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\begin{document}
\section*{Math 332 Winter 2023, Lecture 25: Polynomials}

\textbf{website:}
\texttt{\href{https://www.cip.ifi.lmu.de/~grinberg/t/23wa}{\texttt{https://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/t/23wa}}}

\setcounter{section}{2}

\section{Monoid algebras and polynomials}

Recall: For this entire chapter, we fix a \textbf{commutative} ring $R$.

\setcounter{subsection}{1}

\subsection{Polynomial rings (cont'd)}

\setcounter{subsubsection}{2}

\setcounter{theo}{3}

\subsubsection{Evaluation of univariate polynomials}

Polynomials can not only be added, scaled and multiplied, but also
\textbf{evaluated} (i.e., we can substitute elements of an $R$-algebra for the
indeterminates). This is what makes polynomial rings special (as compared to
arbitrary monoid algebras).

Let us first define evaluation for univariate polynomials.

\begin{definition}
\label{def.polring.univar-sub}Let $p\in R\left[  x\right]  $ be a univariate
polynomial. Let $A$ be any $R$-algebra. Let $a\in A$.

We define the element $p\left(  a\right)  \in A$ as follows: Write $p$ as%
\[
p=\sum_{i\in\mathbb{N}}p_{i}x^{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{where
}p_{i}\in R\text{, and all but finitely many }i\text{ satisfy }p_{i}=0\right)
,
\]
and set%
\[
p\left(  a\right)  :=\sum_{i\in\mathbb{N}}p_{i}a^{i}.
\]


This element $p\left(  a\right)  $ is called the \textbf{evaluation} of $p$ at
$a$; we also say that it is obtained by \textbf{substituting} $a$ for $x$ in
$p$.

Instead of $p\left(  a\right)  $, we also write $p\left[  a\right]  $. (See
the warning below for the reason why.)
\end{definition}

Here, $A$ can be any $R$-algebra (not necessarily commutative): for example,
it can be $R$ itself, or a matrix ring $R^{n\times n}$, or a polynomial ring
like $R\left[  x\right]  $. Thus, polynomials are \textquotedblleft like
functions but better\textquotedblright\ (since a function can only be
evaluated at the elements of its domain, whereas a polynomial can be evaluated
at any element of any $R$-algebra).

Evaluation is also the reason why polynomials are better-behaved than formal
power series. For example, you cannot evaluate the formal power series
$1+x+x^{2}+x^{3}+\cdots$ at $1$ (since $1+1+1^{2}+1^{3}+\cdots$ is ill-defined).

Note that $p\left(  x\right)  =p$ for any $p\in R\left[  x\right]  $. Indeed,
$p\left(  x\right)  $ is the result of substituting $x$ for $x$, and clearly
this substitution doesn't actually change anything in the polynomial.

\begin{warning}
The notation $p\left(  a\right)  $ can be quite ambiguous. For example, is
$p\left(  p+1\right)  $ the evaluation of $p$ at $p+1$ or rather the product
of $p$ with $p+1$ ? Thus, I recommend writing $p\cdot\left(  p+1\right)  $ for
the product and $p\left[  p+1\right]  $ for the evaluation. Generally, I will
use the $p\left[  a\right]  $ notation as often as possible, but sometimes I
will use $p\left(  a\right)  $ for the sake of familiarity.
\end{warning}

Here is a slightly surprising example of evaluation:

\begin{example}
Let $R=\mathbb{Z}/2$, and let $p\in R\left[  x\right]  $ be the polynomial
$x^{2}+x=x\cdot\left(  x+\overline{1}\right)  $. Let us evaluate $p$ at
elements of $R$:%
\begin{align*}
p\left[  \overline{0}\right]   &  =\overline{0}^{2}+\overline{0}=\overline
{0};\\
p\left[  \overline{1}\right]   &  =\overline{1}^{2}+\overline{1}=\overline
{2}=\overline{0}.
\end{align*}
Thus, the polynomial $p$ gives $\overline{0}$ whenever it is evaluated at any
element of $\mathbb{Z}/2$. This does not mean that $p$ is the zero polynomial!
Evaluating $p$ on $2\times2$-matrices reveals that $p$ can take nonzero values
as well:%
\[
p\left[  \left(
\begin{array}
[c]{cc}%
\overline{0} & \overline{1}\\
\overline{1} & \overline{0}%
\end{array}
\right)  \right]  =\left(
\begin{array}
[c]{cc}%
\overline{0} & \overline{1}\\
\overline{1} & \overline{0}%
\end{array}
\right)  ^{2}+\left(
\begin{array}
[c]{cc}%
\overline{0} & \overline{1}\\
\overline{1} & \overline{0}%
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
\overline{1} & \overline{1}\\
\overline{1} & \overline{1}%
\end{array}
\right)  \neq0_{2\times2}.
\]
(And of course, evaluating $p$ at $x$ itself gives $p\left[  x\right]  =p$,
which is nonzero as well.)
\end{example}

Given an $R$-algebra $A$ and an element $a\in A$, the operation of evaluating
polynomials $p\in R\left[  x\right]  $ at $a$ is rather well-behaved:

\begin{theorem}
\label{thm.polring.univar-sub-hom}Let $A$ be an $R$-algebra. Let $a\in A$.
Then, the map%
\begin{align*}
R\left[  x\right]   &  \rightarrow A,\\
p  &  \mapsto p\left[  a\right]
\end{align*}
is an $R$-algebra morphism. In other words:

\begin{itemize}
\item This map respects multiplication and addition. In other words: For any
two polynomials $p,q\in R\left[  x\right]  $, we have%
\begin{align*}
\left(  pq\right)  \left[  a\right]   &  =p\left[  a\right]  \cdot q\left[
a\right]  \ \ \ \ \ \ \ \ \ \ \text{and}\\
\left(  p+q\right)  \left[  a\right]   &  =p\left[  a\right]  +q\left[
a\right]  .
\end{align*}


\item This map respects scaling. In other words: For any $\lambda\in R$ and
$p\in R\left[  x\right]  $, we have%
\[
\left(  \lambda p\right)  \left[  a\right]  =\lambda\cdot p\left[  a\right]
.
\]


\item This map respects zero and one. In other words: We have $0\left[
a\right]  =0$ and $1\left[  a\right]  =1$.
\end{itemize}
\end{theorem}

The proof of this becomes easy using the following lemma (which shows that
\textquotedblleft respects multiplication\textquotedblright\ can be proved by
checking it on monomials):

\begin{lemma}
\label{lem.algebras.mor-on-basis}Let $A$ and $B$ be two $R$-algebras. Let
$f:A\rightarrow B$ be an $R$-linear map. Let $\left(  m_{i}\right)  _{i\in I}$
be a family of vectors in $A$ that spans $A$. If we have%
\[
f\left(  m_{i}m_{j}\right)  =f\left(  m_{i}\right)  f\left(  m_{j}\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in I,
\]
then%
\[
f\left(  ab\right)  =f\left(  a\right)  f\left(  b\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }a,b\in A.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.algebras.mor-on-basis}.]By linearity. See Lemma 4.2.9
in the text for details.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.polring.univar-sub-hom}.]The first bullet point is
easy using Lemma \ref{lem.algebras.mor-on-basis}. The other two bullet points
are easy on their own. See Theorem 4.2.8 in the text for details.
\end{proof}

\subsubsection{Evaluation for multivariate polynomials}

So much for evaluation of univariate polynomials. An analogous concept exists
for multivariate polynomials, but it requires that the \textquotedblleft
inputs\textquotedblright\ (i.e., the values at which we evaluate our
polynomial) mutually commute. Here is the precise definition:

\begin{definition}
\label{def.polring.mulvar-sub}Let $n\in\mathbb{N}$. Let $p\in R\left[
x_{1},x_{2},\ldots,x_{n}\right]  $ be a multivariate polynomial. Let $A$ be
any $R$-algebra. Let $a_{1},a_{2},\ldots,a_{n}\in A$ be $n$ elements of $A$
that mutually commute (i.e., that satisfy $a_{i}a_{j}=a_{j}a_{i}$ for all
$i,j\in\left\{  1,2,\ldots,n\right\}  $).

We define the element $p\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in A$ as
follows: Write the polynomial $p$ as%
\[
p=\sum_{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathbb{N}^{n}}%
p_{i_{1},i_{2},\ldots,i_{n}}x_{1}^{i_{1}}x_{2}^{i_{2}}\cdots x_{n}^{i_{n}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{for }p_{i_{1},i_{2},\ldots,i_{n}}\in
R\right)  ,
\]
and set%
\[
p\left(  a_{1},a_{2},\ldots,a_{n}\right)  :=\sum_{\left(  i_{1},i_{2}%
,\ldots,i_{n}\right)  \in\mathbb{N}^{n}}p_{i_{1},i_{2},\ldots,i_{n}}%
a_{1}^{i_{1}}a_{2}^{i_{2}}\cdots a_{n}^{i_{n}}.
\]


This element $p\left(  a_{1},a_{2},\ldots,a_{n}\right)  $ is called the
\textbf{evaluation} of $p$ at $a_{1},a_{2},\ldots,a_{n}$; we also say that it
is obtained by \textbf{substituting} $a_{1},a_{2},\ldots,a_{n}$ for
$x_{1},x_{2},\ldots,x_{n}$ in $p$. We also denote it by $p\left[  a_{1}%
,a_{2},\ldots,a_{n}\right]  $.
\end{definition}

There is an analogue of Theorem \ref{thm.polring.univar-sub-hom}:

\begin{theorem}
Let $n\in\mathbb{N}$. Let $A$ be any $R$-algebra. Let $a_{1},a_{2}%
,\ldots,a_{n}\in A$ be $n$ elements of $A$ that mutually commute (i.e., that
satisfy $a_{i}a_{j}=a_{j}a_{i}$ for all $i,j\in\left\{  1,2,\ldots,n\right\}
$). Then, the map%
\begin{align*}
R\left[  x_{1},x_{2},\ldots,x_{n}\right]   &  \rightarrow A,\\
p  &  \mapsto p\left(  a_{1},a_{2},\ldots,a_{n}\right)
\end{align*}
is an $R$-algebra morphism.
\end{theorem}

\begin{proof}
See Theorem 4.2.11 in the text. (Here we need the $a_{i}a_{j}=a_{j}a_{i}$ condition.)
\end{proof}

\subsubsection{Constant polynomials}

In Convention 3.1.4 (in Lecture 24), we defined the notion of a constant
element of a monoid ring $R\left[  M\right]  $: This is just an element of the
form $r\cdot1=r\cdot e_{1}$, where $r\in R$ and where $1$ is the neutral
element of $M$.

Since polynomial rings are monoid rings, we obtain in particular the notion of
a constant polynomial. This is exactly what you think: a polynomial of the
form $r\cdot x^{0}$ (or, in the multivariate case, $r\cdot x_{1}^{0}x_{2}%
^{0}\cdots x_{n}^{0}$) for $r\in R$. We identify such a polynomial with the
scalar $r\in R$ itself.

For example, the polynomial $3x^{0}=3\in\mathbb{Z}\left[  x\right]  $ is
constant, but the polynomial $3x$ is not.

\subsubsection{Coefficients}

By their definition, polynomials are $R$-linear combinations of monomials. Let
us introduce a notation for the coefficients in these $R$-linear combinations:

\begin{definition}
\label{def.polring.coeff}Let $p\in R\left[  x_{1},x_{2},\ldots,x_{n}\right]  $
be a polynomial. Let $\mathfrak{m}=x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots
x_{n}^{a_{n}}$ be a monomial. Then, the \textbf{coefficient} of $\mathfrak{m}$
in $p$ is the element $\left[  \mathfrak{m}\right]  p$ of $R$ defined as
follows: If we write $p$ as
\[
p=\sum_{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathbb{N}^{n}}%
p_{i_{1},i_{2},\ldots,i_{n}}x_{1}^{i_{1}}x_{2}^{i_{2}}\cdots x_{n}^{i_{n}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{for }p_{i_{1},i_{2},\ldots,i_{n}}\in
R\right)  ,
\]
then%
\[
\left[  \mathfrak{m}\right]  p:=p_{a_{1},a_{2},\ldots,a_{n}}.
\]

\end{definition}

For example:

\begin{itemize}
\item For univariate polynomials, we have%
\[
\left[  x^{3}\right]  \left(  \left(  1+x\right)  ^{5}\right)  =\dbinom{5}%
{3}=10\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \left[  x^{6}\right]
\left(  \left(  1+x\right)  ^{5}\right)  =0
\]
(since $\left(  1+x\right)  ^{5}=1+5x+10x^{2}+10x^{3}+5x^{4}+x^{5}$).

\item For multivariate polynomials, if we set $n=2$ and rename the variables
$x_{1},x_{2}$ as $x,y$, then we have%
\[
\left[  x^{2}y^{3}\right]  \left(  \left(  x+y\right)  ^{5}\right)  =10
\]
(since $\left(  x+y\right)  ^{5}=x^{5}+5x^{4}y+10x^{3}y^{2}+10x^{2}%
y^{3}+5xy^{4}+y^{5}$) and%
\[
\left[  xy\right]  \left(  \left(  x+y\right)  ^{5}\right)  =0
\]
(for the same reason).
\end{itemize}

\subsubsection{Symbols for indeterminates}

The definition of a multivariate polynomial ring $R\left[  x_{1},x_{2}%
,\ldots,x_{n}\right]  $ that we gave above (Definition 3.2.3 in Lecture 24)
depends on a ring $R$ and the number $n$. The names of the indeterminates are
\textquotedblleft hardcoded\textquotedblright\ to be $x_{1},x_{2},\ldots
,x_{n}$. However, it is actually better to have a more flexible definition,
which allows to arbitrarily specify the names of the indeterminates. Thus, for
example, we should be able to define the polynomial rings $R\left[
x,y\right]  $ and $R\left[  y,z\right]  $, which are each isomorphic to
$R\left[  x_{1},x_{2}\right]  $, but should not be treated as being the same
ring (since the former has indeterminates $x$ and $y$ whereas the latter has
indeterminates $y$ and $z$). Distinguishing between these two isomorphic rings
$R\left[  x,y\right]  $ and $R\left[  y,z\right]  $ offers several advantages,
in particular allowing us to identify them with two \textbf{different}
subrings of $R\left[  x,y,z\right]  $ (in a natural way).

This necessitates some minor changes to our definition of multivariate
polynomial rings (Definition 3.2.3 in Lecture 24). Namely, instead of using
the monoid
\[
C^{\left(  n\right)  }=\left\{  x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{n}^{a_{n}%
}\ \mid\ \left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathbb{N}^{n}\right\}
\]
as the set of monomials, we now need to use the monoid%
\[
C^{\left(  S\right)  }=\left\{  \prod_{s\in S}s^{a_{s}}\ \mid\ a_{s}%
\in\mathbb{N}\text{ for each }s\in S\right\}  ,
\]
where $S$ is our chosen (finite) set of indeterminates (for example,
$S=\left\{  x,y\right\}  $ or $S=\left\{  y,z\right\}  $ or $S=\left\{
\alpha,\mathbf{w},\clubsuit\right\}  $ if we have nothing better to do). A
monomial in this monoid $C^{\left(  S\right)  }$ is a \textquotedblleft
formal\textquotedblright\ product of the form $\prod_{s\in S}s^{a_{s}}$ (with
each factor being a formal power of one of our indeterminates), and stands for
a family $\left(  a_{s}\right)  _{s\in S}\in\mathbb{N}^{S}$ of nonnegative
integers. Two such monomials are multiplied by the rule%
\[
\left(  \prod_{s\in S}s^{a_{s}}\right)  \cdot\left(  \prod_{s\in S}s^{b_{s}%
}\right)  =\prod_{s\in S}s^{a_{s}+b_{s}}.
\]
The polynomial ring in the set $S$ of indeterminates is then defined as the
monoid ring $R\left[  C^{\left(  S\right)  }\right]  $ of this monoid
$C^{\left(  S\right)  }$. We shall refer to such a ring as a
\textbf{multivariate polynomial ring with named variables}, and just call it
$R\left[  S\right]  $.

Thus, for a three-element set $S=\left\{  x,y,z\right\}  $, we have $R\left[
S\right]  =R\left[  x,y,z\right]  $, which is the polynomial ring over $R$ in
three variables that are named $x,y,z$. For instance, $x^{2}+7y^{3}z-xyz$ is a
polynomial in this ring $R\left[  x,y,z\right]  $. As we said, $R\left[
x,y,z\right]  $ is isomorphic to $R\left[  x_{1},x_{2},x_{3}\right]  $ as an
$R$-algebra (via the isomorphism that sends each monomial $x^{a}y^{b}z^{c}$ to
$x_{1}^{a}x_{2}^{b}x_{3}^{c}$); thus, named variables do not introduce
anything genuinely new to our theory as long as we are studying a single
polynomial ring at a time. But their flexibility is helpful when working with
several polynomial rings, e.g., by allowing us to treat $R\left[  x\right]  $
and $R\left[  y\right]  $ as two different subrings of $R\left[  x,y\right]  $.

We will be cavalier about all of this in the following, sometimes renaming
variables at will (we will often rename $x_{1},x_{2}$ as $x,y$ or rename
$x_{1},x_{2},x_{3}$ as $x,y,z$ for no other reasons than brevity).

\subsection{Univariate polynomials}

Let us now take a closer look at univariate polynomial rings, as they have
several special properties that multivariate polynomial rings do not share.

\subsubsection{Degrees and coefficients}

By its definition, the univariate polynomial ring $R\left[  x\right]  $ has
the basis $\left(  x^{0},x^{1},x^{2},\ldots\right)  $ (as an $R$-module).

Recall that if $p\in R\left[  x\right]  $ is a polynomial, and if
$i\in\mathbb{N}$, then $\left[  x^{i}\right]  p$ is the coefficient of $x^{i}$
in $p$. That is, if $p$ is written as $p=\sum_{j\in\mathbb{N}}p_{j}x^{j}$ with
$p_{j}\in R$, then $\left[  x^{i}\right]  p=p_{i}$.

\begin{definition}
Let $p\in R\left[  x\right]  $ be a univariate polynomial.

\textbf{(a)} If $p\neq0$, then the \textbf{degree} of $p$ is defined to be the
largest $i\in\mathbb{N}$ such that $\left[  x^{i}\right]  p\neq0$. The degree
of the zero polynomial $0\in R\left[  x\right]  $ is defined to be $-\infty$
(a symbol that is understood to be smaller than any integer).

The degree of $p$ is denoted by $\deg p$.

\textbf{(b)} If $p\neq0$, then the \textbf{leading coefficient} of $p$ is
defined to be $\left[  x^{\deg p}\right]  p\in R$.

\textbf{(c)} The polynomial $p$ is said to be \textbf{monic} if its leading
coefficient is $1$.
\end{definition}

For example, the polynomial
\[
5x^{3}+2x+1\in\mathbb{Q}\left[  x\right]
\]
has degree $3$ and leading coefficient $5$. Hence, it is not monic (since
$5\neq1$). The polynomial
\[
\overline{5}x^{3}+\overline{2}x+\overline{1}\in\left(  \mathbb{Z}/n\right)
\left[  x\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{for a given integer
}n>0\right)
\]
has

\begin{itemize}
\item degree $3$ if $n>5$;

\item degree $1$ if $n=5$ (since the $\overline{5}x^{3}$ term disappears when
$n=5$);

\item degree $3$ if $n=2,3,4$;

\item degree $-\infty$ if $n=1$.
\end{itemize}

\begin{remark}
Let $n\in\mathbb{N}$. Then,
\begin{align*}
&  \left\{  f\in R\left[  x\right]  \ \mid\ \deg f\leq n\right\} \\
&  =\left\{  f\in R\left[  x\right]  \ \mid\ f=a_{0}x^{0}+a_{1}x^{1}%
+\cdots+a_{n}x^{n}\ \text{for some }a_{i}\in R\right\} \\
&  =\operatorname*{span}\left(  x^{0},x^{1},\ldots,x^{n}\right)  .
\end{align*}
In particular, this is an $R$-submodule of $R\left[  x\right]  $.
\end{remark}

\begin{corollary}
Let $p,q\in R\left[  x\right]  $. Then,%
\begin{align*}
\deg\left(  p+q\right)   &  \leq\max\left\{  \deg p,\deg q\right\}
\ \ \ \ \ \ \ \ \ \ \text{and}\\
\deg\left(  p-q\right)   &  \leq\max\left\{  \deg p,\deg q\right\}  .
\end{align*}

\end{corollary}

\begin{remark}
The polynomials of degree $\leq0$ are just the constant polynomials.
\end{remark}

So much for degrees of sums and differences. What can we say about the degree
of a product?

\begin{proposition}
\label{prop.polring.univar-degpq}Let $p,q\in R\left[  x\right]  $. Then:

\textbf{(a)} We have $\deg\left(  pq\right)  \leq\deg p+\deg q$.

\textbf{(b)} We have $\deg\left(  pq\right)  =\deg p+\deg q$ if $p\neq0$ and
the leading coefficient of $p$ is a unit.

\textbf{(c)} We have $\deg\left(  pq\right)  =\deg p+\deg q$ if $R$ is an
integral domain.

\textbf{(d)} If $n,m\in\mathbb{N}$ satisfy $n\geq\deg p$ and $m\geq\deg q$,
then%
\[
\left[  x^{n+m}\right]  \left(  pq\right)  =\left[  x^{n}\right]  \left(
p\right)  \cdot\left[  x^{m}\right]  \left(  q\right)  .
\]


\textbf{(e)} If $pq=0$ and $p\neq0$ and if the leading coefficient of $p$ is a
unit, then $q=0$.
\end{proposition}

\begin{proof}
See Proposition 4.3.5 in the text for proofs.
\end{proof}

\begin{corollary}
If $R$ is an integral domain, then the polynomial ring $R\left[  x\right]  $
is also an integral domain.
\end{corollary}

\begin{proof}
Follows from Proposition \ref{prop.polring.univar-degpq} \textbf{(c)} (since
$\deg0=-\infty$).
\end{proof}

\begin{remark}
If $R$ is not an integral domain, then polynomials over $R$ can behave rather
strangely. For example, if $R=\mathbb{Z}/4$, then%
\[
\left(  \overline{1}+\overline{2}x\right)  ^{2}=\overline{1}+\overline
{4}x+\overline{4}x^{2}=\overline{1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\overline{4}=\overline{0}\right)  .
\]
So the degree of a polynomial can decrease when it is squared!
\end{remark}

\subsubsection{Division with remainder}

Just like integers, univariate polynomials can be divided with remainder, as
long as the polynomial you are dividing by has an invertible (i.e., unit)
leading coefficient:

\begin{theorem}
[Division-with-remainder theorem for polynomials]%
\label{thm.polring.univar-quorem}Let $b\in R\left[  x\right]  $ be a nonzero
polynomial whose leading coefficient is a unit. Let $a\in R\left[  x\right]  $
be any polynomial.

\textbf{(a)} Then, there is a \textbf{unique} pair $\left(  q,r\right)  $ of
polynomials in $R\left[  x\right]  $ such that%
\[
a=qb+r\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \deg r<\deg b.
\]


\textbf{(b)} Moreover, this pair satisfies $\deg q\leq\deg a-\deg b$.
\end{theorem}

\begin{proof}
See Theorem 4.3.7 in the text.
\end{proof}

The polynomials $q$ and $r$ in Theorem \ref{thm.polring.univar-quorem} are
called the \textbf{quotient} and the \textbf{remainder} obtained when dividing
$a$ by $b$. Note that if $\deg a<\deg b$, then the quotient $q$ is $0$ whereas
the remainder $r$ is $a$. The quotient and the remainder become interesting
when $\deg a\geq\deg b$.

Don't forget the condition \textquotedblleft the leading coefficient of $b$ is
a unit\textquotedblright\ in Theorem \ref{thm.polring.univar-quorem}. This
condition is automatically satisfied if $b$ is monic (since $1$ is a unit),
and it is also automatically satisfied if $R$ is a field (since any nonzero
element of a field is a unit). But there are examples for $R=\mathbb{Z}$ where
it is not satisfied. (See Exercises 5 and 6 on homework set \#6 for some
examples.) \medskip

Recall from elementary number theory that a positive integer $b$ divides an
integer $a$ if and only if the remainder that $a$ leaves when divided by $b$
is $0$. Here is an analogue of this fact for univariate polynomials:

\begin{proposition}
\label{prop.polring.univar-rem0}Let $b\in R\left[  x\right]  $ be a nonzero
polynomial whose leading coefficient is a unit. Let $a\in R\left[  x\right]  $
be any polynomial. Let $q$ and $r$ be the quotient and the remainder obtained
when dividing $a$ by $b$. Then, we have $b\mid a$ in $R\left[  x\right]  $ if
and only if $r=0$.
\end{proposition}

\begin{proof}
See Proposition 4.3.12 in the text (or figure it out on your own -- it is easy).
\end{proof}


\end{document}