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\begin{document}
\section*{Math 332 Winter 2023, Lecture 7: 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}{7}

\subsection{Ideals and kernels}

\subsubsection{Kernels}

Here is a definition that we already made last time:

\begin{definition}
Let $R$ and $S$ be rings. Let $f:R\rightarrow S$ be a ring morphism. Then, the
\textbf{kernel} of $f$ is defined to be the set%
\[
\operatorname*{Ker}f:=\left\{  r\in R\ \mid\ f\left(  r\right)  =0_{S}%
\right\}  .
\]
This is a subset of $R$ (but usually not a subring).
\end{definition}

Examples:

\begin{itemize}
\item Let $n\in\mathbb{Z}$. The kernel of the ring morphism%
\begin{align*}
\pi:\mathbb{Z}  &  \rightarrow\mathbb{Z}/n,\\
a  &  \mapsto\overline{a}%
\end{align*}
is $n\mathbb{Z}=\left\{  \text{all multiples of }n\right\}  $.

\item Let $R$ be a ring. Let $S$ be any set. Recall the ring $R^{S}$ of all
functions from $S$ to $R$ (with pointwise operations). Fix an element $s\in
S$. Then, the kernel of the ring morphism%
\begin{align*}
R^{S}  &  \rightarrow R,\\
f  &  \mapsto f\left(  s\right)
\end{align*}
is the set of all functions $f\in R^{S}$ that vanish on $s$.

\item The kernel of an injective ring morphism $f:R\rightarrow S$ is always
$\left\{  0_{R}\right\}  $. Indeed, any ring morphism $f:R\rightarrow S$ sends
$0_{R}$ to $0_{S}$. If $f$ is furthermore injective, then $0_{R}$ must be the
\textbf{only} element of $R$ that $f$ sends to $0_{S}$ (since any other such
element $a$ would cause $f\left(  a\right)  =0_{S}=f\left(  0_{R}\right)  $,
which would violate injectivity), so we obtain $\operatorname*{Ker}f=\left\{
0_{R}\right\}  $.
\end{itemize}

\subsubsection{Ideals}

As we saw, kernels of ring morphisms are usually not subrings. Instead, here
is what they are:

\begin{definition}
Let $R$ be a ring. An \textbf{ideal} of $R$ means a subset $I$ of $R$ such that

\begin{itemize}
\item we have $a+b\in I$ for any $a,b\in I$ (that is, $I$ is closed under addition);

\item we have $ab\in I$ and $ba\in I$ for any $a\in R$ and $b\in I$ (that is,
$I$ is closed under multiplication \textbf{by arbitrary elements of }$R$ --
not just under multiplication within its own elements);

\item we have $0\in I$ (where $0$ means $0_{R}$).
\end{itemize}
\end{definition}

These three requirements are called the \textbf{ideal axioms}. The second is
called \textquotedblleft\textbf{absorption axiom}\textquotedblright, since it
yields that any product that has (at least) one factor in $I$ will belong to
$I$. In other words, it says that lying in $I$ is \textquotedblleft
infectious\textquotedblright\ (at least in products). When $R$ is commutative,
the requirements $ab\in I$ and $ba\in I$ are equivalent.

Here are some easy consequences of the definition:

\begin{proposition}
Let $R$ be a ring. Let $I$ be an ideal of $R$. Then, $I$ is a subgroup of the
additive group $\left(  R,+,0\right)  $.
\end{proposition}

\begin{proof}
By the first ideal axiom, $I$ is closed under addition. By the third, it
contains $0$. By the second axiom, it is closed under multiplying by $-1$,
which means that it is closed under negation.
\end{proof}

\begin{theorem}
Let $R$ and $S$ be two rings. Let $f:R\rightarrow S$ be a ring morphism. Then,
its kernel $\operatorname*{Ker}f$ is an ideal of $R$.
\end{theorem}

\begin{proof}
Let us check the ideal axioms. Specifically, we will check the second, since
the other two are easier.

So we need to show that $ab\in\operatorname*{Ker}f$ and $ba\in
\operatorname*{Ker}f$ for any $a\in R$ and $b\in\operatorname*{Ker}f$.

To that purpose, we let $a\in R$ and $b\in\operatorname*{Ker}f$ be arbitrary.
Since $f$ is a ring morphism, we then have%
\[
f\left(  ab\right)  =f\left(  a\right)  \cdot\underbrace{f\left(  b\right)
}_{\substack{=0_{S}\\\text{(since }b\in\operatorname*{Ker}f\text{)}}}=f\left(
a\right)  \cdot0_{S}=0_{S},
\]
so that $ab\in\operatorname*{Ker}f$. Similarly, $ba\in\operatorname*{Ker}f$.

As I said, the other two axioms are similar but easier.
\end{proof}

We will soon see that the theorem we just proved has a converse: Any ideal of
a ring $R$ can be written as the kernel of a ring morphism from $R$ to another
ring $S$. Thus, ideals and kernels are \textquotedblleft the same thing,
viewed from different angles\textquotedblright.

\subsubsection{Principal ideals}

The simplest way to construct ideals in a commutative ring is by fixing an
element and taking all its multiples:

\begin{proposition}
Let $R$ be a commutative ring. Let $u\in R$. We define $uR$ to be the set
$\left\{  ur\ \mid\ r\in R\right\}  $. The elements of this set $uR$ are
called the \textbf{multiples} of $u$ (in $R$).

Then, $uR$ is an ideal of $R$. This ideal is known as a \textbf{principal
ideal} of $R$. In particular, $0R=\left\{  0\right\}  $ and $1R=R$ are
therefore principal ideals of $R$.
\end{proposition}

\begin{proof}
The only thing we need to prove is that $uR$ is an ideal of $R$. Let us check
the axioms:

\begin{itemize}
\item We have $a+b\in uR$ for any $a\in uR$ and $b\in uR$. Indeed, let $a\in
uR$ and $b\in uR$. Then, we can write $a$ as $a=ux$ for some $x\in R$ (since
$a\in uR$), and likewise we can write $b$ as $b=uy$ for some $y\in R$, and
then we have $a+b=ux+uy=u\left(  x+y\right)  \in uR$.

\item We have $ab\in uR$ and $ba\in uR$ for any $a\in R$ and $b\in uR$.
Indeed, let $a\in R$ and $b\in uR$. Then, we can write $b$ as $b=uy$ for some
$y\in R$ (since $b\in uR$), and then we have $ab=auy=uay\in uR$ and $ba=ab\in
uR$.

\item We have $0\in uR$, since $0=u\cdot0$.
\end{itemize}
\end{proof}

For example, $2\mathbb{Z}=\left\{  \text{all even integers}\right\}  $ is a
principal ideal of $\mathbb{Z}$.

Principal ideals can also be defined for noncommutative rings, but this is
more complicated (for details, see the footnote in \S 2.8.3 of the text).
(Fortunately, $uR$ still works if $u$ is a central element of $R$.)

\subsubsection{Other examples of ideals}

In the classical number rings $\mathbb{Z},\mathbb{Q},\mathbb{R},\mathbb{C}$,
all ideals are principal (this will be proved below). But some other
commutative rings have non-principal ideals as well. For example:

\begin{itemize}
\item Consider the set of all polynomials $f\in\mathbb{Q}\left[  x,y\right]  $
that have constant term $0$ (equivalently: that vanish when you substitute $0$
for $x$ and for $y$). This set is an ideal of $\mathbb{Q}\left[  x,y\right]  $
(why?), but not a principal ideal (why?).

\item Consider the set of all polynomials $f\in\mathbb{Z}\left[  x\right]  $
whose constant term is even. This set is an ideal of $\mathbb{Z}\left[
x\right]  $, but not a principal ideal.
\end{itemize}

We will come back to this later, as we rigorously define polynomials in one of
the next chapters.

See the text (\S 2.8.4) for more examples of ideals. Here is just one:

\begin{example}
Let $R$ be any ring. Recall that
\[
R^{2\leq2}=\left\{  \left(
\begin{array}
[c]{cc}%
a & b\\
0 & d
\end{array}
\right)  \ \mid\ a,b,d\in R\right\}
\]
is the ring of all upper-triangular $2\times2$-matrices over $R$. Then,%
\begin{align*}
I  &  :=\left\{  \left(
\begin{array}
[c]{cc}%
0 & b\\
0 & d
\end{array}
\right)  \ \mid\ b,d\in R\right\}  ,\\
J  &  :=\left\{  \left(
\begin{array}
[c]{cc}%
a & b\\
0 & 0
\end{array}
\right)  \ \mid\ a,b\in R\right\}  ,\\
K  &  :=\left\{  \left(
\begin{array}
[c]{cc}%
0 & b\\
0 & 0
\end{array}
\right)  \ \mid\ b\in R\right\}
\end{align*}
are ideals of $R^{2\leq2}$, whereas%
\[
L:=\left\{  \left(
\begin{array}
[c]{cc}%
a & 0\\
0 & d
\end{array}
\right)  \ \mid\ a,d\in R\right\}
\]
is not. Checking this will be a problem on homework set \#3.
\end{example}

\subsection{Quotient rings}

What follows is one of the most abstract topics of this course: the definition
of quotient rings, and their basic properties.

Recall the idea behind modular arithmetic: By passing from the integers to
their residue classes modulo $n$ (for a given $n\in\mathbb{Z}$), we are
essentially equating $n$ with $0$: Two integers become \textquotedblleft
equal\textquotedblright\ if they differ by a multiple of $n$. Thus, the
residue classes modulo $n$ are \textquotedblleft what
remains\textquotedblright\ of the integers if we equate $n$ with $0$.

The same passage can be made in greater generality: We can start with any ring
$R$ and any ideal $I$ of $R$, and we can equate all elements of $I$ with $0$
(so that two elements of $R$ become \textquotedblleft equal\textquotedblright%
\ if they differ by an element of $I$). What remains is again called
\textquotedblleft residue classes\textquotedblright\ (now modulo $I$ rather
than modulo $n$), and we can again define addition and multiplication on these
classes. The result is a new ring, called the \textbf{quotient ring} of $R$ by
the ideal $I$, and denoted by $R/I$. Working in this quotient ring $R/I$ is a
natural generalization of modular arithmetic to things that are not
necessarily integers (but can be matrices or polynomials or any other objects instead).

This was the rough idea behind quotient rings. We will now define them rigorously.

\subsubsection{Quotient groups}

We don't need to reinvent the wheel: You have already seen residue classes in
a first course on groups; in that context, they are known as \textquotedblleft%
\textbf{cosets}\textquotedblright. We will only have to turn them into a ring.

Let me recall the definition of cosets:

\begin{itemize}
\item If $H$ is a subgroup of a group $G$, then the \textbf{left cosets} of
$H$ in $G$ are the subsets
\[
gH:=\left\{  gh\ \mid\ h\in H\right\}  \ \ \ \ \ \ \ \ \ \ \text{for all }g\in
G.
\]
There is one left coset $gH$ for each $g\in G$; but different $g$'s can often
lead to the same coset. Thus, there are usually fewer left cosets than
elements of $G$. The set of all left cosets of $H$ is called $G/H$.

\item If $H$ is merely a subgroup of a group $G$, then $G/H$ is not a group,
but just a \textquotedblleft$G$-set\textquotedblright\ (i.e., a set equipped
with an action of $G$). However, when $H$ is a \textbf{normal} subgroup of a
group $G$, then $G/H$ becomes a group itself, with group operation defined by%
\begin{equation}
\left(  g_{1}H\right)  \cdot\left(  g_{2}H\right)  =g_{1}g_{2}%
H\ \ \ \ \ \ \ \ \ \ \text{for all }g_{1},g_{2}\in G. \label{eq.cosets.prod}%
\end{equation}
This group $G/H$ is called the \textbf{quotient group} of $G$ by $H$. The left
cosets of $H$ in $G$ are just called the \textbf{cosets} of $H$ in $G$ in this case.

\item If $G$ is an abelian group, then any subgroup $H$ of $G$ is normal, so
$G/H$ is always a group.

\item Now, assume that $G$ is an \textbf{additive} group (which means that its
binary operation is written as $+$ instead of $\cdot$). This presupposes that
$G$ is abelian. Then, the cosets of $H$ in $G$ are denoted by $g+H$ instead of
$gH$. The quotient group $G/H$ is also written additively (i.e., we denote its
operation by $+$ instead of $\cdot$). The equality (\ref{eq.cosets.prod}) thus
turns into%
\[
\left(  g_{1}+H\right)  +\left(  g_{2}+H\right)  =\left(  g_{1}+g_{2}\right)
+H\ \ \ \ \ \ \ \ \ \ \text{for all }g_{1},g_{2}\in G.
\]
Thus, on the quotient group $G/H$, we have an addition operation, not a
multiplication operation (but mathematically, the definition is the same as
before; it's just the notation that has changed). Note that $G/H$ is again an
abelian group.

\item The most famous example of a quotient group is when $G=\mathbb{Z}$ and
$H=n\mathbb{Z}=\left\{  \text{all multiples of }n\right\}  $ for some fixed
integer $n$. (Here, the group operation is addition.) Then, the quotient group
$\mathbb{Z}/n\mathbb{Z}$ is the cyclic group $\mathbb{Z}/n$, and this is
precisely how $\mathbb{Z}/n$ is defined.

(See
\href{http://homepages.warwick.ac.uk/~maseap/teaching/aa/aanotes.pdf}{Samir
Siksek, \textit{Introduction to Algebra}}, Chapter XII, for more examples of
quotient groups.)
\end{itemize}

\subsubsection{Quotient rings}

Now, piggybacking on the construction of quotient groups we just recalled, we
shall define a similar quotient structure for rings instead of groups. Instead
of normal subgroups, we will use ideals this time:

\begin{definition}
Let $I$ be an ideal of a ring $R$. Thus, $I$ is a subgroup of the additive
group $\left(  R,+,0\right)  $, hence a normal subgroup (since $\left(
R,+,0\right)  $ is abelian). Thus, the quotient group $R/I$ is a well-defined
abelian group. Its elements are the cosets $r+I$ of $I$ in $R$. These cosets
are called the \textbf{residue classes} modulo $I$. A coset $r+I$ is also
denoted by $\overline{r}$ or $\left[  r\right]  $ or $\left[  r\right]  _{I}$
or $r\operatorname{mod}I$. (We will only use the notations $\overline{r}$ and
$r+I$.)

Note that the addition on $R/I$ is given by%
\[
\left(  a+I\right)  +\left(  b+I\right)  =\left(  a+b\right)
+I\ \ \ \ \ \ \ \ \ \ \text{for all }a,b\in R.
\]


We now define a multiplication operation on $R/I$ by setting%
\[
\left(  a+I\right)  \left(  b+I\right)  =ab+I\ \ \ \ \ \ \ \ \ \ \text{for all
}a,b\in R.
\]
(See below for a proof that this is well-defined.)

The set $R/I$, equipped with the addition and the multiplication we just
defined, and with the elements $0+I$ and $1+I$ playing the roles of zero and
unity, is a ring (as we will soon see). This ring is called the
\textbf{quotient ring} of $R$ by the ideal $I$, and is denoted by $R/I$. It is
pronounced \textquotedblleft$R$ \textbf{modulo }$I$\textquotedblright.
\end{definition}

\begin{theorem}
Let $R$ and $I$ be as in this definition. Then, the multiplication on $R/I$ is
well-defined, and $R/I$ becomes a ring when endowed with the operations we
just introduced.
\end{theorem}

We will prove this next time. First, a few examples:

\begin{itemize}
\item Let $n\in\mathbb{Z}$. Then, the set $n\mathbb{Z}=\left\{  \text{all
multiples of }n\right\}  $ is an ideal of $\mathbb{Z}$ (a principal ideal, in
fact). The quotient ring $\mathbb{Z}/n\mathbb{Z}$ is exactly the ring
$\mathbb{Z}/n$ of residue classes modulo $n$ that we introduced a while ago.
\end{itemize}


\end{document}