overline{4}$.
(Indeed, this is equivalent to the answer to Exercise
7 \textbf{(a)} on homework set \#0.)

From now on, assume that $2 \cdot 1_F \neq 0_F$ (that is,
$1_F + 1_F \neq 0_F$). Note that this is
satisfied whenever $F = \ZZ / p \ZZ$ for a prime $p > 2$ (but also for
various other finite fields), but fails when $F = \ZZ / 2 \ZZ$.

\begin{enumerate}

\item[\textbf{(b)}] Prove that $a \neq -a$ for every
nonzero $a \in F$.

\end{enumerate}

From now on, assume that $F$ is finite.

\begin{enumerate}

\item[\textbf{(c)}] Prove that the number of squares in $F$ is
$\dfrac{1}{2} \tup{ \abs{F} + 1 }$.

\item[\textbf{(d)}] Conclude that $\abs{F}$ is odd.

\end{enumerate}

[\textbf{Hint:} For part \textbf{(c)}, argue that each nonzero square in
$F$ can be written as $\alpha^2$ for exactly two $\alpha \in F$.]

\subsection{Solution}

...


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

\begin{thebibliography}{99999999}                                                                                         %

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

\end{thebibliography}

\end{document}

