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\begin{document}
\section*{Math 332 Winter 2023, Lecture 10: 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}{4}

\setcounter{theo}{5}

\subsubsection{The universal property of quotient rings}

When trying to understand a quotient ring $R/I$, it is important to construct
ring morphisms into and out of it.

Constructing morphisms $\alpha:S\rightarrow R/I$ into a quotient ring $R/I$ is
generally easy. (For example, we did so in the proof of Theorem 1.9.5 in
Lecture 9.)

Constructing morphisms $\beta:R/I\rightarrow S$ out of a quotient ring $R/I$
is harder: Not only do you have to define what $\beta\left(  \overline
{r}\right)  $ is for any given residue class $\overline{r}$, but you also need
to make sure that this value $\beta\left(  \overline{r}\right)  $ will depend
not on $r$ but only on the residue class $\overline{r}$. This is called
\textquotedblleft well-definedness\textquotedblright, and is usually not
immediately obvious, because one and the same residue class in $R/I$ can be
written as $\overline{r}$ for different values of $r\in R$. In other words,
the well-definedness of $\beta$ does not come for free if we just define
$\beta\left(  \overline{r}\right)  $ by some formula for all $r\in R$; we also
need to check that the value that this formula gives is unambiguous.

This can be done by hand, but it is extra work (adding to the trouble of
proving that $\beta$ is a ring morphism).

The \textbf{universal property of quotient rings} is a theorem that does some
of this work for you. It gives a way to define a ring morphism $\beta
:R/I\rightarrow S$ by providing a ring morphism $f:R\rightarrow S$ and showing
that $f\left(  I\right)  =0$ (that is, $f$ sends all elements of $I$ to $0$).
Once you have done this part, the theorem automatically gives you a ring
morphism $f^{\prime}:R/I\rightarrow S$ that sends each residue class
$\overline{r}\in R/I$ to $f\left(  r\right)  $. Here is the precise statement:

\begin{theorem}
[Universal property of quotient rings]\label{thm.quotring.uniprop1}Let $R$ be
a ring. Let $I$ be an ideal of $R$.

Let $S$ be a ring. Let $f:R\rightarrow S$ be a ring morphism. Assume that
$f\left(  I\right)  =0$ (by this we mean that $f\left(  i\right)  =0$ for each
$i\in I$). Then, the map%
\begin{align*}
f^{\prime}:R/I  &  \rightarrow S,\\
\overline{r}  &  \mapsto f\left(  r\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{for all }r\in R\right)
\end{align*}
is well-defined (i.e., the value $f\left(  r\right)  $ depends only on the
residue class $\overline{r}$, not on $r$ itself) and is a ring morphism.
\end{theorem}

Here is an example:

\begin{itemize}
\item Consider the canonical projections%
\begin{align*}
\pi_{6}:\mathbb{Z}  &  \rightarrow\mathbb{Z}/6,\\
r  &  \mapsto r+6\mathbb{Z}%
\end{align*}
and%
\begin{align*}
\pi_{3}:\mathbb{Z}  &  \rightarrow\mathbb{Z}/3,\\
r  &  \mapsto r+3\mathbb{Z}.
\end{align*}
(Each of these two projections sends each integer $r$ to its residue class
$\overline{r}$, but the residue class is a modulo-$6$ class for $\pi_{6}$ and
a modulo-$3$ class for $\pi_{3}$.)

Then, $\pi_{3}\left(  6\mathbb{Z}\right)  =0$ (because any $j\in6\mathbb{Z}$
is a multiple of $6$, thus a multiple of $3$, and therefore its residue class
$j+3\mathbb{Z}$ is $\overline{0}$, and thus $\pi_{3}\left(  j\right)
=j+3\mathbb{Z}=\overline{0}=0_{\mathbb{Z}/3}$). Thus, by Theorem
\ref{thm.quotring.uniprop1} (applied to $R=\mathbb{Z}$, $I=6\mathbb{Z}$,
$S=\mathbb{Z}/3$ and $f=\pi_{3}$), we see that the map%
\begin{align*}
\pi_{3}^{\prime}:\mathbb{Z}/6  &  \rightarrow\mathbb{Z}/3,\\
\overline{r}  &  \mapsto\pi_{3}\left(  r\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{that is, }r+6\mathbb{Z}\mapsto r+3\mathbb{Z}\right)
\end{align*}
is well-defined and is a ring morphism. Explicitly, this morphism $\pi
_{3}^{\prime}$ sends%
\begin{align*}
\text{the modulo-}6\text{ residue classes\ \ }  &  \overline{0},\ \ \overline
{1},\ \ \overline{2},\ \ \overline{3},\ \ \overline{4},\ \ \overline{5}\\
\text{to the modulo-}3\text{ residue classes\ }  &  \overline{0}%
,\ \ \overline{1},\ \ \overline{2},\ \ \overline{3},\ \ \overline
{4},\ \ \overline{5}.
\end{align*}
In other words, it sends%
\begin{align*}
\text{the modulo-}6\text{ residue classes\ \ }  &  \overline{0},\ \ \overline
{1},\ \ \overline{2},\ \ \overline{3},\ \ \overline{4},\ \ \overline{5}\\
\text{to the modulo-}3\text{ residue classes\ }  &  \overline{0}%
,\ \ \overline{1},\ \ \overline{2},\ \ \overline{0},\ \ \overline
{1},\ \ \overline{2}%
\end{align*}
(because in $\mathbb{Z}/3$, we have $\overline{3}=\overline{0}$ and
$\overline{4}=\overline{1}$ and $\overline{5}=\overline{2}$). If you want, you
can easily check that this is actually a morphism.

More generally, if $n$ and $m$ are two integers such that $m\mid n$, then
there is a ring morphism%
\begin{align}
\mathbb{Z}/n  &  \rightarrow\mathbb{Z}/m,\nonumber\\
\overline{r}  &  \mapsto\overline{r}\ \ \ \ \ \ \ \ \ \ \left(  \text{that is,
}r+n\mathbb{Z}\mapsto r+m\mathbb{Z}\right)  .
\label{eq.exa.quotring.Z/n-to-Z/m}%
\end{align}
This follows from Theorem \ref{thm.quotring.uniprop1}, applied to
$R=\mathbb{Z}$, $I=n\mathbb{Z}$, $S=\mathbb{Z}/m$ and $f=\pi_{m}$ (the
canonical projection from $\mathbb{Z}$ to $\mathbb{Z}/m$).

Incidentally, this accounts for all ring morphisms that go between two
quotient rings of $\mathbb{Z}$. That is:

\begin{itemize}
\item If $m$ and $n$ are two integers such that $m\mid n$, then there is only
one ring morphism $\mathbb{Z}/n\rightarrow\mathbb{Z}/m$, and it is the one we
just constructed (sending each $r+n\mathbb{Z}$ to $r+m\mathbb{Z}$).

\item If $m$ and $n$ are two integers such that $m\nmid n$, then there is no
ring morphism $\mathbb{Z}/n\rightarrow\mathbb{Z}/m$.
\end{itemize}

Proving this is a nice (and easy) exercise!
\end{itemize}

\begin{proof}
[Proof of Theorem \ref{thm.quotring.uniprop1}.]First, we must prove that
$f^{\prime}$ is well-defined (i.e., that the formula $f^{\prime}\left(
\overline{r}\right)  =f\left(  r\right)  $ does not assign two different
output values to one and the same residue class).

In other words, we must show that if two elements $a$ and $b$ of $R$ have the
same residue class $\overline{a}=\overline{b}$ (modulo $I$), then $f\left(
a\right)  =f\left(  b\right)  $.

So let us do this. Let $a$ and $b$ be two elements of $R$ such that
$\overline{a}=\overline{b}$. From $\overline{a}=\overline{b}$, we obtain
$a-b\in I$. Hence, $f\left(  a-b\right)  =0$ (because $f\left(  I\right)
=0$). However, $f$ is a ring morphism and thus respects differences. Hence,
$f\left(  a-b\right)  =f\left(  a\right)  -f\left(  b\right)  $. Thus,
$f\left(  a\right)  -f\left(  b\right)  =f\left(  a-b\right)  =0$, and
therefore $f\left(  a\right)  =f\left(  b\right)  $, as desired.

Thus, we have proved that the map $f^{\prime}$ is well-defined, i.e., exists.

It remains to prove that $f^{\prime}$ is a ring morphism. We have four axioms
to check, but we will only check the \textquotedblleft respects
multiplication\textquotedblright\ axiom, since the other three axioms are
proved similarly.

To check the \textquotedblleft respects multiplication\textquotedblright%
\ axiom, we must prove that $f^{\prime}\left(  xy\right)  =f^{\prime}\left(
x\right)  \cdot f^{\prime}\left(  y\right)  $ for all $x,y\in R/I$.

Fix $x,y\in R/I$. Then, $x=\overline{a}$ and $y=\overline{b}$ for some $a,b\in
R$ (since any element of $R/I$ is a residue class, i.e., has the form
$\overline{r}$ for some $r\in R$). Using these $a$ and $b$, we then have%
\[
xy=\overline{a}\cdot\overline{b}=\overline{ab}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of the multiplication on }R/I\right)  ,
\]
so that%
\begin{align*}
f^{\prime}\left(  xy\right)   &  =f^{\prime}\left(  \overline{ab}\right)
=f\left(  ab\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}f^{\prime}\right)  \\
&  =f\left(  a\right)  \cdot f\left(  b\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }f\text{ is a ring morphism}\right)
\end{align*}
and%
\[
f^{\prime}\left(  \underbrace{x}_{=\overline{a}}\right)  \cdot f^{\prime
}\left(  \underbrace{y}_{=\overline{b}}\right)  =\underbrace{f^{\prime}\left(
\overline{a}\right)  }_{\substack{=f\left(  a\right)  \\\text{(by
the}\\\text{definition of }f^{\prime}\text{)}}}\cdot\underbrace{f^{\prime
}\left(  \overline{b}\right)  }_{\substack{=f\left(  b\right)  \\\text{(by
the}\\\text{definition of }f^{\prime}\text{)}}}=f\left(  a\right)  \cdot
f\left(  b\right)  .
\]
Comparing the last two equalities, we find $f^{\prime}\left(  xy\right)
=f^{\prime}\left(  x\right)  \cdot f^{\prime}\left(  y\right)  $. Thus, we
have shown that $f^{\prime}$ respects multiplication. As we said, the other
axioms can be checked in the same way, so we have shown that $f^{\prime}$ is a
ring morphism. This concludes the proof of Theorem \ref{thm.quotring.uniprop1}.
\end{proof}

So we have proved the universal property of quotient rings.

For various reasons, it is helpful to have an alternative formulation of this
property, which does not refer to specific elements but instead
\textquotedblleft implicitly\textquotedblright\ describes the morphism
$f^{\prime}$ by an equality:

\begin{theorem}
[Universal property of quotient rings, abstract form]%
\label{thm.quotring.uniprop2}Let $R$ be a ring. Let $I$ be an ideal of $R$.
Let $\pi:R\rightarrow R/I$ be the canonical projection (sending each $r\in R$
to its residue class $\overline{r}$).

Let $S$ be a ring. Let $f:R\rightarrow S$ be a ring morphism. Assume that
$f\left(  I\right)  =0$ (that is, $f\left(  i\right)  =0$ for each $i\in I$).
Then, there is a unique ring morphism $f^{\prime}:R/I\rightarrow S$ that
satisfies%
\[
f=f^{\prime}\circ\pi.
\]

\end{theorem}

\begin{proof}
Theorem \ref{thm.quotring.uniprop1} shows that there is a unique ring morphism
$f^{\prime}:R/I\rightarrow S$ that satisfies
\begin{equation}
f^{\prime}\left(  \overline{r}\right)  =f\left(  r\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }r\in R.
\label{pf.thm.quotring.uniprop2.old-cond}%
\end{equation}


We shall now prove that the equality $f=f^{\prime}\circ\pi$ is just an
equivalent restatement of the condition
(\ref{pf.thm.quotring.uniprop2.old-cond}).

Indeed, we have the following chain of equivalences:%
\begin{align*}
&  \ \left(  f=f^{\prime}\circ\pi\right) \\
&  \Longleftrightarrow\ \left(  f\left(  r\right)  =\left(  f^{\prime}\circ
\pi\right)  \left(  r\right)  \text{ for all }r\in R\right)
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since two maps are equal}\\
\text{if and only if they}\\
\text{agree on each input}%
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  f\left(  r\right)  =f^{\prime}\left(
\pi\left(  r\right)  \right)  \text{ for all }r\in R\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  f^{\prime}\circ\pi\right)
\left(  r\right)  =f^{\prime}\left(  \pi\left(  r\right)  \right)  \text{ for
each }r\right) \\
&  \Longleftrightarrow\ \left(  f\left(  r\right)  =f^{\prime}\left(
\overline{r}\right)  \text{ for all }r\in R\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\left(  r\right)  =\overline
{r}\text{ for each }r\right) \\
&  \Longleftrightarrow\ \left(  f^{\prime}\left(  \overline{r}\right)
=f\left(  r\right)  \text{ for all }r\in R\right)  .
\end{align*}
In other words, the equality $f=f^{\prime}\circ\pi$ is equivalent to the
condition (\ref{pf.thm.quotring.uniprop2.old-cond}).

Now, recall that there is a unique ring morphism $f^{\prime}:R/I\rightarrow S$
that satisfies the condition (\ref{pf.thm.quotring.uniprop2.old-cond}). In
view of the previous sentence, we can reformulate this as follows: There is a
unique ring morphism $f^{\prime}:R/I\rightarrow S$ that satisfies
$f=f^{\prime}\circ\pi$. This proves Theorem \ref{thm.quotring.uniprop2}.
\end{proof}

The equality $f=f^{\prime}\circ\pi$ in Theorem \ref{thm.quotring.uniprop2} can
be restated in yet another way: Namely, it is restated as \textquotedblleft
the diagram%
\[%
%TCIMACRO{\TeXButton{TeX field}{\xymatrix@C=4pc{
%R \ar[d]_{\pi} \ar[dr]^f \\
%R/I \ar@{-->}[r]_{f'} & S
%}}}%
%BeginExpansion
\xymatrix@C=4pc{
R \ar[d]_{\pi} \ar[dr]^f \\
R/I \ar@{-->}[r]_{f'} & S
}%
%EndExpansion
\]
commutes\textquotedblright. In general, a \textbf{diagram} is a bunch of sets
(drawn as nodes) and a bunch of maps between these sets (drawn as arrows
between these nodes). In our case, the sets are the rings $R,\ R/I,\ S$ and
the maps are the ring morphisms $f,\ \pi,\ f^{\prime}$. We say that a diagram
\textbf{commutes} (or \textbf{is commutative}) if, for any two nodes in the
diagram, all ways of going from the first node to the second (walking along
the arrows) result in the same composed map. In our case, there are two ways
to go from $R$ to $S$: one direct way, which gives the map $f$, and one
indirect way (via $R/I$), which gives the map $f^{\prime}\circ\pi$. Thus, the
commutativity of the above diagram is saying that $f=f^{\prime}\circ\pi$ (and
this is all that it is saying, since there are no two other ways that go
between the same two nodes).

In general, diagrams are a good way to visualize situations in which there are
several maps going between the same sets. For example, here is a diagram that
shows the rings $\mathbb{Z}$, $\mathbb{Z}/12$, $\mathbb{Z}/6$, $\mathbb{Z}/4$,
$\mathbb{Z}/3$, $\mathbb{Z}/2$ and $\mathbb{Z}/1$ (the latter ring is trivial)
as well as various morphisms between them:
\[%
%TCIMACRO{\TeXButton{big diagram}{\xymatrix{
%& \ZZ\ar@/_1pc/[ddl] \ar@/_5pc/[dddl] \ar@/_13pc/[dddd] \ar[d] \ar@
%/^1pc/[dddr] \\
%& \ZZ/12 \ar[dl] \ar[d] \ar[ddr] \\
%\ZZ/4 \ar[d] & \ZZ/6 \ar[dl] \ar[dr] \ar[dd] \\
%\ZZ/2 \ar[dr] & & \ZZ/3 \ar[dl] \\
%& \ZZ/1
%}}}%
%BeginExpansion
\xymatrix{
& \ZZ\ar@/_1pc/[ddl] \ar@/_5pc/[dddl] \ar@/_13pc/[dddd] \ar[d] \ar@
/^1pc/[dddr] \\
& \ZZ/12 \ar[dl] \ar[d] \ar[ddr] \\
\ZZ/4 \ar[d] & \ZZ/6 \ar[dl] \ar[dr] \ar[dd] \\
\ZZ/2 \ar[dr] & & \ZZ/3 \ar[dl] \\
& \ZZ/1
}%
%EndExpansion
\ \ .
\]
In this diagram, all arrows coming out of the $\mathbb{Z}$ node are canonical
projections $\mathbb{Z}\rightarrow\mathbb{Z}/n$ (sending each $r\in\mathbb{Z}$
to $\overline{r}\in\mathbb{Z}/n$), whereas all the other arrows are instances
of the morphisms (\ref{eq.exa.quotring.Z/n-to-Z/m}) constructed above. Note
that we have not drawn all possible morphisms (e.g., the morphism
$\mathbb{Z}/4\rightarrow\mathbb{Z}/1$ is missing) to avoid crowding the
diagram. This diagram commutes, as you can easily see directly by convincing
yourself that each of the arrows sends each residue class $\overline{r}$ (or,
in the case of $\mathbb{Z}$, each integer $r$) to the corresponding residue
class $\overline{r}$ modulo the respective number.

We do not have any particularly strong need for diagrams in this course, but
they can be helpful as visual aids. More advanced courses such as homological
algebra and category theory involve diagrams in far more substantial ways.

\subsubsection{Injectivity means zero kernel}

Let us take a break from abstractions and notations, and prove a simple lemma
about injectivity of ring morphisms:

\begin{lemma}
\label{lem.ringmor.inj}Let $R$ and $S$ be two rings. Let $f:R\rightarrow S$ be
a ring morphism. Then, $f$ is injective if and only if $\operatorname*{Ker}%
f=\left\{  0_{R}\right\}  $.
\end{lemma}

\begin{proof}
You may have seen analogous results about groups or vector spaces; the proof
is essentially the same. Nevertheless, let me prove Lemma
\ref{lem.ringmor.inj} from scratch:

$\Longrightarrow:$ If $f$ is injective, then
\begin{align*}
\operatorname*{Ker}f  &  =\left\{  r\in R\ \mid\ f\left(  r\right)
=0_{S}\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of a
kernel}\right) \\
&  =\left\{  r\in R\ \mid\ f\left(  r\right)  =f\left(  0_{R}\right)
\right\}  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }0_{S}=f\left(  0_{R}\right) \\
\text{(because }f\text{ is a ring morphism)}%
\end{array}
\right) \\
&  =\left\{  r\in R\ \mid\ r=0_{R}\right\}  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }f\text{ is injective, and thus }\ f\left(  r\right)  =f\left(
0_{R}\right) \\
\text{can only happen if }r=0_{R}%
\end{array}
\right) \\
&  =\left\{  0_{R}\right\}  .
\end{align*}


$\Longleftarrow:$ Assume that $\operatorname*{Ker}f=\left\{  0_{R}\right\}  $.
We must show that $f$ is injective. In other words, we must show that any
$a,b\in R$ that have the same value under $f$ (that is, satisfy $f\left(
a\right)  =f\left(  b\right)  $) must be equal (that is, $a=b$).

So let $a,b\in R$ be two elements satisfying $f\left(  a\right)  =f\left(
b\right)  $. We must prove that $a=b$.

The map $f$ is a ring morphism, hence respects differences. Thus
\[
f\left(  a-b\right)  =f\left(  a\right)  -f\left(  b\right)  =0_{S}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\left(  a\right)  =f\left(
b\right)  \right)  .
\]
In other words, $a-b\in\operatorname*{Ker}f=\left\{  0_{R}\right\}  $, which
means that $a-b=0_{R}$. In other words, $a=b$. This proves that $f$ is injective.
\end{proof}


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