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\begin{document}
\section*{Math 332 Winter 2023, Lecture 9: 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}{2}

\setcounter{theo}{4}

\subsubsection{More examples of quotient rings}

\textbf{Recall:} If $R$ is a ring, and $I$ is an ideal of $R$, then $R/I$ is
the \textbf{quotient ring} of $R$ by $I$.

Its elements are the \textbf{cosets} $r+I$, also called \textbf{residue
classes} and denoted by $\overline{r}$ (if $I$ is clear from the context).

A residue class $\overline{r}=r+I$ is (formally speaking) the set of all $r+i$
where $i$ ranges over $I$.

Two residue classes $\overline{r}$ and $\overline{s}$ are identical if and
only if $r-s\in I$. (Thus, passing from $R$ to $R/I$ amounts to
\textquotedblleft equating\textquotedblright\ any two elements that differ
only by an element of $I$.)

Addition and multiplication of residue classes are defined \textquotedblleft
by representatives\textquotedblright:%
\begin{align*}
\overline{a}+\overline{b} &  =\overline{a+b};\\
\overline{a}\cdot\overline{b} &  =\overline{ab}\ \ \ \ \ \ \ \ \ \ \text{for
all }a,b\in R.
\end{align*}


The most basic example of a quotient ring is $\mathbb{Z}/n\mathbb{Z}%
=\mathbb{Z}/n$. For instance, if $n=2$, then $\mathbb{Z}/n=\mathbb{Z}/2$ has
only two elements: $\overline{0}=\left\{  \text{all even integers}\right\}  $
and $\overline{1}=\left\{  \text{all odd integers}\right\}  $, and the rule
for addition in $\mathbb{Z}/2$ says (e.g.) that $\overline{1}+\overline
{1}=\overline{1+1}=\overline{2}=\overline{0}$ (or, in colloquial terms,
\textquotedblleft odd + odd = even\textquotedblright).

In Lecture 8, we saw a few examples of quotient rings. Here are two more:

\begin{itemize}
\item As we recall, if $R$ is a ring and $n\in\mathbb{N}$ an integer, then%
\begin{align*}
R^{n\leq n}  &  =\left\{  \text{all upper-triangular }n\times n\text{-matrices
with entries in }R\right\} \\
&  =\left\{  \left(
\begin{array}
[c]{cccc}%
a_{1,1} & a_{1,2} & \cdots & a_{1,n}\\
& a_{2,2} & \cdots & a_{2,n}\\
&  & \ddots & \vdots\\
&  &  & a_{n,n}%
\end{array}
\right)  \ \mid\ a_{i,j}\in R\text{ for all }i\leq j\right\}
\end{align*}
(where empty spaces stand for entries that are $0$) is a ring.

Let us consider the special case $R=\mathbb{Q}$ and $n=3$ for simplicity
(although everything we are doing can be generalized to arbitrary $R$ and
$n$). Thus,%
\[
R^{n\leq n}=\mathbb{Q}^{3\leq3}=\left\{  \left(
\begin{array}
[c]{ccc}%
a & b & c\\
0 & d & e\\
0 & 0 & f
\end{array}
\right)  \ \mid\ a,b,c,d,e,f\in\mathbb{Q}\right\}  .
\]


I claim that the subset%
\begin{align*}
\mathbb{Q}^{3<3}:=  &  \left\{  \text{all strictly upper-triangular }%
3\times3\text{-matrices with entries in }\mathbb{Q}\right\} \\
=  &  \left\{  \left(
\begin{array}
[c]{ccc}%
0 & b & c\\
0 & 0 & e\\
0 & 0 & 0
\end{array}
\right)  \ \mid\ b,c,e\in\mathbb{Q}\right\}
\end{align*}
is an ideal of $\mathbb{Q}^{3\leq3}$. To prove this, we need to show that it
satisfies the three ideal axioms. The first and the third axioms are obvious
(e.g., a sum of two strictly upper-triangular matrices is again strictly
upper-triangular). To prove the second ideal axiom, we need to show that if
$A\in\mathbb{Q}^{3<3}$ is strictly upper-triangular and $B\in\mathbb{Q}%
^{3\leq3}$ is just upper-triangular, then $AB$ and $BA$ are strictly
upper-triangular. This can just be checked by hand: Both products%
\begin{align*}
\left(
\begin{array}
[c]{ccc}%
0 & x & y\\
0 & 0 & z\\
0 & 0 & 0
\end{array}
\right)  \left(
\begin{array}
[c]{ccc}%
a & b & c\\
0 & d & e\\
0 & 0 & f
\end{array}
\right)   &  =\left(
\begin{array}
[c]{ccc}%
0 & dx & ex+fy\\
0 & 0 & fz\\
0 & 0 & 0
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\\
\left(
\begin{array}
[c]{ccc}%
a & b & c\\
0 & d & e\\
0 & 0 & f
\end{array}
\right)  \left(
\begin{array}
[c]{ccc}%
0 & x & y\\
0 & 0 & z\\
0 & 0 & 0
\end{array}
\right)   &  =\left(
\begin{array}
[c]{ccc}%
0 & ax & ay+bz\\
0 & 0 & dz\\
0 & 0 & 0
\end{array}
\right)
\end{align*}
are strictly upper-triangular.\footnote{The analogous claim holds for
arbitrary $R$ and $n$ (not just for $R=\mathbb{Q}$ and $n=3$), and can be
proved using the definition of a matrix product.} More generally, when you
multiply two upper-triangular matrices, the diagonal entries get multiplied
entrywise: e.g.,%
\[
\left(
\begin{array}
[c]{ccc}%
a & b & c\\
0 & d & e\\
0 & 0 & f
\end{array}
\right)  \left(
\begin{array}
[c]{ccc}%
a^{\prime} & b^{\prime} & c^{\prime}\\
0 & d^{\prime} & e^{\prime}\\
0 & 0 & f^{\prime}%
\end{array}
\right)  =\left(
\begin{array}
[c]{ccc}%
aa^{\prime} & bd^{\prime}+ab^{\prime} & be^{\prime}+cf^{\prime}+ac^{\prime}\\
0 & dd^{\prime} & de^{\prime}+ef^{\prime}\\
0 & 0 & ff^{\prime}%
\end{array}
\right)  .
\]
Thus, if one of the two matrices has a zero diagonal, then the product will
also have a zero diagonal. So we see again that $\mathbb{Q}^{3<3}$ is an ideal
of $\mathbb{Q}^{3\leq3}$.

What is the quotient ring $\mathbb{Q}^{3\leq3}/\mathbb{Q}^{3<3}$ ? Any element
of this quotient ring has the form%
\[
\overline{A}=A+\mathbb{Q}^{3<3}\ \ \ \ \ \ \ \ \ \ \text{for some }%
A\in\mathbb{Q}^{3\leq3}.
\]
Such a residue class $\overline{A}$ is a coset of $\mathbb{Q}^{3<3}$, so it
consists of all matrices that can be obtained from $A$ by adding a strictly
upper-triangular matrix. For example, if $A=\left(
\begin{array}
[c]{ccc}%
1 & 2 & 3\\
0 & 4 & 5\\
0 & 0 & 6
\end{array}
\right)  $, then%
\begin{align*}
\overline{A}  &  =A+\mathbb{Q}^{3<3}=\left(
\begin{array}
[c]{ccc}%
1 & 2 & 3\\
0 & 4 & 5\\
0 & 0 & 6
\end{array}
\right)  +\left\{  \left(
\begin{array}
[c]{ccc}%
0 & a & b\\
0 & 0 & c\\
0 & 0 & 0
\end{array}
\right)  \ \mid\ a,b,c\in\mathbb{Q}\right\} \\
&  =\left\{  \left(
\begin{array}
[c]{ccc}%
1 & 2 & 3\\
0 & 4 & 5\\
0 & 0 & 6
\end{array}
\right)  +\left(
\begin{array}
[c]{ccc}%
0 & a & b\\
0 & 0 & c\\
0 & 0 & 0
\end{array}
\right)  \ \mid\ a,b,c\in\mathbb{Q}\right\} \\
&  =\left\{  \left(
\begin{array}
[c]{ccc}%
1 & 2+a & 3+b\\
0 & 4 & 5+c\\
0 & 0 & 6
\end{array}
\right)  \ \mid\ a,b,c\in\mathbb{Q}\right\} \\
&  =\left\{  \left(
\begin{array}
[c]{ccc}%
1 & x & y\\
0 & 4 & z\\
0 & 0 & 6
\end{array}
\right)  \ \mid\ x,y,z\in\mathbb{Q}\right\}  ,
\end{align*}
We shall thus denote this coset by%
\[
\left(
\begin{array}
[c]{ccc}%
1 & \mathbb{Q} & \mathbb{Q}\\
0 & 4 & \mathbb{Q}\\
0 & 0 & 6
\end{array}
\right)  ,
\]
where the $\mathbb{Q}$'s mean \textquotedblleft you can put an arbitrary
element of $\mathbb{Q}$ here\textquotedblright. So you can think of
$\overline{A}$ as a \textquotedblleft matrix\textquotedblright\ in which the
three entries above the diagonal are undetermined. (Formally, this is a set of matrices.)

The rules for adding and multiplying such \textquotedblleft partly determined
matrices\textquotedblright\ are just as you would expect: e.g.,%
\[
\left(
\begin{array}
[c]{ccc}%
a & \mathbb{Q} & \mathbb{Q}\\
0 & b & \mathbb{Q}\\
0 & 0 & c
\end{array}
\right)  \left(
\begin{array}
[c]{ccc}%
d & \mathbb{Q} & \mathbb{Q}\\
0 & e & \mathbb{Q}\\
0 & 0 & f
\end{array}
\right)  =\left(
\begin{array}
[c]{ccc}%
ad & \mathbb{Q} & \mathbb{Q}\\
0 & be & \mathbb{Q}\\
0 & 0 & cf
\end{array}
\right)  .
\]


You might observe that the undetermined $\mathbb{Q}$-entries are just
\textquotedblleft acting like zeroes\textquotedblright\ here: The formula we
just saw looks exactly like the formula%
\[
\left(
\begin{array}
[c]{ccc}%
a & 0 & 0\\
0 & b & 0\\
0 & 0 & c
\end{array}
\right)  \left(
\begin{array}
[c]{ccc}%
d & 0 & 0\\
0 & e & 0\\
0 & 0 & f
\end{array}
\right)  =\left(
\begin{array}
[c]{ccc}%
ad & 0 & 0\\
0 & be & 0\\
0 & 0 & cf
\end{array}
\right)
\]
for multiplying diagonal matrices, except that we have $\mathbb{Q}$'s instead
of $0$'s above the diagonal. The same holds for addition instead of
multiplication. Thus, the residue classes in $\mathbb{Q}^{3\leq3}%
/\mathbb{Q}^{3<3}$ are behaving like diagonal matrices. In essence, we are
saying that%
\[
\mathbb{Q}^{3\leq3}/\mathbb{Q}^{3<3}\cong\mathbb{Q}^{3=3},
\]
where $\mathbb{Q}^{3=3}$ is the ring of diagonal $3\times3$-matrices (yes,
they do form a ring -- to be precise, a subring of $\mathbb{Q}^{3\times3}$).
To be more specific: The map%
\begin{align*}
\mathbb{Q}^{3\leq3}/\mathbb{Q}^{3<3}  &  \rightarrow\mathbb{Q}^{3=3},\\
\overline{\left(
\begin{array}
[c]{ccc}%
a & b & c\\
0 & d & e\\
0 & 0 & f
\end{array}
\right)  }  &  \mapsto\left(
\begin{array}
[c]{ccc}%
a & 0 & 0\\
0 & d & 0\\
0 & 0 & f
\end{array}
\right)
\end{align*}
is a ring isomorphism. (To prove this, you need to show that this map is
well-defined, is a ring morphism, and is invertible.)

This means that we have been wasting time defining the quotient ring
$\mathbb{Q}^{3\leq3}/\mathbb{Q}^{3<3}$. After all, we want to find something
new, not to recover an isomorphic copy of a known ring like $\mathbb{Q}^{3=3}$ !

\item Let us try again. Again consider the ring $\mathbb{Q}^{3\leq3}$ of
upper-triangular $3\times3$-matrices over $\mathbb{Q}$. Now, consider the set%
\[
\mathbb{Q}^{3<<3}:=\left\{  \left(
\begin{array}
[c]{ccc}%
0 & 0 & y\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}
\right)  \ \mid\ y\in\mathbb{Q}\right\}  .
\]
This set $\mathbb{Q}^{3<<3}$ is again an ideal of $\mathbb{Q}^{3\leq3}$,
because%
\begin{align*}
\left(
\begin{array}
[c]{ccc}%
a & b & c\\
0 & d & e\\
0 & 0 & f
\end{array}
\right)  \left(
\begin{array}
[c]{ccc}%
0 & 0 & y\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}
\right)   &  =\left(
\begin{array}
[c]{ccc}%
0 & 0 & ay\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\\
\left(
\begin{array}
[c]{ccc}%
0 & 0 & y\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}
\right)  \left(
\begin{array}
[c]{ccc}%
a & b & c\\
0 & d & e\\
0 & 0 & f
\end{array}
\right)   &  =\left(
\begin{array}
[c]{ccc}%
0 & 0 & fy\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}
\right)
\end{align*}
(the other ideal axioms are again easy).

What is the quotient ring $\mathbb{Q}^{3\leq3}/\mathbb{Q}^{3<<3}$ ? A residue
class $\overline{A}=A+\mathbb{Q}^{3<<3}$ in this quotient ring looks as
follows:%
\begin{align*}
\overline{A}  &  =\left(
\begin{array}
[c]{ccc}%
a & b & c\\
0 & d & e\\
0 & 0 & f
\end{array}
\right)  +\mathbb{Q}^{3<<3}\\
&  =\left(
\begin{array}
[c]{ccc}%
a & b & c\\
0 & d & e\\
0 & 0 & f
\end{array}
\right)  +\left\{  \left(
\begin{array}
[c]{ccc}%
0 & 0 & y\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}
\right)  \ \mid\ y\in\mathbb{Q}\right\} \\
&  =\left\{  \left(
\begin{array}
[c]{ccc}%
a & b & c\\
0 & d & e\\
0 & 0 & f
\end{array}
\right)  +\left(
\begin{array}
[c]{ccc}%
0 & 0 & y\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}
\right)  \ \mid\ y\in\mathbb{Q}\right\} \\
&  =\left\{  \left(
\begin{array}
[c]{ccc}%
a & b & c+y\\
0 & d & e\\
0 & 0 & f
\end{array}
\right)  \ \mid\ y\in\mathbb{Q}\right\} \\
&  =\left\{  \left(
\begin{array}
[c]{ccc}%
a & b & z\\
0 & d & e\\
0 & 0 & f
\end{array}
\right)  \ \mid\ z\in\mathbb{Q}\right\} \\
&  =\left(
\begin{array}
[c]{ccc}%
a & b & \mathbb{Q}\\
0 & d & e\\
0 & 0 & f
\end{array}
\right)  .
\end{align*}
This is again a matrix \textquotedblleft with an undetermined
entry\textquotedblright, but this time it has $8$ determined entries
(including the zeroes under the diagonal) and only one undetermined entry. In
particular, it is no longer \textquotedblleft just a diagonal matrix in
disguise\textquotedblright. Here is how we multiply such \textquotedblleft
partly determined matrices\textquotedblright:%
\[
\left(
\begin{array}
[c]{ccc}%
a & b & \mathbb{Q}\\
0 & d & e\\
0 & 0 & f
\end{array}
\right)  \cdot\left(
\begin{array}
[c]{ccc}%
a^{\prime} & b^{\prime} & \mathbb{Q}\\
0 & d^{\prime} & e^{\prime}\\
0 & 0 & f^{\prime}%
\end{array}
\right)  =\left(
\begin{array}
[c]{ccc}%
aa^{\prime} & bd^{\prime}+ab^{\prime} & \mathbb{Q}\\
0 & dd^{\prime} & f^{\prime}e+de^{\prime}\\
0 & 0 & ff^{\prime}%
\end{array}
\right)  .
\]
For comparison, if we had zeroes instead of undetermined $\mathbb{Q}$-entries,
multiplication would look as follows:%
\[
\left(
\begin{array}
[c]{ccc}%
a & b & 0\\
0 & d & e\\
0 & 0 & f
\end{array}
\right)  \cdot\left(
\begin{array}
[c]{ccc}%
a^{\prime} & b^{\prime} & 0\\
0 & d^{\prime} & e^{\prime}\\
0 & 0 & f^{\prime}%
\end{array}
\right)  =\left(
\begin{array}
[c]{ccc}%
aa^{\prime} & bd^{\prime}+ab^{\prime} & be^{\prime}\\
0 & dd^{\prime} & f^{\prime}e+de^{\prime}\\
0 & 0 & ff^{\prime}%
\end{array}
\right)  .
\]
Note the $be^{\prime}$ entry (which, in general, is not $0$) in the top right
cell! Thus, the set of all matrices of the form $\left(
\begin{array}
[c]{ccc}%
\ast & \ast & 0\\
0 & \ast & \ast\\
0 & 0 & \ast
\end{array}
\right)  $ (where a $\ast$ stands for an arbitrary entry) is not a subring of
$\mathbb{Q}^{3\leq3}$. Thus, unlike in our previous example, our
\textquotedblleft partly determined matrices\textquotedblright\ cannot be
emulated by regular matrices by putting $0$'s in the positions of the
undetermined $\mathbb{Q}$-entries. Thus, $\mathbb{Q}^{3\leq3}/\mathbb{Q}%
^{3<<3}$ is not isomorphic to a subring of $\mathbb{Q}^{3\leq3}$ (at least not
in an obvious way). So we have obtained a genuinely new ring this time.
\end{itemize}

\subsubsection{The canonical projection}

Theorem 1.8.4 in Lecture 7 says that the kernel of any ring morphism is an
ideal of its domain. Now we will prove the converse: Any ideal is a kernel.
Better yet:

\begin{theorem}
\label{thm.quotring.pi}Let $R$ be a ring. Let $I$ be an ideal of $R$. Consider
the map%
\begin{align*}
\pi:R  &  \rightarrow R/I,\\
r  &  \mapsto r+I=\overline{r}.
\end{align*}
This map $\pi$ is a surjective ring morphism with kernel $I$.
\end{theorem}

This morphism $\pi$ is called the \textbf{canonical projection} from $R$ onto
$R/I$.

\begin{proof}
[Proof of Theorem \ref{thm.quotring.pi}.]To prove that $\pi$ is a ring
morphism, we need to check that $\pi$ respects addition, multiplication, zero
and unity. This is all straightforward. For example, $\pi$ respects
multiplication because all $a,b\in R$ satisfy%
\begin{align*}
\underbrace{\pi\left(  a\right)  }_{=\overline{a}}\cdot\underbrace{\pi\left(
b\right)  }_{=\overline{b}}  &  =\overline{a}\cdot\overline{b}=\overline
{ab}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of multiplication on
}R/I\right) \\
&  =\pi\left(  ab\right)  .
\end{align*}
Thus, $\pi$ is a ring morphism.

Why is $\pi$ surjective? Because (by definition) each element of $R/I$ is a
residue class, i.e., has the form $\overline{r}$ for some $r\in R$.

Finally, $\pi$ has kernel $I$, since%
\begin{align*}
\operatorname*{Ker}\pi &  =\left\{  r\in R\ \mid\ \pi\left(  r\right)
=0_{R/I}\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of a
kernel}\right) \\
&  =\left\{  r\in R\ \mid\ \pi\left(  r\right)  =\overline{0}\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }0_{R/I}=\overline{0}\right) \\
&  =\left\{  r\in R\ \mid\ \overline{r}=\overline{0}\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\left(  r\right)  =\overline
{r}\text{ by definition of }\pi\right) \\
&  =\left\{  r\in R\ \mid\ r-0\in I\right\} \\
&  =\left\{  r\in R\ \mid\ r\in I\right\}  =I.
\end{align*}

\end{proof}

For example, if we take $R=\mathbb{Z}$ and $I=2\mathbb{Z}$ in Theorem
\ref{thm.quotring.pi}, then the canonical projection $\pi$ is the map%
\begin{align*}
\pi:\mathbb{Z}  &  \rightarrow\mathbb{Z}/2,\\
r  &  \mapsto r+2\mathbb{Z}=\overline{r}.
\end{align*}
This map $\pi$ sends each even integer to $\overline{0}$ and each odd integer
to $\overline{1}$. In other words, $\pi$ assigns to each integer its parity
(as an element of $\mathbb{Z}/2$).


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