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\begin{document}
\section*{Math 332 Winter 2023, Lecture 18: Modules}

\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}}}

\setcounter{section}{1}

\section{Modules}

So far, we have been studying rings. Now we shall move on to studying modules
over these rings.

Roughly speaking, a ring is a system of \textquotedblleft number-like
objects\textquotedblright\ that can be \textquotedblleft
added\textquotedblright\ (with one another) and \textquotedblleft
multiplied\textquotedblright\ (with one another).

In contrast, a \textbf{module} (over a given ring $R$) is a system of
\textquotedblleft vector-like objects\textquotedblright\ that can be
\textquotedblleft added\textquotedblright\ (with one another) and
\textquotedblleft scaled\textquotedblright\ (by elements of $R$). If this
sounds familiar to you, then you are right: Modules generalize vector spaces.
In many ways, modules are simpler and better-behaved than rings; but of
course, they rely on rings, whence we have had to introduce rings before
introducing modules.

\subsection{Definitions and examples}

\begin{convention}
We shall fix a ring $R$ for the rest of this section.
\end{convention}

\subsubsection{Defining modules}

There are two notions of an \textquotedblleft$R$-module\textquotedblright: the
\textquotedblleft left $R$-modules\textquotedblright\ and the
\textquotedblleft right $R$-modules\textquotedblright. Let us define the left ones:

\begin{definition}
\label{def.mod.leftmod}Let $R$ be a ring. A \textbf{left }$R$\textbf{-module}
(or a \textbf{left module over }$R$) means a set $M$ equipped with

\begin{itemize}
\item a binary operation $+$ (that is, a map from $M\times M$ to $M$) that is
called \textbf{addition};

\item an element $0_{M}$ of $M$ that is called the \textbf{zero element} or
the \textbf{zero vector} or just the \textbf{zero}, and will be just denoted
by $0$ if there is no danger of confusion;

\item a map from $R\times M$ to $M$ that is called the \textbf{action of }$R$
\textbf{on }$M$, and is written as multiplication (i.e., we denote the image
of a pair $\left(  r,m\right)  \in R\times M$ under this map by $rm$ or
$r\cdot m$)
\end{itemize}

\noindent such that the following properties (the \textquotedblleft%
\textbf{module axioms}\textquotedblright) hold:

\begin{itemize}
\item $\left(  M,+,0\right)  $ is an abelian group.

\item The \textbf{right distributivity law} holds: We have $\left(
r+s\right)  m=rm+sm$ for all $r,s\in R$ and $m\in M$.

\item The \textbf{left distributivity law} holds: We have $r\left(
m+n\right)  =rm+rn$ for all $r\in R$ and $m,n\in M$.

\item The \textbf{associativity law} holds: We have $\left(  rs\right)
m=r\left(  sm\right)  $ for all $r,s\in R$ and $m\in M$.

\item We have $0_{R}m=0_{M}$ for every $m\in M$.

\item We have $r\cdot0_{M}=0_{M}$ for every $r\in R$.

\item We have $1m=m$ for every $m\in M$.
\end{itemize}

When $M$ is a left $R$-module, the elements of $M$ are called \textbf{vectors}%
, and the elements of $R$ are called \textbf{scalars}.
\end{definition}

As the name \textquotedblleft left $R$-module\textquotedblright\ suggests,
there is an analogous notion of a \textbf{right }$R$\textbf{-module}. In this
latter notion, the action is not a map from $R\times M$ to $M$ but a map from
$M\times R$ to $M$, and we accordingly use the notation $mr$ rather than $rm$
for its values. The associativity law for right $R$-modules requires that
$m\left(  rs\right)  =\left(  mr\right)  s$ for all $r,s\in R$ and $m\in M$.

When the ring $R$ is commutative, any left $R$-module $M$ becomes a right
$R$-module by setting%
\[
mr=rm\ \ \ \ \ \ \ \ \ \ \text{for all }r\in R\text{ and }m\in M,
\]
and conversely, any right $R$-module $M$ becomes a left $R$-module by setting
\[
rm=mr\ \ \ \ \ \ \ \ \ \ \text{for all }r\in R\text{ and }m\in M.
\]
In other words, when the ring $R$ is commutative, we can \textquotedblleft
translate\textquotedblright\ any left action\footnote{By a \textquotedblleft
left action\textquotedblright, I mean an action of the form $R\times
M\rightarrow M$, as in the definition of a left $R$-module. Correspondingly, a
\textquotedblleft right action\textquotedblright\ means an action of the form
$M\times R\rightarrow M$, as in the definition of a right $R$-module.} of $R$
on $M$ into a right action, and vice versa.

This mutual \textquotedblleft translatability\textquotedblright\ allows us to
treat left $R$-modules and right $R$-modules as the same thing when $R$ is a
commutative ring. Thus, when $R$ is commutative, we will just speak of
\textquotedblleft$R$\textbf{-modules}\textquotedblright\ and view them as left
or right as we please.

However, if $R$ is not commutative, then our \textquotedblleft
translation\textquotedblright\ between left and right $R$-modules will usually
destroy associativity: For example, if $M$ is a left $R$-module, then its
associativity law says that $\left(  rs\right)  m=r\left(  sm\right)  $ for
all $r,s\in R$ and $m\in M$. If we translate its left action into a right
action, then this equality becomes $m\left(  rs\right)  =\left(  ms\right)
r$, which is \textbf{not} the associativity law for a right $R$-module. There
is a way to salvage this using the \textbf{opposite ring }%
$R^{\operatorname*{op}}$ of $R$; this is the same ring as $R$ but with the
order of multiplication swapped (i.e., what is $rs$ in $R$ becomes $sr$ in
$R^{\operatorname*{op}}$). Then, a left $R$-module becomes a right
$R^{\operatorname*{op}}$-module, and vice versa. (See \cite[homework set \#2,
Exercise 2 \textbf{(d)}]{21w} for the details.) Thus, the notion of a left
$R$-module is not equivalent to the notion of a right $R$-module for a
\textbf{given} noncommutative ring $R$, but the general notion of a
\textquotedblleft left module over a ring\textquotedblright\ is equivalent to
the general notion of a \textquotedblleft right module over a
ring\textquotedblright.

This allows us to focus on left $R$-modules and sleep well knowing that
everything we prove about them has analogue for right $R$-modules (and the
latter can be proved in the same way as the former).

\bigskip

When $R$ is a field, the $R$-modules are known as the $R$\textbf{-vector
spaces}. These are precisely the vector spaces you have seen in an advanced
linear algebra class. Vector spaces have a fairly rigid structure: In
particular, a vector space is uniquely determined (up to isomorphism) by its
dimension. In contrast, modules can be rather wild (although the
\textquotedblleft nice\textquotedblright\ families of modules, such as $R^{n}$
for all $n\in\mathbb{N}$, still exist for every ring $R$). The wilder the ring
is, the more diverse are its modules.

\bigskip

One more remark about Definition \ref{def.mod.leftmod}: The \textquotedblleft%
$0_{R}m=0_{M}$\textquotedblright\ and \textquotedblleft$r\cdot0_{M}=0_{M}%
$\textquotedblright\ axioms are redundant (i.e., follow from the other
axioms). Do you see why?

\bigskip

Another piece of terminology:

\begin{definition}
Let $M$ be a left $R$-module, and let $r\in R$ be a scalar. Then, the map%
\begin{align*}
M  &  \rightarrow M,\\
m  &  \mapsto rm
\end{align*}
is called \textbf{scaling} by $r$. This map is a group homomorphism from the
additive group $\left(  M,+,0\right)  $ to itself. For instance, scaling by
$1$ is the identity map on $M$, whereas scaling by $0$ sends every vector in
$M$ to the zero vector.
\end{definition}

\subsubsection{Defining submodules}

We will soon see some examples of $R$-modules; but let us first define
$R$-submodules. This is just the natural generalization of vector subspaces:

\begin{definition}
\label{def.mod.submod}Let $M$ be a left $R$-module. An $R$\textbf{-submodule}
(or, to be more precise, a \textbf{left }$R$\textbf{-submodule}) of $M$ means
a subset $N$ of $M$ such that

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

\item we have $ra\in N$ for all $r\in R$ and $a\in N$ (that is, $N$ is
\textbf{closed under scaling});

\item we have $0_{M}\in N$ (that is, $N$ \textbf{contains zero}).
\end{itemize}
\end{definition}

In \S 2.2.1 (in Lecture 19), we will see that any $R$-submodule of a left
$R$-module $M$ is also closed under negation (i.e., under taking additive
inverses), and thus becomes a left $R$-module in its own
right.\footnote{Proving this is actually an easy exercise you can solve right
now.} Of course, all of this applies likewise to right $R$-modules.

\subsubsection{Examples}

Some examples:

\begin{itemize}
\item Let $R$ be a ring. Then, $R$ itself becomes a left $R$-module: Just
define the action to be the multiplication of $R$. Thus, the elements of $R$
serve as both vectors and scalars. Scaling a vector $m$ by a scalar $r$ just
means multiplying $m$ by $r$ (that is, taking the product $rm$ inside $R$).

The $R$-submodules of this left $R$-module $R$ are the subsets $L$ of $R$ that
are closed under addition and contain $0$ and satisfy $ra\in L$ for all $r\in
R$ and $a\in L$. These subsets $L$ are called the \textbf{left ideals} of $R$.
They differ from the ideals of $R$ in that we only require $ra\in L$, not
$ar\in L$. Correspondingly, many rings have more left ideals than ideals. For
example: If $R$ is the matrix ring $\mathbb{Q}^{2\times2}$, then $R$ has only
two ideals ($\left\{  0_{R}\right\}  $ and $R$) but plenty of left ideals
(e.g., the set $\left\{  \left(
\begin{array}
[c]{cc}%
0 & a\\
0 & b
\end{array}
\right)  \ \mid\ a,b\in\mathbb{Q}\right\}  $).

When $R$ is commutative, the left ideals of $R$ are precisely the ideals of
$R$, so the notion of $R$-submodules then becomes a generalization of ideals.

\item Let $R$ be any ring, and let $n\in\mathbb{N}$. Then,%
\[
R^{n}=\left\{  \left(  a_{1},a_{2},\ldots,a_{n}\right)  \ \mid\ \text{all
}a_{i}\text{ belong to }R\right\}
\]
is a left $R$-module, where addition and action are defined entrywise, i.e.,
by setting%
\[
\left(  a_{1},a_{2},\ldots,a_{n}\right)  +\left(  b_{1},b_{2},\ldots
,b_{n}\right)  =\left(  a_{1}+b_{1},\ a_{2}+b_{2},\ \ldots,\ a_{n}%
+b_{n}\right)
\]
and%
\[
r\cdot\left(  a_{1},a_{2},\ldots,a_{n}\right)  =\left(  ra_{1},ra_{2}%
,\ldots,ra_{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{for }r\in R\right)
.
\]
The zero vector of this $R$-module is $\left(  0,0,\ldots,0\right)  $.

\item Let $R$ be any ring, and let $n,m\in\mathbb{N}$. Consider the set
$R^{n\times m}$ of all $n\times m$-matrices with entries in $R$. This set
$R^{n\times m}$ is not a ring unless $n=m$, but it always is a left
$R$-module, where addition and action are defined entrywise. For instance, the
action is given for $2\times2$-matrices by%
\[
r\cdot\left(
\begin{array}
[c]{cc}%
a & b\\
c & d
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
ra & rb\\
rc & rd
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{for }r\in R\right)  ,
\]
and similarly for matrices of all other sizes. The zero vector of $R^{n\times
m}$ is the zero matrix $0_{n\times m}$.

This set $R^{n\times m}$ is also a right $R$-module in a similarly obvious way.

According to Definition \ref{def.mod.leftmod}, this allows us to refer to the
matrices in $R^{n\times m}$ as \textquotedblleft vectors\textquotedblright.
This shows that our notion of vectors is much more general than just row
vectors and column vectors. Numbers are vectors; matrices are vectors;
everything that lives in a module is a vector. This general interpretation of
the word \textquotedblleft vector\textquotedblright\ will soon reveal its
usefulness, as it allows us to apply various linear-algebraic notions (such as
span and linear independence) to anything that looks, swims and quacks like a vector.

\item Just as we defined the left $R$-module $R^{n}$ (consisting of
$n$-tuples) for any $n\in\mathbb{N}$, we can define a left $R$-module
\textquotedblleft$R^{\infty}$\textquotedblright\ consisting of all infinite
sequences of elements of $R$. The proper name of this $R$-module is
$R^{\mathbb{N}}$ (since \textquotedblleft$\infty$\textquotedblright\ is
imprecise: there are many infinities in mathematics). Explicitly,
$R^{\mathbb{N}}$ is defined to be the left $R$-module
\[
\left\{  \left(  a_{0},a_{1},a_{2},\ldots\right)  \ \mid\ \text{all }%
a_{i}\text{ belong to }R\right\}  ,
\]
whose addition and action are defined entrywise.

This left $R$-module $R^{\mathbb{N}}$ has an $R$-submodule%
\[
R^{\left(  \mathbb{N}\right)  }:=\left\{  \left(  a_{0},a_{1},a_{2}%
,\ldots\right)  \in R^{\mathbb{N}}\ \mid\ \text{only finitely many }%
i\in\mathbb{N}\text{ satisfy }a_{i}\neq0\right\}  .
\]
You can check that this is indeed an $R$-submodule of $R^{\mathbb{N}}$. For
instance, for $R=\mathbb{Q}$, we have%
\begin{align*}
\left(  1,1,1,\ldots\right)   &  \in R^{\mathbb{N}}\setminus R^{\left(
\mathbb{N}\right)  };\\
\left(  0,0,0,\ldots\right)   &  \in R^{\left(  \mathbb{N}\right)  };\\
\left(  3,2,0,5,\underbrace{0,0,0,0,0,\ldots}_{\text{only zeroes here}%
}\right)   &  \in R^{\left(  \mathbb{N}\right)  };\\
\left(  \underbrace{1,0,1,0,1,0,\ldots}_{\text{ones and zeroes alternate}%
}\right)   &  \in R^{\mathbb{N}}\setminus R^{\left(  \mathbb{N}\right)  }.
\end{align*}

\end{itemize}

Note that the zero vector of an $R$-module is uniquely determined by its
addition, so we don't have to provide it in a definition.

\subsubsection{Direct products}

Fix a ring $R$.

Most of our above examples of $R$-modules involve tuples on which addition and
action work entrywise. There is a general concept for this:

\begin{definition}
Let $n\in\mathbb{N}$, and let $M_{1},M_{2},\ldots,M_{n}$ be any $n$ left
$R$-modules. Then, the Cartesian product $M_{1}\times M_{2}\times\cdots\times
M_{n}$ becomes a left $R$-module as well, where addition and action are
defined entrywise: e.g., the action is defined by%
\[
r\cdot\left(  m_{1},m_{2},\ldots,m_{n}\right)  =\left(  rm_{1},rm_{2}%
,\ldots,rm_{n}\right)  \text{ for all }r\in R\text{ and }m_{i}\in M_{i}.
\]
This left $R$-module $M_{1}\times M_{2}\times\cdots\times M_{n}$ is called the
\textbf{direct product} of $M_{1},M_{2},\ldots,M_{n}$.
\end{definition}

This notion of direct product can be generalized further to direct products of
arbitrarily (not necessarily finitely) many left $R$-modules. The resulting
products are called $\prod_{i\in I}M_{i}$. See \S 3.3.1 in the text for more
about those.

A particular case of direct products is of particular importance:

\begin{definition}
Let $M$ be any left $R$-module. Let $n\in\mathbb{N}$. Then, we set%
\[
M^{n}:=\underbrace{M\times M\times\cdots\times M}_{n\text{ times}}.
\]

\end{definition}

In particular, when $M$ is the natural left $R$-module $R$ (that is, the
$R$-module $R$ whose action is just multiplication), then $M^{n}$ is the left
$R$-module $R^{n}$ that was discussed above.

\subsubsection{Restriction of scalars}

Here are some more ways to construct modules:

\begin{itemize}
\item If $R$ is a subring of a ring $S$, then $S$ is a left $R$-module (where
the action of $R$ on $S$ is defined by restricting the multiplication map
$S\times S\rightarrow S$ to $R\times S$) and a right $R$-module (in a similar way).

Let me restate this in a more down-to-earth way: If $R$ is a subring of a ring
$S$, then we can multiply any element of $R$ with any element of $S$ (since
both elements lie in the ring $S$); this makes $S$ into a left $R$-module (and
likewise, $S$ becomes a right $R$-module). Explicitly, the action of $R$ on
the left $R$-module $S$ is given by
\[
rs=rs\ \ \ \ \ \ \ \ \ \ \text{for all }r\in R\text{ and }s\in S
\]
(where the \textquotedblleft$rs$\textquotedblright\ on the left hand side
means the image of $\left(  r,s\right)  $ under the action, whereas the
\textquotedblleft$rs$\textquotedblright\ on the right hand side means the
product of $r$ and $s$ in the ring $S$).

Thus, for example, $\mathbb{C}$ is an $\mathbb{R}$-module (since $\mathbb{R}$
is a subring of $\mathbb{C}$) and also a $\mathbb{Q}$-module (for similar
reasons). In a linear algebra class, you would say \textquotedblleft vector
space\textquotedblright\ instead of \textquotedblleft module\textquotedblright%
\ here, but this is the same thing.

\item More generally, if $R$ and $S$ are any two rings, and if $f:R\rightarrow
S$ is a ring morphism, then $S$ becomes a left $R$-module, with the action of
$R$ on $S$ being defined by%
\[
rs=f\left(  r\right)  \cdot s\ \ \ \ \ \ \ \ \ \ \text{for all }r\in R\text{
and }s\in S.
\]
In a similar way, $S$ becomes a right $R$-module (with action defined by
$sr=s\cdot f\left(  r\right)  $). This is easy to prove (the module axioms
follow from the fact that $f$ is a ring morphism). These $R$-module structures
on $S$ are said to be \textbf{induced} by the morphism $f$.

Our previous example (where $R$ is a subring of $S$) is the particular case of
this where $f$ is the canonical inclusion\footnote{Recall what
\textquotedblleft canonical inclusion\textquotedblright\ means:
\par
If $U$ is a subset of a set $V$, then the map%
\begin{align*}
U  &  \rightarrow V,\\
u  &  \mapsto u
\end{align*}
is called the \textbf{canonical inclusion} of $U$ into $V$.
\par
If $U$ is a subring of a ring $V$, then the canonical inclusion of $U$ into
$V$ is furthermore a ring morphism. (This follows trivially from the
definition of a subring.)} (i.e., the map $R\rightarrow S,\ r\mapsto r$).

Here are some other particular cases:

\begin{itemize}
\item Any quotient ring $R/I$ of a ring $R$ (by some ideal $I$) becomes a left
$R$-module, because the canonical projection $\pi:R\rightarrow R/I$ is a ring
morphism. Explicitly, the action of $R$ on $R/I$ is given by%
\[
r\cdot\overline{u}=\underbrace{\pi\left(  r\right)  }_{=\overline{r}}%
\cdot\,\overline{u}=\overline{r}\cdot\overline{u}=\overline{ru}%
\ \ \ \ \ \ \ \ \ \ \text{for any }r,u\in R.
\]


\item Another (less transparent) particular case: I claim that the abelian
group $\mathbb{Z}/5$ becomes a $\mathbb{Z}\left[  i\right]  $-module if we
define the action by%
\[
\left(  a+bi\right)  \cdot m=\overline{a+2b}\cdot
m\ \ \ \ \ \ \ \ \ \ \text{for all }a+bi\in\mathbb{Z}\left[  i\right]  \text{
and }m\in\mathbb{Z}/5.
\]
This $\mathbb{Z}\left[  i\right]  $-module structure is actually induced by
the ring morphism%
\begin{align*}
f:\mathbb{Z}\left[  i\right]   &  \rightarrow\mathbb{Z}/5,\\
a+bi  &  \mapsto\overline{a+2b},
\end{align*}
which is a ring morphism thanks to the fact that $\overline{2}^{2}%
=-\overline{1}$ in $\mathbb{Z}/5$ (check this!). There is also another
$\mathbb{Z}\left[  i\right]  $-module structure on $\mathbb{Z}/5$, given by%
\[
\left(  a+bi\right)  \cdot m=\overline{a+3b}\cdot
m\ \ \ \ \ \ \ \ \ \ \text{for all }a+bi\in\mathbb{Z}\left[  i\right]  \text{
and }m\in\mathbb{Z}/5.
\]


When we speak of the $\mathbb{Z}\left[  i\right]  $-module $\mathbb{Z}/5$, we
have to specify which of these two structures we are meaning. They are not the
same: One structure has $i\cdot\overline{1}=\overline{2}$; the other has
$i\cdot\overline{1}=\overline{3}$. So to speak, $\mathbb{Z}/5$ is
\textquotedblleft a $\mathbb{Z}\left[  i\right]  $-module in two different
ways\textquotedblright.
\end{itemize}

\item Even more generally: If $R$ and $S$ are two rings, and if
$f:R\rightarrow S$ is a ring morphism, then any left $S$-module $M$ (not just
$S$ itself) naturally becomes a left $R$-module, with the action defined by%
\[
rm=f\left(  r\right)  m\ \ \ \ \ \ \ \ \ \ \text{for all }r\in R\text{ and
}m\in M.
\]
(You can think of this as letting $R$ act on $M$ \textquotedblleft by
proxy\textquotedblright: In order to scale a vector $m\in M$ by a scalar $r\in
R$, you just scale it by the scalar $f\left(  r\right)  \in S$.)

This method of turning $S$-modules into $R$-modules is called
\textbf{restriction of scalars} (or, more specifically, \textbf{restricting}
an $S$-module to $R$ via $f$).

If we apply this method to a canonical inclusion (i.e., if $R$ is a subring of
$S$ and if $f:R\rightarrow S$ is the canonical inclusion), then we conclude
that any module over a ring naturally becomes a module over any
subring.\footnote{You can think of it as forgetting how to scale vectors by
scalars that don't belong to the subring.} For example, any $\mathbb{C}%
$-module naturally becomes an $\mathbb{R}$-module (this is known as
\textquotedblleft decomplexification\textquotedblright\ in linear algebra) and
a $\mathbb{Q}$-module and a $\mathbb{Z}$-module.
\end{itemize}

\begin{thebibliography}{999}                                                                                              %


\bibitem[21w]{21w}Darij Grinberg, \textit{Math 533: Abstract Algebra I, Winter
2021}.\newline\url{https://www.cip.ifi.lmu.de/~grinberg/t/21w/}
\end{thebibliography}


\end{document}