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\newcommand{\myname}{Darij Grinberg} % ENTER YOUR NAME 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 533: Abstract Algebra I, \\
Winter 2021:
Homework \psetnumber\\% The assignment title
\horrule{2pt} \\[0.5cm] % Thick bottom horizontal rule
Please solve \textbf{5 of the 10 problems!}
\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}

Let $F$ be a finite field.
Prove the following:

\begin{itemize}

\item[\textbf{(a)}]
We have $a^{\abs{F}} = a$ for each $a \in F$.

\item[\textbf{(b)}]
If $i > 1$ is an integer such that each $a \in F$
satisfies $a^i = a$, then $i \geq \abs{F}$.

\end{itemize}

\subsection{Remark}

Note that part \textbf{(a)} generalizes Fermat's Little
Theorem (the part that says that $a^p = a$ for any
prime $p$ and any $a \in \ZZ / p$).

\subsection{Solution}

...

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

\section{Exercise 2}

\subsection{Problem}

Let $p$ be a prime number.
Let $F$ be a finite field of characteristic $p$.
Let $f$ be the map $F \to F,\  a \mapsto a^p$.
We know (from Lecture 14) that this map $f$ is a
ring morphism from $F$ to $F$ (known as the
\emph{Frobenius endomorphism} of $F$).

\begin{itemize}

\item[\textbf{(a)}]
Prove that $f$ is a ring isomorphism from $F$ to
$F$ (so it is invertible).

\item[\textbf{(b)}]
Now, replace the words ``field of characteristic $p$''
by the (more general) ``commutative $\ZZ / p$-algebra''
in the above.
Find an example where the claim of part \textbf{(a)}
becomes false.

\end{itemize}

\subsection{Solution}

...

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

\section{Exercise 3}

\subsection{Problem}

Let $S$ and $F$ be two fields such that
$S$ is a subring of $F$.
(For example, we can take $S = \QQ$ and $F = \RR$.)

For any $a \in F$, we define an
\emph{annihilating polynomial of $a$} to be a
polynomial $f \in S\ive{x}$ such that $a$
is a root of $f$.
For instance:
\begin{itemize}
\item If $a \in S$, then $x - a$
is an annihilating polynomial of $a$.
\item
If $a$ is a square root of an element
$v \in S$, then $x^2 - v$ is an annihilating
polynomial of $a$.
\item
If $S = \QQ$ and $F = \RR$, then
$x^4 - 10 x^2 + 1$ is an annihilating polynomial
of $\sqrt{2} + \sqrt{3}$.
\item
The real number $\pi$ is known to be
transcendental; this means that there exists
no nonzero annihilating polynomial of $\pi$
(for $S = \QQ$ and $F = \RR$).
\end{itemize}

The \emph{minimal polynomial} of an element
$a \in F$ is defined to be the monic
annihilating polynomial of $a$ of smallest
possible degree.

Prove the following:

\begin{itemize}

\item[\textbf{(a)}]
The minimal polynomial of an element
$a \in F$ is unique whenever it exists.
That is, if there is at least one monic
annihilating polynomial of $a$, then only one
of these polynomials has smallest possible degree.

\item[\textbf{(b)}]
The minimal polynomial of an element
$a \in F$ is always irreducible if it exists.

\item[\textbf{(c)}]
If $f$ is the minimal polynomial of an element
$a \in F$, then the map
\begin{align*}
S \ive{x} / f &\to F, \\
\overline{g} &\mapsto g\ive{a}
\end{align*}
is a (well-defined) ring morphism, and is injective.

\item[\textbf{(d)}]
If $f$ is the minimal polynomial of an element
$a \in F$, then the annihilating polynomials of $a$
are precisely the polynomials $g \in S\ive{x}$
that are divisible by $f$.

\end{itemize}

\subsection{Remark}

The minimal polynomial of an $a \in F$ is defined
in the exact same way as the minimal polynomial of a
square matrix was defined in linear algebra.
However, the minimal polynomial of a matrix is not
always irreducible, so that part \textbf{(b)} of
this exercise is specific to fields.

\subsection{Solution}

...

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

\section{Exercise 4}

\subsection{Problem}

Set $S = \QQ$ and $F = \RR$ in Exercise 3.
Let $p$ and $q$ be two positive integers
such that none of $p$, $q$ and $pq$ is
a perfect square (i.e., a square in $\ZZ$).
(For example, we can take $p = 5$ and
$q = 8$.)
Let $a = \sqrt{p} + \sqrt{q} \in F$.

Let $f$ denote the polynomial
\[
\tup{x^2 - p - q}^2 - 4pq
= x^4 - 2\tup{p+q} x^2 + \tup{p-q}^2
\in S\ive{x} .
\]

\begin{itemize}

\item[\textbf{(a)}]
Show that $f$ is an annihilating polynomial
of $a$ (that is, $f\ive{a} = 0$).

\item[\textbf{(b)}]
Show that $f$ has no rational root.

\item[\textbf{(c)}]
Show that $f$ is irreducible (in $S\ive{x}$).

\item[\textbf{(d)}]
Conclude that $f$
is the minimal polynomial of $a$.

\end{itemize}

[\textbf{Hint:} Part \textbf{(c)} is the
tricky one. The polynomial $f$ is
\emph{even} -- meaning that $f\ive{-x} = f$,
or, equivalently (since $S$ has characteristic
$0$) that no odd powers of $x$ appear in $f$.
Use this to argue that if
$f = g_1 g_2 \cdots g_k$ is the factorization
of $f$ into monic irreducible polynomials, then
substituting $-x$ for $x$ into it must yield
another factorization
$f = f\ive{-x} = g_1\ive{-x} g_2\ive{-x} \cdots g_k\ive{-x}$
of $f$ into monic irreducible polynomials
(why are they still monic?).
Since $S\ive{x}$ is a UFD, the two factorizations
must be identical (up to the order of the factors).
This narrows down the possibilities for
$g_1, g_2, \ldots, g_k$ substantially.]

\subsection{Remark}

The assumption that none of $p$, $q$ and $pq$ is
a square is not just sufficient, but also
necessary for the claim of part \textbf{(d)}.
For example, if $p = 3$ and $q = 12$ (so that
$pq = 36$ is a square), then
$\sqrt{p} + \sqrt{q} = \sqrt{3} + \sqrt{12}
= 3 \sqrt{3}$ has minimal polynomial
$x^2 - 27$, and the polynomial $f$ fails to be
irreducible.

\subsection{Solution}

...

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

\section{Exercise 5}

\subsection{Problem}

Let $p$ be a prime number.
Let $F$ be a finite field of characteristic $p$.
As we know from Lecture 14, this entails
that $F$ contains ``a copy of $\ZZ / p$''
(that is, a subring isomorphic to $\ZZ / p$).
We identify this copy with $\ZZ / p$ itself, so
that $\ZZ / p$ is a subring of $F$.
We write $S$ for the field $\ZZ / p$, and we shall
use the terminology from Exercise 3.

Let $m$ be the positive integer satisfying
$\abs{F} = p^m$.
(We know from Lecture 14 that this $m$ exists.)

Prove the following:

\begin{itemize}

\item[\textbf{(a)}]
Each $a \in F$ has a minimal polynomial
(i.e., there is always at least one monic
annihilating polynomial of $a$).

\item[\textbf{(b)}]
If the minimal polynomial of an element $a \in F$
has degree $k$, then $a^{p^k} = a$.

\item[\textbf{(c)}]
If the minimal polynomial of an element $a \in F$
has degree $k$, then $k \leq m$.

\end{itemize}

[\textbf{Hint:} Exercise 3 \textbf{(c)}
can help with parts \textbf{(b)} and
\textbf{(c)} of the present exercise.
A size argument might also be useful in
\textbf{(c)}.]

\subsection{Solution}

...

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

\section{Exercise 6}

\subsection{Problem}

We continue in the setting of Exercise 5.
Prove the following:

\begin{itemize}

\item[\textbf{(a)}]
There exists at least one $a \in F$ such that
none of the $m-1$ powers $a^{p^1}, a^{p^2},
\ldots, a^{p^{m-1}}$ equals $a$.

\item[\textbf{(b)}]
There exists at least one monic irreducible polynomial
$f \in S\ive{x} = \tup{\ZZ/p}\ive{x}$ of degree
$m$ that satisfies $F \cong S\ive{x} / f$.

\item[\textbf{(c)}]
Any monic irreducible polynomial
$g \in S\ive{x}$ of degree $m$
divides $x^{p^m} - x \in S\ive{x}$.

\end{itemize}

[\textbf{Hint:}
For part \textbf{(a)}, it helps
to notice that $p^m > p^1 + p^2 + \cdots + p^{m-1}$.

For part \textbf{(b)}, try to find an element
$a \in F$ whose minimal polynomial has degree $m$.

Part \textbf{(c)} is entirely self-contained.]

\subsection{Remark}

Part \textbf{(b)} of this exercise shows that any
finite field can be constructed (up to isomorphism)
by adjoining a (single) root of an irreducible
polynomial to a field of the form $\ZZ / p$.
It also shows that for any prime $p$ and any
positive integer $m$, there exists an irreducible
polynomial of degree $m$ over $\ZZ/p$.

Parts \textbf{(b)} and \textbf{(c)} can be used
in showing that the finite field of a given size
is unique up to isomorphism (i.e., any two finite
fields of the same size are isomorphic).

\subsection{Solution}

...

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

\section{Exercise 7}

\subsection{Problem}

Let $p$ be a prime number.

\begin{itemize}

\item[\textbf{(a)}]
Prove that
$\tup{1+x}^{ap+c} = \tup{1+x^p}^a \tup{1+x}^c$
in the polynomial ring $\tup{\ZZ/p}\ive{x}$
for any $a, c \in \NN$.

\item[\textbf{(b)}]
Prove \emph{Lucas's congruence}:
Any $a, b \in \NN$ and any
$c, d \in \set{0, 1, \ldots, p-1}$ satisfy
\[
\dbinom{ap+c}{bp+d} \equiv \dbinom{a}{b} \dbinom{c}{d} \mod p .
\]

\end{itemize}

\subsection{Solution}

...

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

\section{Exercise 8}

\subsection{Problem}

Let $p$ be an odd prime.
Let $\zeta$ be a root of the polynomial
$x^4 + 1 \in \tup{\ZZ/p}\ive{x}$ in a
commutative $\ZZ/p$-algebra $A$.
Thus, $\zeta^4 = -1$, so that $\zeta$
is a unit (with inverse $-\zeta^3$).
Let $\tau \in A$ be defined by
$\tau = \zeta + \zeta^{-1}$.
Prove the following:

\begin{itemize}

\item[\textbf{(a)}]
We have $\tau^2 = 2$.
(Here, $2$ stands for $2 \cdot 1_A \in A$.)

\item[\textbf{(b)}]
We have $\tau^p = \legendre{2}{p} \tau$,
where $\legendre{2}{p}$ means a
Legendre symbol.

\item[\textbf{(c)}]
If $p \equiv \pm 1 \mod 8$ (that is, if $p$
is congruent to $1$ or to $-1$ modulo $8$),
then $\tau^p = \tau$.

\item[\textbf{(d)}]
If $p \equiv \pm 3 \mod 8$ (that is, if $p$
is congruent to $3$ or to $-3$ modulo $8$),
then $\tau^p = - \tau$.

\item[\textbf{(e)}]
Conclude the \emph{Second Supplementary Law
to Quadratic Reciprocity}, which says that
\begin{align}
\legendre{2}{p}
= \begin{cases}
 1, & \text{if $p \equiv \pm 1 \mod 8$;} \\
-1, & \text{if $p \equiv \pm 3 \mod 8$.}
\end{cases}
\label{eq.exe.QR2.2/p}
\end{align}

\end{itemize}

[\textbf{Hint:} For part \textbf{(b)}, start
out by writing $\tau^p = \tup{\tau^2}^{\tup{p-1}/2} \tau$.]

\subsection{Remark}
The right hand side of \eqref{eq.exe.QR2.2/p}
is also commonly written as
$\tup{-1}^{\tup{p^2-1}/8}$.
\subsection{Solution}

...

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

\section{Exercise 9}

\subsection{Problem}

Let $R$ be a commutative ring.
Let $P$ be the polynomial ring $R\ive{x, y}$.

Fix $N \in \NN$.
Let $P_N$ be the $R$-submodule
\begin{align*}
\set{f \in P \ \mid \ f = 0 \text{ or } \deg f < N}
\end{align*}
of $P$.
(This is an $R$-submodule, since it is the
span of the family
$\tup{x^i y^j}_{\tup{i, j} \in \NN^2;\ i + j < N}$.)

\begin{itemize}

\item[\textbf{(a)}]
Consider the $R$-algebra morphism
\begin{align*}
S : P &\to     R\ive{x}, \\
    f &\mapsto f\tup{x, x^N} .
\end{align*}
(This is the map that substitutes $x^N$ for $y$
in any polynomial $f \in P$.
It is an $R$-algebra morphism, as we know from
Lecture 11.)

Prove that the restriction of $S$ to $P_N$ is injective.

\end{itemize}

From now on, assume that $R$ is a field.

\begin{itemize}

\item[\textbf{(b)}]
Let $f \in P_N$ be such that the
polynomial $S\tup{f} \in R\ive{x}$ is irreducible.
Show that $f \in P = R\ive{x, y}$
is irreducible.

% \item[\textbf{(c)}]
% Use this to show that the irreducible factorization
% of a polynomial $f \in P$ is unique up to associates
% if $R$ is a field.

\end{itemize}

% [\textbf{Hint:} For part \textbf{(c)}, you are allowed
% to use the fact that $R\ive{x}$ is a UFD
% (when $R$ is a field).
% It also helps to observe that (again when $R$ is a
% field) two polynomials in $P$ (or in $R\ive{x}$) are
% associate to each other if and only if they arise
% from each other by scaling by a nonzero element of
% $R$. This is a consequence of the fact that
% multiplying by a nonconstant polynomial always
% increases the degree.]

\subsection{Remark}

The converse of part \textbf{(b)} does not hold.
For example, if $R = \QQ$ and $N = 2$,
then the polynomial
$f := 1 + 2x + y \in P$ is irreducible, but the
polynomial
$S\tup{f} = 1 + 2x + x^2 = \tup{1+x}^2 \in R\ive{x}$
is not.

\subsection{Solution}

...

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

\section{Exercise 10}

\subsection{Problem}

Let $R$ be a commutative ring.
Let $f \in R\ive{x}$ be a polynomial.
Prove that $f$ is a unit of the ring $R\ive{x}$
if and only if\footnote{Recall that
$\ive{x^i}f$ denotes the coefficient of the
monomial $x^i$ in $f$.}
\begin{itemize}
\item the coefficient $\ive{x^0}f$ is a unit of $R$,
       and
\item all the remaining coefficients $\ive{x^1}f,
       \ive{x^2}f, \ive{x^3}f, \ldots$ of $f$
       are nilpotent.
\end{itemize}

[\textbf{Hint:} Recall the result (from
Exercise 7 \textbf{(c)} on homework set \#1)
that the nilpotent elements of a commutative
ring form an ideal, as well as the result
(from Exercise 1 on homework set \#2) that
the difference of a unit and a nilpotent
element is always a unit (in a commutative
ring).
This should help with the ``if'' direction.
For the ``only if'' direction, let
$f = f_0 x^0 + f_1 x^1 + \cdots + f_n x^n \in R\ive{x}$
be a unit and
$g = g_0 x^0 + g_1 x^1 + \cdots + g_m x^m \in R\ive{x}$
be its inverse.
Use induction on $r$ to show that
$f_n^{r+1} g_{m-r} = 0$ for each
$r \in \set{0, 1, \ldots, m}$.
Use this to conclude that $f_n$ is nilpotent.]

\subsection{Remark}

This exercise precisely delineates the
phenomenon of nonconstant polynomials that
have inverses (such as the polynomial
$\overline{1} + \overline{2} x \in \tup{\ZZ/4}\ive{x}$).
In particular, it shows that this requires
$R$ not only to have zero-divisors (i.e.,
not be an integral domain), but also to
have nonzero nilpotent elements (so, e.g.,
it cannot happen for $R = \ZZ/6$).

\subsection{Solution}

...


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\begin{thebibliography}{99999999}                                                                                         %

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