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\begin{document}
\section*{Math 332 Winter 2023, Lecture 3: 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)}

\subsection{Defining rings (cont'd)}

\setcounter{subsubsection}{1}

\subsubsection{Some examples of rings (cont'd)}

\textbf{Warm-up:}

Let's practice computations in $\mathbb{Z}/n$ (\textquotedblleft modular
arithmetic\textquotedblright).

Recall that $\overline{a}$ denotes the residue class of an integer $a$ modulo
$n$. Note that this notation does not mention $n$, but of course the meaning
of $\overline{a}$ does depend on $n$. So we will rely on context to clarify
what the $n$ is. Less ambiguously, we can denote the same residue class
$\overline{a}$ as $a+n\mathbb{Z}$, which stands for \textquotedblleft coset of
$a$ with respect to the subgroup $n\mathbb{Z}$ of the abelian group
$\mathbb{Z}$\textquotedblright. The residue classes modulo $n$ are precisely
the cosets of the subgroup $n\mathbb{Z}$ in $\mathbb{Z}$.

Let us now compute some products of residue classes.

\begin{itemize}
\item In $\mathbb{Z}/12$, we have%
\[
\overline{3}\cdot\overline{7}=\overline{3\cdot7}=\overline{21}=\overline
{9}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }21\equiv9\operatorname{mod}%
12\right)
\]
and%
\[
\overline{6}\cdot\overline{8}=\overline{6\cdot8}=\overline{48}=\overline
{0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }48\equiv0\operatorname{mod}%
12\right)  .
\]


\item In $\mathbb{Z}/15$, we have%
\[
\overline{6}\cdot\overline{5}=\overline{6\cdot5}=\overline{30}=\overline
{0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }30\equiv0\operatorname{mod}%
15\right)  .
\]
Note that the $\overline{6}$ here is not the same as the $\overline{6}$ in the
previous example. Indeed, the $\overline{6}$ here is $6+15\mathbb{Z}$, whereas
the previous $\overline{6}$ was $6+12\mathbb{Z}$.

Note that there is no such thing as $\overline{6}/\overline{5}$. Nor is there
such a thing as $\overline{20}/\overline{5}$. Indeed, it sounds reasonable to
define $\overline{20}/\overline{5}=\overline{20/5}=\overline{4}$. By the same
logic, it sounds reasonable to define $\overline{50}/\overline{5}%
=\overline{50/5}=\overline{10}$. However, these two equalities would
contradict one another, since their LHSs are equal (since $\overline
{20}=\overline{50}$) but their RHSs are not. So division in $\mathbb{Z}/15$
cannot be defined unambiguously in general.
\end{itemize}

\bigskip

Let's go back to examples of rings. Here is one more:

\begin{itemize}
\item Consider a $4$-element set with four elements $0,1,a,b$. We endow this
set with two operations $+$ and $\cdot$ defined by the following tables of
values:%
\[%
\begin{tabular}
[c]{|c||c|c|c|c|}\hline
$x+y$ & $0$ & $1$ & $a$ & $b$\\\hline\hline
$0$ & $0$ & $1$ & $a$ & $b$\\\hline
$1$ & $1$ & $0$ & $b$ & $a$\\\hline
$a$ & $a$ & $b$ & $0$ & $1$\\\hline
$b$ & $b$ & $a$ & $1$ & $0$\\\hline
\end{tabular}
\ \ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \
\begin{tabular}
[c]{|c||c|c|c|c|}\hline
$xy$ & $0$ & $1$ & $a$ & $b$\\\hline\hline
$0$ & $0$ & $0$ & $0$ & $0$\\\hline
$1$ & $0$ & $1$ & $a$ & $b$\\\hline
$a$ & $0$ & $a$ & $b$ & $1$\\\hline
$b$ & $0$ & $b$ & $1$ & $a$\\\hline
\end{tabular}
\ \ \ \ \ .
\]
This turns our set into a ring, which we call $F_{4}$. You can check all the
ring axioms for it by brute force; it's not fun. We will later see a
\textquotedblleft conceptual\textquotedblright\ way to construct this ring
$F_{4}$, which will render this busywork unnecessary.
\end{itemize}

We will see many more examples through this course.

\subsection{Calculating in rings}

\subsubsection{What works}

The intuition for a commutative ring is essentially \textquotedblleft a system
of numbers, but built of something other than numbers\textquotedblright. So we
expect all the standard rules for calculating with numbers to also hold in any
commutative ring. For a noncommutative ring, we need to be more careful, since
commutativity of multiplication ($ab=ba$) is not required, and therefore any
rule that relies on commutativity (e.g., the binomial formula) doesn't have to
hold either.

Let us be more precise about what rules we expect to hold.

If $a_{1},a_{2},\ldots,a_{n}$ are $n$ elements of a ring, then the sum
$a_{1}+a_{2}+\cdots+a_{n}$ is well-defined (i.e., you can add its addends from
the front or from the back or in any other order, and the results will all be
the same). More generally, any finite sum of the form $\sum_{s\in S}a_{s}$
(with $S$ being a finite set) is well-defined whenever the $a_{s}$ belong to a
ring. Such finite sums behave like usual finite sums of numbers. This is
called \textbf{generalized commutativity}. (For rigorous proofs, you can find
some references in the text.) An empty sum is defined to be the zero of the
underlying ring.

If our ring is commutative, then the same is true for finite products of the
form $\prod_{s\in S}a_{s}$. However, this is not the case if our ring is
noncommutative, because the order in a product matters even for just $2$
factors. But a product with a well-defined order, such as $a_{1}a_{2}\cdots
a_{n}$, is well-defined in any ring (commutative or not). For example, $abcde$
is well-defined, i.e., the result does not depend on whether you read it as
$\left(  a\left(  bc\right)  \right)  \left(  de\right)  $ or as $\left(
\left(  \left(  ab\right)  c\right)  d\right)  e$ or in any of the other
possible ways. This is called \textbf{generalized associativity}. An empty
product is defined to be the unity of the underlying ring.

In any ring, subtraction satisfies the rules you would expect: For any two
elements $a,b$ of a ring, we have%
\begin{align*}
\left(  -a\right)  b  &  =a\left(  -b\right)  =-\left(  ab\right)  ;\\
\left(  -a\right)  \left(  -b\right)   &  =ab;\\
\left(  -1\right)  a  &  =-a.
\end{align*}


Furthermore, for any three elements $a,b,c$ of a ring, we have%
\[
a\left(  b-c\right)  =ab-ac\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \left(  a-b\right)  c=ac-bc
\]
(\textquotedblleft\textbf{distributivity of subtraction}\textquotedblright).

\bigskip

Next, some more definitions.

If $n$ is an integer, and $a$ is an element of a ring $R$, then we define an
element $na$ of $R$ by%
\[
na=%
\begin{cases}
\underbrace{a+a+\cdots+a}_{n\text{ times}}, & \text{if }n\geq0;\\
-\left(  \underbrace{a+a+\cdots+a}_{-n\text{ times}}\right)  , & \text{if
}n<0.
\end{cases}
\]
Note that this defines multiplying (aka \textbf{scaling}) an element of $R$ by
an integer. This is not the same as multiplying two elements of $R$ with each
other. (However, if $R$ does contain $\mathbb{Z}$ as a subset, then usually
the two operations will give the same result, unless the multiplication of $R$
has been rigged to make this false\footnote{An example of such a rigged
multiplication is the ring $\mathbb{Z}^{\prime}$ from Lecture 1.}.)

If $n$ is a nonnegative integer, and $a$ is an element of a ring $R$, then we
define an element $a^{n}$ of $R$ by%
\[
a^{n}=\underbrace{aa\cdots a}_{n\text{ factors}}.
\]
In particular,%
\[
a^{0}=\left(  \text{empty product}\right)  =1_{R}\ \ \ \ \ \ \ \ \ \ \text{by
definition.}%
\]


\bigskip

Now, we have learned to scale elements of a ring by integers, and take them to
nonnegative integer powers. These operations satisfy some of the rules that
you would expect. For example, if $a,b\in R$ (with $R$ being a ring), and
$n,m\in\mathbb{Z}$, then%
\begin{align*}
\left(  n+m\right)  a  &  =na+ma;\\
n\left(  a+b\right)   &  =na+nb;\\
\left(  nm\right)  a  &  =n\left(  ma\right)  ;\\
\left(  -1\right)  a  &  =-a.
\end{align*}
Furthermore, if $a\in R$ and $n,m\in\mathbb{N}$, then%
\begin{align*}
a^{n+m}  &  =a^{n}\cdot a^{m};\\
a^{nm}  &  =\left(  a^{n}\right)  ^{m}.
\end{align*}


Also,%
\begin{align*}
1_{R}^{n}  &  =1_{R}\ \ \ \ \ \ \ \ \ \ \text{for any }n\in\mathbb{N};\\
0_{R}^{n}  &  =0_{R}\ \ \ \ \ \ \ \ \ \ \text{for any integer }n>0;\\
0_{R}^{0}  &  =1_{R}.
\end{align*}


Moreover, if $a,b\in R$ satisfy $ab=ba$, then we have%
\[
a^{i}b^{j}=b^{j}a^{i}\ \ \ \ \ \ \ \ \ \ \text{for }i,j\in\mathbb{N}%
\]
and%
\[
\left(  ab\right)  ^{n}=a^{n}b^{n}\ \ \ \ \ \ \ \ \ \ \text{for }%
n\in\mathbb{N}%
\]
and the binomial formula%
\[
\left(  a+b\right)  ^{n}=\sum_{k=0}^{n}\dbinom{n}{k}a^{k}b^{n-k}%
\ \ \ \ \ \ \ \ \ \ \text{for }n\in\mathbb{N}.
\]
All of this is proved just as for numbers (with the exception of $a^{i}%
b^{j}=b^{j}a^{i}$, which is unnecessary for numbers, but can be easily proved
by a double induction on $i$ and on $j$).

\begin{warning}
None of the latter three identities can be expected to hold if $ab\neq ba$. It
is easy to find examples of $2\times2$-matrices $a,b\in\mathbb{Q}^{2\times2}$
which violate them all.
\end{warning}

\subsubsection{What doesn't work}

Here are some less familiar features of rings:

\begin{itemize}
\item It is not always true that $a\neq0$ and $b\neq0$ imply $ab\neq0$ (when
$a$ and $b$ are elements of a ring $R$). For example, $\overline{6}%
\cdot\overline{5}=\overline{0}=0$ in $\mathbb{Z}/15$, but neither
$\overline{6}$ nor $\overline{5}$ is $0$.

\item It is not always true that $ab=1$ implies $ba=1$. Counterexamples,
however, are hard to find, since \textquotedblleft$ab=1\Longrightarrow
ba=1$\textquotedblright\ holds not just for numbers and residue classes of
integers, but also (e.g.) for square matrices with real entries. It will take
several weeks until we meet a ring where the \textquotedblleft%
$ab=1\Longrightarrow ba=1$\textquotedblright\ implication is false.
\end{itemize}

\subsection{Subrings}

\subsubsection{Definition}

Groups have subgroups; vector spaces have subspaces. Not surprisingly, rings
have their substructures too:

\begin{definition}
Let $R$ be a ring. A \textbf{subring} of $R$ is a subset $S$ of $R$ such that

\begin{itemize}
\item we have $a+b\in S$ for all $a,b\in S$;

\item we have $ab\in S$ for all $a,b\in S$;

\item we have $-a\in S$ for all $a\in S$;

\item we have $0\in S$ (where the $0$ means the zero of $R$);

\item we have $1\in S$ (where the $1$ means the unity of $R$).
\end{itemize}
\end{definition}

These five conditions are called the \textquotedblleft\textbf{subring
axioms}\textquotedblright. They are called \textquotedblleft$S$ is closed
under addition\textquotedblright, \textquotedblleft$S$ is closed under
multiplication\textquotedblright, \textquotedblleft$S$ is closed under
negation\textquotedblright, \textquotedblleft$S$ contains $0$%
\textquotedblright\ and \textquotedblleft$S$ contains $1$\textquotedblright,
respectively. Altogether, they ensure that the following proposition holds:

\begin{proposition}
Let $S$ be a subring of a ring $R$. Then, $S$ automatically is a ring in its
own right (with its operations $+$ and $\cdot$ obtained by restricting the
corresponding operations of $R$, and with its zero and unity passed down from
$R$).
\end{proposition}

\subsubsection{Examples}

Here are some examples of subrings:

\begin{itemize}
\item From the classical construction of the number systems, you know that
$\mathbb{Z}\subseteq\mathbb{Q}\subseteq\mathbb{R}\subseteq\mathbb{C}$. Each of
these three \textquotedblleft$\subseteq$\textquotedblright\ signs can be
strengthened to \textquotedblleft is a subring of\textquotedblright. For
instance, $\mathbb{Z}$ is a subring of $\mathbb{Q}$.

\item You can extend this chain further to the right: $\mathbb{C}$ is a
subring of $\mathbb{H}$ (the Hamilton quaternions).

\item However, we \textbf{cannot} extend this chain to the left: The only
subring of $\mathbb{Z}$ is $\mathbb{Z}$ itself. Indeed, if $S$ is a subring of
$\mathbb{Z}$, then $1\in S$ (by the last subring axiom), therefore
$\underbrace{1+1+\cdots+1}_{n\text{ times}}\in S$ for any $n>0$ (since $S$ is
closed under addition), so that $1,2,3,\ldots\in S$, therefore
$-1,-2,-3,\ldots\in S$ (since $S$ is closed under negation), and finally $0\in
S$ (by an axiom), so that $S$ contains all the integers $\ldots
,-2,-1,0,1,2,\ldots$, and therefore $S=\mathbb{Z}$.

\item There are lots of rings between $\mathbb{Z}$ and $\mathbb{Q}$ (that is,
rings $\mathbb{B}$ such that $\mathbb{Z}$ is a subring of $\mathbb{B}$ and
$\mathbb{B}$ is a subring of $\mathbb{Q}$). For example, the set of all
rational numbers of the form%
\[
\dfrac{a}{b}\ \ \ \ \ \ \ \ \ \ \text{with }a\in\mathbb{Z}\text{ being
arbitrary and }b\in\mathbb{Z}\text{ being odd}%
\]
is such a ring\footnote{To prove this, you need to show the subring axioms for
this set. For example, why is it closed under addition? Well, if we add two
such numbers $\dfrac{a}{b}$ and $\dfrac{c}{d}$ (with $a,b,c,d\in\mathbb{Z}$
and with $b,d$ odd), then we get $\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}%
$, which again has an odd denominator. Thus, our set is closed under addition.
Closedness under multiplication is similar, and the remaining axioms are
almost obvious.}. More such rings you can find on Homework Set \#1 exercise 5.
\end{itemize}


\end{document}