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\begin{document}
\section*{Math 332 Winter 2023, Lecture 8: 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}{8}

\subsection{Quotient rings (cont'd)}

\setcounter{subsubsection}{1}

\setcounter{theo}{2}

\subsubsection{Quotient rings (cont'd)}

Last time, we made the following definition and stated the theorem that comes
after it:

\begin{definition}
\label{def.quotring.quotring-repeat}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%
\begin{equation}
\left(  a+I\right)  +\left(  b+I\right)  =\left(  a+b\right)
+I\ \ \ \ \ \ \ \ \ \ \text{for all }a,b\in R.
\label{eq.def.quotring.quotring-repeat.+}%
\end{equation}


We now define a multiplication operation on $R/I$ by setting%
\begin{equation}
\left(  a+I\right)  \left(  b+I\right)  =ab+I\ \ \ \ \ \ \ \ \ \ \text{for all
}a,b\in R. \label{eq.def.quotring.quotring-repeat.*}%
\end{equation}
(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}
\label{thm.quotring.wd}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}

Note that the rules (\ref{eq.def.quotring.quotring-repeat.+}) and
(\ref{eq.def.quotring.quotring-repeat.*}), by which we defined addition and
multiplication on $R/I$, can be rewritten as%
\begin{equation}
\overline{a}+\overline{b}=\overline{a+b}\ \ \ \ \ \ \ \ \ \ \text{for all
}a,b\in R \label{eq.def.quotring.quotring.+bar}%
\end{equation}
and%
\begin{equation}
\overline{a}\cdot\overline{b}=\overline{ab}\ \ \ \ \ \ \ \ \ \ \text{for all
}a,b\in R. \label{eq.def.quotring.quotring.*bar}%
\end{equation}


\bigskip

We will prove Theorem \ref{thm.quotring.wd} later today. First, however, 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.
In fact, the above definition of $R/I$ is just the natural generalization of
the definition of the ring $\mathbb{Z}/n$, where we replaced integers by
elements of $R$ and multiples of $n$ by elements of $I$.

\item Two stupid general examples:

Recall that every ring $R$ has at least the ideals $\left\{  0_{R}\right\}  $
and $R$. What are the respective quotient rings?

\begin{itemize}
\item The quotient ring $R/\left\{  0_{R}\right\}  $ is isomorphic to $R$.
Indeed, each residue class modulo $\left\{  0_{R}\right\}  $ has the form
$r+\left\{  0_{R}\right\}  =\left\{  r\right\}  $, which is a $1$-element set.
Thus, the elements of $R/\left\{  0_{R}\right\}  $ are just the elements of
$R$ \textquotedblleft stuck in set braces\textquotedblright, with the same
rules for adding and multiplying as in $R$ (that is, $\left\{  a\right\}
+\left\{  b\right\}  =\left\{  a+b\right\}  $ and $\left\{  a\right\}
\cdot\left\{  b\right\}  =\left\{  ab\right\}  $).

\item The quotient ring $R/R$ is trivial. Indeed, there is only one residue
class modulo $R$, and this class contains all elements of $R$. (In fact, for
any $r\in R$, the corresponding residue class $r+R$ is $R$ itself.)
\end{itemize}

These are the most boring quotient rings you can imagine. Interesting things
happen when the ideal $I$ is \textquotedblleft between\textquotedblright%
\ $\left\{  0_{R}\right\}  $ and $R$.

\item Let $R$ be the ring $\mathbb{Z}\left[  i\right]  =\left\{
a+bi\ \mid\ a,b\in\mathbb{Z}\right\}  $ of Gaussian integers. Consider its
principal ideal%
\begin{align*}
3R  &  =\left\{  3r\ \mid\ r\in R\right\} \\
&  =\left\{  3r\ \mid\ r\in\mathbb{Z}\left[  i\right]  \right\} \\
&  =\left\{  3a+3bi\ \mid\ a,b\in\mathbb{Z}\right\} \\
&  =\left\{  c+di\ \mid\ c,d\in\mathbb{Z}\text{ are multiples of }3\right\}  .
\end{align*}


What is the quotient ring $R/\left(  3R\right)  $ ? The elements of this ring
have the form
\[
\overline{a+bi}\ \ \ \ \ \ \ \ \ \ \text{with }a,b\in\left\{  0,1,2\right\}
\]
(do not confuse the line over the $a+bi$ with the identical-looking notation
for \textquotedblleft complex conjugate\textquotedblright; we are not using
complex conjugates anywhere in this example). In fact, any Gaussian integer
can be reduced to a Gaussian integer of the form $a+bi$ with $a,b\in\left\{
0,1,2\right\}  $ by subtracting an appropriate Gaussian-integer multiple of
$3$ (because we can subtract a multiple of $3$ to turn its real part into one
of $0,1,2$, and then subtract a multiple of $3i$ to turn its imaginary part
into one of $0,1,2$). In other words,%
\[
R/\left(  3R\right)  =\left\{  \overline{0},\ \ \overline{1},\ \ \overline
{2},\ \ \overline{i},\ \ \overline{1+i},\ \ \overline{2+i},\ \ \overline
{2i},\ \ \overline{1+2i},\ \ \overline{2+2i}\right\}  .
\]
It is easy to see that this is a $9$-element ring (i.e., the residue classes
$\overline{0},\ \ \overline{1},\ \ \overline{2},\ \ \overline{i}%
,\ \ \overline{1+i},\ \ \overline{2+i},\ \ \overline{2i},\ \ \overline
{1+2i},\ \ \overline{2+2i}$ are distinct), and a field (i.e., all the nonzero
residue classes are invertible). So we have found a little finite field with
$9$ elements.

Let us do some computations in this field: We have%
\[
\overline{2+i}+\overline{2+2i}=\overline{\left(  2+i\right)  +\left(
2+2i\right)  }=\overline{4+3i}=\overline{1}%
\]
since $\left(  4+3i\right)  -1=3\left(  1+i\right)  \in3R$. Also,%
\begin{align*}
\overline{2+i}\cdot\overline{2+2i}  &  =\overline{\left(  2+i\right)  \left(
2+2i\right)  }=\overline{2\cdot2+2\cdot2i+i\cdot2+i\cdot2i}\\
&  =\overline{4+4i+2i-2}=\overline{2+6i}=\overline{2},
\end{align*}
since $\left(  2+6i\right)  -2=3\cdot2i\in3R$. Similarly,%
\[
\overline{2+i}\cdot\overline{1+i}=\overline{1},
\]
which shows that the elements $\overline{2+i}$ and $\overline{1+i}$ are
inverse to each other in $R/\left(  3R\right)  $.

For the curious: If we replace $3$ by any other positive integer $n$, then
$R/\left(  nR\right)  $ will be a finite ring with $n^{2}$ elements. Depending
on the value of $n$, it will or won't be a field. For instance, we found that
it is a field for $n=3$. However, it is not a field for $n=5$, because in
$R/\left(  5R\right)  $, we have%
\[
\overline{1+2i}\cdot\overline{1-2i}=\overline{\left(  1+2i\right)  \left(
1-2i\right)  }=\overline{1+4}=\overline{5}=\overline{0}.
\]
We will learn more about when $R/\left(  nR\right)  $ is a field later on.

\item Again take $R=\mathbb{Z}\left[  i\right]  $, but now consider the
quotient ring $R/\left(  \left(  1+i\right)  R\right)  $. How many elements
does it have? The answer is $2$, but this is not that obvious any more,
because how can we tell which Gaussian integers belong to $\left(  1+i\right)
R$ (that is, are Gaussian-integer multiples of $1+i$) ?

Here is one way to prove that $R/\left(  \left(  1+i\right)  R\right)  $ has
$2$ elements (and to find these elements):

\begin{itemize}
\item Observe that $2\in\left(  1+i\right)  R$ (because $2=\left(  1+i\right)
\left(  1-i\right)  $). Thus, every Gaussian integer can be reduced to a
Gaussian integer of the form $a+bi$ with $a,b\in\left\{  0,1\right\}  $ by
adding an element of $\left(  1+i\right)  R$.

\item Thus, $R/\left(  \left(  1+i\right)  R\right)  =\left\{  \overline
{0},\ \overline{1},\ \overline{i},\ \overline{1+i}\right\}  $.

\item Furthermore, $\overline{0}=\overline{1+i}$ and $\overline{1}%
=\overline{i}$ (why?).

\item Thus, $R/\left(  \left(  1+i\right)  R\right)  =\left\{  \overline
{0},\ \overline{1}\right\}  $ (why?).

\item Finally, we have $\overline{0}\neq\overline{1}$, since $0-1$ is not a
multiple of $1+i$ (because $\dfrac{0-1}{1+i}=\dfrac{-1}{1+i}=\dfrac{-1\left(
1-i\right)  }{\left(  1+i\right)  \left(  1-i\right)  }=\dfrac{-1+i}{2}%
=\dfrac{-1}{2}+\dfrac{1}{2}i$ is not a Gaussian integer).

\item Consequently, $R/\left(  \left(  1+i\right)  R\right)  $ consists of the
two distinct elements $\overline{0}$ and $\overline{1}$.
\end{itemize}

Can we analyze $R/\left(  \left(  7+9i\right)  R\right)  $ likewise? What
about $R/\left(  \left(  a+bi\right)  R\right)  $ for general $a$ and $b$ ?
This will be a ring of size $a^{2}+b^{2}$ (unless $a=b=0$), but we don't quite
have the tools to prove this yet.
\end{itemize}

See the text for a few more examples (some of which will be on homework set \#3).

Let us now make good on our debts and prove Theorem \ref{thm.quotring.wd}:

\begin{proof}
[Proof of Theorem \ref{thm.quotring.wd}.]We must prove that the operations $+$
and $\cdot$ on $R/I$ are well-defined, and that $R/I$ is a ring when equipped
with these operations.

The latter is very easy: All the ring axioms are inherited from $R$. For
example, to see that $\cdot$ on $R/I$ is associative, we must show that
$\overline{a}\cdot\left(  \overline{b}\cdot\overline{c}\right)  =\left(
\overline{a}\cdot\overline{b}\right)  \cdot\overline{c}$ for all $a,b,c\in R$;
but this is clear since the LHS is $\overline{a\left(  bc\right)  }$ by
definition and the RHS is $\overline{\left(  ab\right)  c}$ by definition and
since associativity of multiplication in $R$ yields $a\left(  bc\right)
=\left(  ab\right)  c$.

The harder\ part is the first part: We must show that $+$ and $\cdot$ on $R/I$
are well-defined. For $+$, this has already been done in group theory (it is
part of what it means for the quotient \textbf{group} $R/I$ to be
well-defined). Thus, we only need to do it for $\cdot$.

Well-definedness\ for $\cdot$ means that the product $\overline{a}%
\cdot\overline{b}$ of two residue classes $\overline{a}=a+I$ and $\overline
{b}=b+I$ depends \textbf{only on these residue classes} and not on the
elements $a,b\in R$ themselves. In other words, it means that if $x$ and $y$
are two residue classes modulo $I$, and if we compute their product $xy$ using
the formula (\ref{eq.def.quotring.quotring.*bar}) by writing $x$ as
$\overline{a}$ and $y$ as $\overline{b}$ for some $a,b\in R$, then the exact
choices of $a$ and $b$ do not affect the resulting value $xy=\overline{ab}$.

To prove this, we must therefore show the following: If one and the same
residue class in $R/I$ can be written both as $\overline{a}$ and as
$\overline{a^{\prime}}$, and if one and the same residue class in $R/I$ can be
written both as $\overline{b}$ and as $\overline{b^{\prime}}$, then
$\overline{ab}=\overline{a^{\prime}b^{\prime}}$.

In other words, we must show the following: If four elements $a,b,a^{\prime
},b^{\prime}\in R$ satisfy $\overline{a}=\overline{a^{\prime}}$ and
$\overline{b}=\overline{b^{\prime}}$, then $\overline{ab}=\overline{a^{\prime
}b^{\prime}}$.

This is the ring-theoretical generalization of the well-known fact that if
five integers $a,b,a^{\prime},b^{\prime},n\in\mathbb{Z}$ satisfy $a\equiv
a^{\prime}\operatorname{mod}n$ and $b\equiv b^{\prime}\operatorname{mod}n$,
then $ab\equiv a^{\prime}b^{\prime}\operatorname{mod}n$. As we recall, this
classical fact can be proved by rewriting $a\equiv a^{\prime}%
\operatorname{mod}n$ as $a=a^{\prime}+nx$ for some integer $x$, and likewise
rewriting $b\equiv b^{\prime}\operatorname{mod}n$ as $b=b^{\prime}+ny$ for
some integer $y$, and then multiplying these two equalities to find
\[
ab=\left(  a^{\prime}+nx\right)  \left(  b^{\prime}+ny\right)  =a^{\prime
}b^{\prime}+\underbrace{a^{\prime}ny+nxb^{\prime}+nxny}_{\text{divisible by
}n}\equiv a^{\prime}b^{\prime}\operatorname{mod}n.
\]


The proof of the ring-theoretical generalization is not much different: Let
$a,b,a^{\prime},b^{\prime}\in R$ satisfy $\overline{a}=\overline{a^{\prime}}$
and $\overline{b}=\overline{b^{\prime}}$. Then, from $\overline{a}%
=\overline{a^{\prime}}$, we obtain $a-a^{\prime}\in I$. In other words,
$a-a^{\prime}=i$ for some $i\in I$. Similarly, $b-b^{\prime}=j$ for some $j\in
I$. Consider these $i$ and $j$. From $a-a^{\prime}=i$, we obtain $a=a^{\prime
}+i$. Similarly, $b=b^{\prime}+j$. Multiplying the latter two equalities, we
obtain%
\[
ab=\left(  a^{\prime}+i\right)  \left(  b^{\prime}+j\right)  =a^{\prime
}b^{\prime}+a^{\prime}j+ib^{\prime}+ij.
\]
Hence, we can conclude that $\overline{ab}=\overline{a^{\prime}b^{\prime}}$ if
we can show that $a^{\prime}j+ib^{\prime}+ij\in I$. But this follows from the
ideal axioms, since $i$ and $j$ belong to $I$. (In more detail: The second
ideal axiom yields $a^{\prime}j\in I$ and $ib^{\prime}\in I$ and $ij\in I$;
then, the first ideal axiom yields $a^{\prime}j+ib^{\prime}+ij\in I$.)

Thus, we have proved that $\overline{ab}=\overline{a^{\prime}b^{\prime}}$
(since $ab=a^{\prime}b^{\prime}+\underbrace{a^{\prime}j+ib^{\prime}+ij}_{\in
I}$ shows that $ab$ and $a^{\prime}b^{\prime}$ belong to the same coset of $I$
in $R$). This concludes the proof that the operation $\cdot$ on $R/I$ is
well-defined. As we said, this also completes our proof of Theorem
\ref{thm.quotring.wd}.
\end{proof}


\end{document}