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\begin{document}
\section*{Math 332 Winter 2023, Lecture 5: Rings}

\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}}}

\section{Rings and ideals (cont'd)}

\setcounter{subsection}{4}

\subsection{Units and fields (cont'd)}

\subsubsection{Units and inverses (cont'd)}

Recall the last definition we made in Lecture 4:

\begin{definition}
Let $R$ be a ring.

\textbf{(a)} An element $a\in R$ is said to be a \textbf{unit} of $R$ (or
\textbf{invertible} in $R$) if there exists a $b\in R$ such that $ab=ba=1$. In
this case, $b$ is unique and is known as the \textbf{inverse} (or
\textbf{reciprocal}, or \textbf{multiplicative inverse}) of $a$, and is
denoted by $a^{-1}$.

\textbf{(b)} We let $R^{\times}$ denote the set of all units of $R$.
\end{definition}

We then gave some examples for units (and non-units).

Our next example we state as a proposition:

\begin{proposition}
\label{prop.ringunits.Z/n}Let $n\in\mathbb{Z}$. Then:

\textbf{(a)} The units of the ring $\mathbb{Z}/n$ are precisely the residue
classes $\overline{a}$ where $a\in\mathbb{Z}$ is coprime to $n$.

\textbf{(b)} Let $a\in\mathbb{Z}$. Then, $\overline{a}$ is a unit of
$\mathbb{Z}/n$ if and only if $a$ is coprime to $n$.
\end{proposition}

\begin{proof}
Clearly, it suffices to prove part \textbf{(b)}.

\textbf{(b)} We prove the \textquotedblleft if\textquotedblright%
\ ($\Longleftarrow$) and \textquotedblleft only if\textquotedblright%
\ ($\Longrightarrow$) direction separately:

$\Longleftarrow:$ Assume that $a\in\mathbb{Z}$ is coprime to $n$. We must
prove that $\overline{a}$ is a unit of $\mathbb{Z}/n$.

Since $a$ is coprime to $n$, we have $\gcd\left(  a,n\right)  =1$. But
Bezout's theorem yields that there exist $x,y\in\mathbb{Z}$ such that
$xa+yn=\gcd\left(  a,n\right)  $. Consider these $x,y$.

We have $xa+yn=\gcd\left(  a,n\right)  =1$. In other words, $xa-1=-yn$, which
is a multiple of $n$. Therefore, $xa\equiv1\operatorname{mod}n$. In terms of
residue classes, this is saying that $\overline{xa}=\overline{1}$. In other
words, $\overline{x}\cdot\overline{a}=\overline{1}$. Since $\mathbb{Z}/n$ is
commutative, this entails $\overline{a}\cdot\overline{x}=\overline{1}$ as
well. Thus, $\overline{x}$ is an inverse of $\overline{a}$ in $\mathbb{Z}/n$.
Therefore, $\overline{a}$ is a unit of $\mathbb{Z}/n$.

$\Longrightarrow:$ Assume that $\overline{a}$ is a unit of $\mathbb{Z}/n$. We
must prove that $a$ is coprime to $n$.

Since $\overline{a}$ is a unit of $\mathbb{Z}/n$, it has an inverse
$\overline{x}$. This inverse $\overline{x}$ satisfies $\overline{xa}%
=\overline{x}\cdot\overline{a}=\overline{1}$, which means that $xa\equiv
1\operatorname{mod}n$. In other words, $xa$ differs from $1$ by a multiple of
$n$. Hence,\footnote{We are using the following fact here: If $\alpha
,\beta,\gamma$ are three integers satisfying $\alpha\equiv\beta
\operatorname{mod}\gamma$, then $\gcd\left(  \alpha,\gamma\right)
=\gcd\left(  \beta,\gamma\right)  $. In other words, when we compute the
greatest common divisor of two integers, we can add any multiple of one
integer to the other.} $\gcd\left(  xa,n\right)  =\gcd\left(  1,n\right)  =1$.
This shows that $xa$ and $n$ are coprime. Since $a$ divides $xa$, this also
entails that $a$ and $n$ are coprime (since any common divisor of $a$ and $n$
would also divide $xa$ and thus would be a common divisor of $xa$ and $n$). In
other words, $a$ is coprime to $n$.
\end{proof}

Here are some examples of Proposition \ref{prop.ringunits.Z/n}:

\begin{itemize}
\item The units of the ring $\mathbb{Z}/12$ are $\overline{1},\overline
{5},\overline{7},\overline{11}$ (because among the integers $0,1,\ldots,11$,
it is the four numbers $1,5,7,11$ that are coprime to $12$).

\item The units of the ring $\mathbb{Z}/5$ are $\overline{1},\overline
{2},\overline{3},\overline{4}$.

\item The only unit of the ring $\mathbb{Z}/2$ is $\overline{1}$.
\end{itemize}

\bigskip

Now here is a general property of units in any ring:

\begin{theorem}
\label{thm.ringunits.group}Let $R$ be a ring. Then, the set $R^{\times}$ is a
multiplicative group. More precisely: $\left(  R^{\times},\cdot,1\right)  $ is
a group.
\end{theorem}

\begin{proof}
It suffices to show the following facts:

\begin{enumerate}
\item The unity $1$ of $R$ belongs to $R^{\times}$.

\item If $a,b\in R^{\times}$, then $ab\in R^{\times}$.

\item If $a\in R^{\times}$, then $a$ has an inverse in $R^{\times}$.
\end{enumerate}

Once these three facts are proved, all other group axioms for $R^{\times}$
follow from the ring axioms for $R$. So let us prove these three facts:

\textit{Proof of Fact 1:} The element $1$ has an inverse, namely $1$ itself.

\textit{Proof of Fact 2:} Let $a,b\in R^{\times}$. Why is $ab\in R^{\times}$ ?

Since $a,b\in R^{\times}$, there are inverses $a^{-1}$ and $b^{-1}$ for $a$
and $b$.

I claim that $b^{-1}a^{-1}$ is an inverse for $ab$. Indeed, this follows from%
\[
b^{-1}\underbrace{a^{-1}\cdot a}_{=1}b=b^{-1}b=1\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ a\underbrace{b\cdot b^{-1}}_{=1}a^{-1}=aa^{-1}=1.
\]
Thus, the element $ab$ is a unit (since it has an inverse), i.e., we have
$ab\in R^{\times}$.

\textit{Proof of Fact 3:} Let $a\in R^{\times}$. Then, $a$ has an inverse
$a^{-1}$ (by the definition of a unit). We need to check that this inverse
$a^{-1}$ also belongs to $R^{\times}$.

But $a^{-1}$ has an inverse, namely $a$ (since $aa^{-1}=1$ and $a^{-1}a=1$).
Thus, $a^{-1}\in R^{\times}$ follows.

Theorem \ref{thm.ringunits.group} is now proved.
\end{proof}

\bigskip

Thus, every ring $R$ produces \textbf{two} groups: the additive group $\left(
R,+,0\right)  $ and the multiplicative group $\left(  R^{\times}%
,\cdot,1\right)  $ (standardly called the \textbf{group of units} of $R$). The
latter usually has fewer elements than the former, since it only contains the
units of $R$.

\bigskip

\begin{theorem}
[Shoe-sock theorem]\label{thm.ringunits.shoesock}Let $R$ be a ring. Let $a,b$
be two units of $R$. Then, $ab$ is a unit of $R$, and its inverse is%
\[
\left(  ab\right)  ^{-1}=b^{-1}a^{-1}.
\]

\end{theorem}

\begin{proof}
See the proof of Fact 2 in the proof of Theorem \ref{thm.ringunits.group}.
\end{proof}

\begin{theorem}
Let $R$ be a ring. Let $a$ be a unit of $R$. Then, $a^{-1}$ is a unit of $R$,
and its inverse is $\left(  a^{-1}\right)  ^{-1}=a$.
\end{theorem}

\begin{proof}
See the proof of Fact 3 in the proof of Theorem \ref{thm.ringunits.group}.
\end{proof}

\bigskip

\subsubsection{Fields}

As we saw, many rings (such as $\mathbb{Z}$) have few units, but many other
rings (such as $\mathbb{Q}$ or $\mathbb{R}$) have many. The rings of the
latter kind are known as \textquotedblleft fields\textquotedblright:

\begin{definition}
\label{def.fields.field}Let $R$ be a commutative ring. Assume that $0\neq1$ in
$R$. We say that $R$ is a \textbf{field} if every nonzero element of $R$ is a unit.
\end{definition}

Examples:

\begin{itemize}
\item The rings $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{C}$ are fields. The
ring $\mathbb{Z}$ is not (e.g., since $2$ is not a unit).

\item The ring $\mathbb{S}=\mathbb{Q}\left[  \sqrt{5}\right]  =\left\{
a+b\sqrt{5}\ \mid\ a,b\in\mathbb{Q}\right\}  $ (from Lecture 2) is a field.
Indeed, if $a+b\sqrt{5}$ is a nonzero element of $\mathbb{S}$, then
$a+b\sqrt{5}$ is a unit, since its inverse is%
\begin{align*}
\left(  a+b\sqrt{5}\right)  ^{-1}  &  =\dfrac{1}{a+b\sqrt{5}}=\dfrac
{a-b\sqrt{5}}{\left(  a+b\sqrt{5}\right)  \left(  a-b\sqrt{5}\right)  }\\
&  =\dfrac{a-b\sqrt{5}}{a^{2}-5b^{2}}=\dfrac{a}{a^{2}-5b^{2}}+\dfrac{-b}%
{a^{2}-5b^{2}}\sqrt{5}\in\mathbb{S}.
\end{align*}
(Strictly speaking, we need to make sure that $a-b\sqrt{5}$ is nonzero. The
reason why this is true is that $\sqrt{5}$ is irrational, so $a+b\sqrt{5}%
\neq0$ entails that $a$ and $b$ are not both $0$ and therefore we easily
obtain $a-b\sqrt{5}\neq0$.)

\item The Hamilton quaternions $\mathbb{H}$ would be a field if they were
commutative. Indeed, it is not hard to see that every nonzero quaternion
$a+bi+cj+dk\in\mathbb{H}$ is a unit. However, $\mathbb{H}$ is not commutative,
thus does not qualify as a field.

A noncommutative ring $R$ with $0\neq1$ whose all nonzero elements are units
is called a \textbf{division ring} or a \textbf{skew-field}. So $\mathbb{H}$
is a skew-field.

\item Let $n$ be a positive integer. Then, $\mathbb{Z}/n$ is a field if and
only if $n$ is prime. (See below for a proof.)
\end{itemize}

\subsection{Fields and integral domains: some connections}

The notions of fields and integral domains are related:

\begin{proposition}
\label{prop.fields.intdom}\textbf{(a)} Every field is an integral domain.

\textbf{(b)} Every \textbf{finite} integral domain is a field. (Of course,
\textquotedblleft finite\textquotedblright\ means \textquotedblleft finite as
a set\textquotedblright.)
\end{proposition}

\begin{proof}
\textbf{(a)} Let $F$ be a field. Why is $F$ an integral domain?

Let $a,b\in F$ be nonzero. We must prove that $ab\neq0$.

Since $a$ is nonzero, $a$ is a unit (since $F$ is a field), thus has an
inverse $a^{-1}$. If we had $ab=0$, then we would have $a^{-1}\cdot
\underbrace{ab}_{=0}=a^{-1}\cdot0=0$, which would contradict
$\underbrace{a^{-1}\cdot a}_{=1}b=b\neq0$. So we have $ab\neq0$.

Thus, $F$ is an integral domain. \medskip

\textbf{(b)} Let $R$ be a \textbf{finite} integral domain. We must show that
$R$ is a field.

Let $a\in R$ be nonzero. Our goal is to show that $a$ is a unit, i.e., has an inverse.

Consider the map%
\begin{align*}
R  &  \rightarrow R,\\
x  &  \mapsto ax.
\end{align*}
This map is injective (because if $x,y\in R$ satisfy $ax=ay$, then $a\left(
x-y\right)  =ax-ay=0$, so that $x-y=0$ since $R$ is an integral
domain\footnote{Indeed, if we had $x-y\neq0$, then we would obtain $a\left(
x-y\right)  \neq0$ (since $a\neq0$ and $x-y\neq0$, and since $R$ is an
integral domain), which would contradict $a\left(  x-y\right)  =0$.}, and
therefore $x=y$). However, the Pigeonhole Principle for Injections says that
if a map between two finite sets of the same size is injective, then it is
bijective. Hence, our map%
\begin{align*}
R  &  \rightarrow R,\\
x  &  \mapsto ax
\end{align*}
is bijective. In particular, it must take the unity $1$ as a value. In other
words, there exists some $x\in R$ such that $ax=1$. This $x$ must therefore
also satisfy $xa=1$ (since $R$ is an integral domain, thus commutative), and
thus is an inverse of $a$. So $a$ is a unit, and we are done.
\end{proof}

\bigskip

Without the word \textquotedblleft finite\textquotedblright, Proposition
\ref{prop.fields.intdom} \textbf{(b)} would fail, since $\mathbb{Z}$ is an
integral domain but not a field. Other examples of this nature are polynomial rings.

\bigskip

\begin{corollary}
Let $n$ be a positive integer. Then, the following equivalences hold:%
\[
\left(  \mathbb{Z}/n\text{ is an integral domain}\right)
\ \Longleftrightarrow\ \left(  \mathbb{Z}/n\text{ is a field}\right)
\ \Longleftrightarrow\ \left(  n\text{ is prime}\right)  .
\]

\end{corollary}

\begin{proof}
The first of these two $\Longleftrightarrow$ signs follows from Proposition
\ref{prop.fields.intdom}. So we only need to prove the second
$\Longleftrightarrow$ sign.

$\Longrightarrow:$ Assume that $\mathbb{Z}/n$ is a field. Then, all its
nonzero elements are units. In other words, the residue classes $\overline
{1},\overline{2},\ldots,\overline{n-1}$ are units. Equivalently, the numbers
$1,2,\ldots,n-1$ are coprime to $n$ (because of Proposition
\ref{prop.ringunits.Z/n} \textbf{(b)}). Hence, $n$ is prime (why?).

$\Longleftarrow:$ Assume that $n$ is prime. Then, the only positive divisors
of $n$ are $1$ and $n$. Hence, the numbers $1,2,\ldots,n-1$ are coprime to $n$
(because if $a$ is any of these numbers, then $\gcd\left(  a,n\right)  $ must
be a positive divisor of $n$, but the only positive divisors of $n$ are $1$
and $n$; hence, we must have either $\gcd\left(  a,n\right)  =1$ or
$\gcd\left(  a,n\right)  =n$; but the second possibility is ruled out by the
fact that $\gcd\left(  a,n\right)  \leq a<n$). In other words, the residue
classes $\overline{1},\overline{2},\ldots,\overline{n-1}$ are units (by
Proposition \ref{prop.ringunits.Z/n} \textbf{(b)}). This yields that
$\mathbb{Z}/n$ is a field (since $n>1$ entails that $\overline{0}\neq
\overline{1}$ in $\mathbb{Z}/n$).
\end{proof}

\bigskip

The group of units $\left(  \mathbb{Z}/p\right)  ^{\times}$ of the field
$\mathbb{Z}/p$ (where $p$ is a prime) has a nice application: Fermat's Little
Theorem. See \S 2.6.3 in the text for details.

\subsubsection{Division}

As we know, rings have addition, subtraction and multiplication, but not
always division. Nevertheless, when $b$ is a unit of a ring, it makes sense to
define $\dfrac{a}{b}$ to be the product $ab^{-1}$. Unfortunately, it makes
just as much sense to define it to be $b^{-1}a$ instead. Usually, $ab^{-1}\neq
b^{-1}a$. Thus, even if $b$ is a unit, it is ill-advised to define $\dfrac
{a}{b}$ for arbitrary rings $R$. (If you really want to, you can talk about
\textquotedblleft left division\textquotedblright\ and \textquotedblleft right
division\textquotedblright, but you should distinguish between the two.)

However, when $R$ is commutative, this ambiguity disappears, and the notation
$\dfrac{a}{b}$ becomes useful. Thus, we do introduce it:

\begin{definition}
Let $R$ be a commutative ring. Let $a\in R$ and $b\in R^{\times}$. Then,
$\dfrac{a}{b}$ means the element $ab^{-1}=b^{-1}a\in R$. This element is also
written $a/b$, and is called the \textbf{quotient} of $a$ by $b$. The
operation $\left(  a,b\right)  \mapsto\dfrac{a}{b}$ is called
\textbf{division}.
\end{definition}

In particular, in a field, we can divide by any nonzero element.

Division satisfies the rules that you would expect: If $R$ is a commutative
ring, and if $a,c\in R$ and $b,d\in R^{\times}$, then%
\begin{align*}
\dfrac{a}{b}+\dfrac{c}{d}  &  =\dfrac{ad+bc}{bd};\\
\dfrac{a}{b}\cdot\dfrac{c}{d}  &  =\dfrac{ac}{bd};\\
\dfrac{a}{b}\diagup\dfrac{c}{d}  &  =\dfrac{ad}{bc}\ \ \ \ \ \ \ \ \ \ \left(
\text{if }c\in R^{\times}\right)  ;
\end{align*}
etc.. And of course, division undoes multiplication: When $b\in R^{\times}$,
we have the equivalence%
\[
\left(  \dfrac{a}{b}=c\right)  \Longleftrightarrow\left(  a=bc\right)  .
\]



\end{document}