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\begin{document}
\section*{Math 332 Winter 2023, Lecture 23: 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}

\setcounter{subsection}{9}

\subsection{Defining algebras: the case of $\mathbb{H}$}

An $R$-algebra carries more information than a ring: namely, it has the extra
structure of an action. Thus, in order to define an $R$-algebra, it is natural
to start by defining a ring and then putting the action on it (and showing
that it satisfies the $R$-module axioms and scale-invariance).

Often, however, it is easier to proceed differently: First, define an
$R$-module, and then define the multiplication and the unity to turn it into
an $R$-algebra. If you do things in this order, you can use the $R$-module
structure as scaffolding for defining the multiplication.

Here is an example of how this can work:

Recall the ring $\mathbb{H}$ of Hamilton quaternions, which were
\textquotedblleft defined\textquotedblright\ (in \S 1.1.2 in Lecture 2) to be
\textquotedblleft numbers\textquotedblright\ of the form $a+bi+cj+dk$ with
$a,b,c,d\in\mathbb{R}$ and equipped with the multiplication rules%
\[
i^{2}=j^{2}=k^{2}%
=-1,\ \ \ \ \ \ \ \ \ \ ij=-ji=k,\ \ \ \ \ \ \ \ \ \ jk=-kj=i,\ \ \ \ \ \ \ \ \ \ ki=-ik=j.
\]
It is clear how to calculate in $\mathbb{H}$ using these rules. But why does
this ring $\mathbb{H}$ exist in the first place?

This is not a vacuous question. For instance, if we replace the rule
$k^{2}=-1$ by $k^{2}=1$ in the above definition (but still require $i^{2}$ and
$j^{2}$ to be $-1$), then we get%
\[
j^{2}\underbrace{k^{2}}_{=1}=j^{2}=-1
\]
and thus%
\[
-1=j^{2}k^{2}=j\underbrace{jk}_{=i}k=j\underbrace{ik}_{=-j}=j\left(
-j\right)  =-\underbrace{j^{2}}_{=-1}=-\left(  -1\right)  =1.
\]
Adding $1$ to this equality, we obtain $0=2$. Upon multiplication by $1/2$,
this becomes $0=1$, so that our new ring is actually trivial. Instead of
expanding our number system with new numbers, we have inadvertently collapsed
it to a trivial ring!

Thus, if we wantonly define rings by inventing new \textquotedblleft
numbers\textquotedblright\ and declaring new rules, then we cannot expect our
new rings to contain our initial rings (such as $\mathbb{R}$ in the above
example) as subrings; instead, they may end up trivial or otherwise smaller
than expected. This makes them rather useless.

This is similar to why you cannot divide by $0$: If you introduce a new
\textquotedblleft number\textquotedblright\ $\infty=\dfrac{1}{0}$, then
$0\cdot\infty$ equals both $0$ and $1$, so you get $0=1$ and thus your ring is trivial.

Of course, this cautionary tale does not always have to become reality. What
it shows is just that we need to be cautious: When creating new
\textquotedblleft numbers\textquotedblright, we must make sure we don't
accidentally collapse the old ones.

But how can we make this sure? For instance, how do we know that the ring
$\mathbb{H}$ we defined by the above rules actually contains $\mathbb{R}$ as a
subring (as opposed, e.g., to being trivial)?

One safe way of defining $\mathbb{H}$ is as follows: We define a quaternion to
be a $4$-tuple $\left(  a,b,c,d\right)  $ of real numbers (this $4$-tuple is
supposed to stand for $a+bi+cj+dk$), and we define addition, multiplication
and scaling of these $4$-tuples explicitly by the formulas%
\[
\left(  a,b,c,d\right)  +\left(  a^{\prime},b^{\prime},c^{\prime},d^{\prime
}\right)  =\left(  a+a^{\prime},\ b+b^{\prime},\ c+c^{\prime},\ d+d^{\prime
}\right)
\]
and%
\begin{align*}
\left(  a,b,c,d\right)  \left(  a^{\prime},b^{\prime},c^{\prime},d^{\prime
}\right)   &  =(\ aa^{\prime}-bb^{\prime}-cc^{\prime}-dd^{\prime},\\
&  \ \ \ ab^{\prime}+ba^{\prime}+cd^{\prime}-dc^{\prime},\\
&  \ \ \ ac^{\prime}-bd^{\prime}+ca^{\prime}+db^{\prime},\\
&  \ \ \ ad^{\prime}+bc^{\prime}-cb^{\prime}+da^{\prime}\ )
\end{align*}
and%
\[
r\left(  a,b,c,d\right)  =\left(  ra,rb,rc,rd\right)
\ \ \ \ \ \ \ \ \ \ \text{for }r\in\mathbb{R}.
\]
This is a valid definition, but you have to check that all the ring axioms
(and module axioms, and scale-invariance) hold. Associativity of
multiplication, in particular, is a lot of work. (The quaternions $\left(
a,b,c,d\right)  $ thus defined can be rewritten as $a+bi+cj+dk$ once we
observe that the unity of the ring is $\left(  1,0,0,0\right)  $ and we
furthermore set $i:=\left(  0,1,0,0\right)  $ and $j:=\left(  0,0,1,0\right)
$ and $k:=\left(  0,0,0,1\right)  $. Pedants will note that the ring of
quaternions thus defined does not literally contain $\mathbb{R}$ as a subring,
but merely contains a subring isomorphic to $\mathbb{R}$: namely, the subring
$\left\{  \left(  a,0,0,0\right)  \ \mid\ a\in\mathbb{R}\right\}  $. But this
is not a serious problem, and the same situation occurs when defining the
complex numbers.)

This definition of $\mathbb{H}$ does its job well, but as we just said, it is
laborious to justify and somewhat inflexible if one is looking to generalize
the construction.

A simpler and slicker way to define $\mathbb{H}$ proceeds as follows: First
define $\mathbb{H}$ as an $\mathbb{R}$-module (which is easy: it will just be
the free $\mathbb{R}$-module $\mathbb{R}^{4}$), and then build the
multiplication on top of it, using the notion of bilinearity. As we saw in
\S 2.9.1 (Lecture 22), the multiplication of an $\mathbb{R}$-algebra is always
$\mathbb{R}$-bilinear, and as we saw in \S 2.7, if we want to define an
$\mathbb{R}$-bilinear map $M\times N\rightarrow P$ where $M$ and $N$ are two
free $\mathbb{R}$-modules, then we can do this simply by specifying its values
on all pairs of basis elements (Theorem 2.7.2 in Lecture 22). Thus, instead of
defining the product of any two quaternions, we will only need to define the
product of two quaternions from a given basis (which we will take to be
$\left(  1,i,j,k\right)  $).

Let us do this. We define $\mathbb{H}$ to be the $\mathbb{R}$-module
$\mathbb{R}^{4}$, which is a free $\mathbb{R}$-module of rank $4$. Thus, a
quaternion is defined to be a $4$-tuple $\left(  a,b,c,d\right)  $ of real
numbers. The addition and the scaling of quaternions are thus entrywise (since
this is how $\mathbb{R}^{4}$ is defined).

We denote the standard basis $\left(  e_{1},e_{2},e_{3},e_{4}\right)  $ of
$\mathbb{H}$ by $\left(  \mathbf{e},\mathbf{i},\mathbf{j},\mathbf{k}\right)
$. The four basis vectors $\mathbf{e},\mathbf{i},\mathbf{j},\mathbf{k}$ here
will eventually be revealed (or renamed) as $1,i,j,k$, but for now I will use
boldface letters to avoid suggesting too much.

Now, we define the multiplication of $\mathbb{H}$ to be the $\mathbb{R}%
$-bilinear map $\mu:\mathbb{H}\times\mathbb{H}\rightarrow\mathbb{H}$ that
satisfies%
\begin{align*}
\mu\left(  \mathbf{e},\mathbf{e}\right)   &  =\mathbf{e}%
,\ \ \ \ \ \ \ \ \ \ \mu\left(  \mathbf{e},\mathbf{i}\right)  =\mathbf{i}%
,\ \ \ \ \ \ \ \ \ \ \mu\left(  \mathbf{e},\mathbf{j}\right)  =\mathbf{j}%
,\ \ \ \ \ \ \ \ \ \ \mu\left(  \mathbf{e},\mathbf{k}\right)  =\mathbf{k},\\
\mu\left(  \mathbf{i},\mathbf{e}\right)   &  =\mathbf{i}%
,\ \ \ \ \ \ \ \ \ \ \mu\left(  \mathbf{i},\mathbf{i}\right)  =-\mathbf{e}%
,\ \ \ \ \ \ \ \ \ \ \mu\left(  \mathbf{i},\mathbf{j}\right)  =\mathbf{k}%
,\ \ \ \ \ \ \ \ \ \ \mu\left(  \mathbf{i},\mathbf{k}\right)  =-\mathbf{j},\\
\mu\left(  \mathbf{j},\mathbf{e}\right)   &  =\mathbf{j}%
,\ \ \ \ \ \ \ \ \ \ \mu\left(  \mathbf{j},\mathbf{i}\right)  =-\mathbf{k}%
,\ \ \ \ \ \ \ \ \ \ \mu\left(  \mathbf{j},\mathbf{j}\right)  =-\mathbf{e}%
,\ \ \ \ \ \ \ \ \ \ \mu\left(  \mathbf{j},\mathbf{k}\right)  =\mathbf{i},\\
\mu\left(  \mathbf{k},\mathbf{e}\right)   &  =\mathbf{k}%
,\ \ \ \ \ \ \ \ \ \ \mu\left(  \mathbf{k},\mathbf{i}\right)  =\mathbf{j}%
,\ \ \ \ \ \ \ \ \ \ \mu\left(  \mathbf{k},\mathbf{j}\right)  =-\mathbf{i}%
,\ \ \ \ \ \ \ \ \ \ \mu\left(  \mathbf{k},\mathbf{k}\right)  =-\mathbf{e}.
\end{align*}
By the universal property of free modules wrt bilinear maps (Theorem 2.7.2 in
Lecture 22), there really is a unique such $\mathbb{R}$-bilinear map
$\mu:\mathbb{H}\times\mathbb{H}\rightarrow\mathbb{H}$; thus, we have defined
our $\mu$.

We claim that the $\mathbb{R}$-module $\mathbb{H}$ becomes an $\mathbb{R}%
$-algebra (and thus a ring) if we endow it with the multiplication $\mu$ and
the unity $\mathbf{e}$. To prove this, we need to show a few axioms. Some of
them (scale-invariance, distributivity and the $0a=a0=0$ axiom) follow from
the $\mathbb{R}$-bilinearity of $\mu$; others follow from the fact that
$\mathbb{H}$ is an $\mathbb{R}$-module. It remains to prove two axioms:

\begin{enumerate}
\item The map $\mu$ is associative (i.e., we have $\mu\left(  \mu\left(
a,b\right)  ,c\right)  =\mu\left(  a,\mu\left(  b,c\right)  \right)  $ for all
$a,b,c\in\mathbb{H}$).

\item The element $\mathbf{e}$ is a neutral element for $\mu$ (i.e., we have
$\mu\left(  a,\mathbf{e}\right)  =\mu\left(  \mathbf{e},a\right)  =a$ for all
$a\in\mathbb{H}$).
\end{enumerate}

Let us prove the first axiom. Here, again, the bilinearity of $\mu$ will make
our life easier. Indeed, we have the following more general fact:

\begin{lemma}
\label{lem.algebras.assoc-on-basis}Let $R$ be a commutative ring. Let $M$ be
an $R$-module. Let $\left(  m_{i}\right)  _{i\in I}$ be a family of vectors in
$M$ that spans $M$. Let $f:M\times M\rightarrow M$ be an $R$-bilinear map.
Assume that%
\[
f\left(  f\left(  m_{i},m_{j}\right)  ,m_{k}\right)  =f\left(  m_{i},f\left(
m_{j},m_{k}\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{for all }i,j,k\in I.
\]
Then,%
\[
f\left(  f\left(  a,b\right)  ,c\right)  =f\left(  a,f\left(  b,c\right)
\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }a,b,c\in M.
\]

\end{lemma}

In other words, this lemma says that in order to prove that an $R$-bilinear
map $f:M\times M\rightarrow M$ is associative, it suffices to prove that it is
associative on a given family that spans $M$ (i.e., that it holds whenever the
three inputs $a,b,c$ are entries of this family).

\begin{proof}
[Proof of Lemma \ref{lem.algebras.assoc-on-basis}.]See Lemma 3.12.1 in the
text. In a nutshell: Expand $a,b,c$ as $R$-linear combinations%
\[
a=\sum_{i\in I}a_{i}m_{i},\ \ \ \ \ \ \ \ \ \ b=\sum_{j\in I}b_{j}%
m_{j},\ \ \ \ \ \ \ \ \ \ c=\sum_{k\in I}c_{k}m_{k},
\]
and plug this into $f\left(  f\left(  a,b\right)  ,c\right)  $ and $f\left(
a,f\left(  b,c\right)  \right)  $. Using the bilinearity of $f$, you find%
\begin{align*}
f\left(  f\left(  a,b\right)  ,c\right)   &  =\sum_{i\in I}\ \ \sum_{j\in
I}\ \ \sum_{k\in I}a_{i}b_{j}c_{k}f\left(  f\left(  m_{i},m_{j}\right)
,m_{k}\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\\
f\left(  a,f\left(  b,c\right)  \right)   &  =\sum_{i\in I}\ \ \sum_{j\in
I}\ \ \sum_{k\in I}a_{i}b_{j}c_{k}f\left(  m_{i},f\left(  m_{j},m_{k}\right)
\right)  .
\end{align*}
The right hand sides of these equalities are equal (by assumption); thus, so
are the left hand sides.

This proof can be summarized in two words: \textquotedblleft by
linearity\textquotedblright.
\end{proof}

Having proved Lemma \ref{lem.algebras.assoc-on-basis}, we can now prove the
associativity law $\mu\left(  \mu\left(  a,b\right)  ,c\right)  =\mu\left(
a,\mu\left(  b,c\right)  \right)  $ for $\mu$ on $\mathbb{H}$: Indeed, by
Lemma \ref{lem.algebras.assoc-on-basis}, it suffices to prove this law in the
case when $a,b,c\in\left\{  \mathbf{e},\mathbf{i},\mathbf{j},\mathbf{k}%
\right\}  $. These are $64$ equalities to prove in total (since there are $64$
triples $\left(  a,b,c\right)  \in\left\{  \mathbf{e},\mathbf{i}%
,\mathbf{j},\mathbf{k}\right\}  ^{3}$). This can be reduced down to $27$
equalities (by realizing that if any of $a,b,c$ is $\mathbf{e}$, then
$\mu\left(  \mu\left(  a,b\right)  ,c\right)  =\mu\left(  a,\mu\left(
b,c\right)  \right)  $ holds for easy reasons), and further down to $9$
equalities (using the cyclic symmetry in $\mathbf{i},\mathbf{j},\mathbf{k}$ in
the definition of $\mu$). One of them is $\mu\left(  \mu\left(  \mathbf{i}%
,\mathbf{k}\right)  ,\mathbf{k}\right)  =\mu\left(  \mathbf{i},\mu\left(
\mathbf{k,k}\right)  \right)  $. By checking this and the $8$ other
equalities, you can convince yourself that $\mu$ is associative, i.e., that
associativity holds for $\mathbb{H}$.

What about the axiom saying that $\mathbf{e}$ is a neutral element for $\mu$ ?
Again, by linearity, it suffices to prove $\mu\left(  \mathbf{e},a\right)
=\mu\left(  a,\mathbf{e}\right)  =a$ only for $a\in\left\{  \mathbf{e}%
,\mathbf{i},\mathbf{j},\mathbf{k}\right\}  $ (why?). But this is obvious from
a look at the definition of $\mu$.

Thus, we have proved that $\mathbb{H}$ is indeed a ring and even an
$\mathbb{R}$-algebra. To connect this definition with the informal one we gave
in \S 1.1.2 (Lecture 2), we just have to set%
\[
i=\mathbf{i},\ \ \ \ \ \ \ \ \ \ j=\mathbf{j},\ \ \ \ \ \ \ \ \ \ k=\mathbf{k}%
\]
(and observe that the unity of $\mathbb{H}$ is $\mathbf{e}$, so that
$1=\mathbf{e}$), and therefore every quaternion $\left(  a,b,c,d\right)
\in\mathbb{H}$ can be rewritten as
\[
\left(  a,b,c,d\right)  =a\underbrace{\mathbf{e}}_{=1}%
+\,b\underbrace{\mathbf{i}}_{=i}+\,c\underbrace{\mathbf{j}}_{=j}%
+\,d\underbrace{\mathbf{k}}_{=k}=a+bi+cj+dk.
\]


\section{Monoid algebras and polynomials}

\begin{convention}
For this entire chapter, we fix a \textbf{commutative} ring $R$.
\end{convention}

In the previous section, we have learned how to define an $R$-algebra the
\textquotedblleft quick\textquotedblright\ way: Define an $R$-module first,
and then define its multiplication as a certain $R$-bilinear map $\mu$ (which
you can specify on the basis elements, if the $R$-module has a basis). Then,
associativity and neutrality of the unity can be proved just by verifying them
for basis elements.

Let me now apply this strategy to the construction of an important class of
algebras: the \textbf{monoid algebras}, and in particular the
\textbf{polynomial rings}.

\subsection{Monoid algebras}

\subsubsection{Definition}

Recall the notion of a \textbf{monoid}: Roughly speaking, it is a
\textquotedblleft group without inverses\textquotedblright. That is, a
\textbf{monoid} is a triple $\left(  M,\cdot,1\right)  $, where $M$ is a set,
$\cdot$ is an associative binary operation on $M$, and $1$ is an element of
$M$ that is neutral for $\cdot$. We will write $mn$ for $m\cdot n$ when
$m,n\in M$, and we will call $mn$ the \textbf{product} of $m$ and $n$. The
monoid $M$ is said to be \textbf{abelian} if $mn=nm$ for all $m,n\in M$. Given
a monoid $\left(  M,\cdot,1\right)  $, the binary operation $\cdot$ is called
the \textbf{operation} of $M$, and the element $1$ is called the
\textbf{neutral element} of $M$. We say that the monoid $M$ is \textbf{written
multiplicatively} (or, for short, \textbf{multiplicative}) when its operation
is denoted by $\cdot$, and we say that it is \textbf{written additively} (or,
for short, \textbf{additive}) when its operation is denoted by $+$. Usually,
the neutral element of a multiplicative monoid is denoted by $1$, whereas the
neutral element of an additive monoid is denoted by $0$.

If $M$ is a monoid written multiplicatively, then we can define the
\textbf{monoid algebra} $R\left[  M\right]  $. Informally, this is the
$R$-algebra obtained by \textquotedblleft throwing\textquotedblright\ the
elements of $M$ \textquotedblleft into\textquotedblright\ the ring $R$. Its
elements are \textquotedblleft formal $R$-linear combinations of elements of
$M$\textquotedblright, i.e., expressions of the form%
\[
r_{1}m_{1}+r_{2}m_{2}+\cdots+r_{k}m_{k}%
\]
with $k\in\mathbb{N}$ and $m_{1},m_{2},\ldots,m_{k}\in M$ and $r_{1}%
,r_{2},\ldots,r_{k}\in R$. These expressions are multiplied by distributivity
and using the multiplications of $R$ and $M$: that is,%
\[
\left(  r_{1}m_{1}+r_{2}m_{2}+\cdots+r_{k}m_{k}\right)  \left(  s_{1}%
n_{1}+s_{2}n_{2}+\cdots+s_{\ell}n_{\ell}\right)  =\sum_{i=1}^{k}\ \ \sum
_{j=1}^{\ell}\underbrace{r_{i}s_{j}}_{\substack{\text{product}\\\text{in }%
R}}\ \ \underbrace{m_{i}n_{j}}_{\substack{\text{product}\\\text{in }M}}.
\]


In order to make this rigorous, let us recall a few concepts:

If $M$ is any set, then $R^{M}$ is the $R$-module%
\[
\left\{  \left(  r_{m}\right)  _{m\in M}\ \mid\ r_{m}\in R\text{ for each
}m\in M\right\}
\]
(consisting of all families $\left(  r_{m}\right)  _{m\in M}$ of elements of
$R$), whereas $R^{\left(  M\right)  }$ is the $R$-submodule%
\[
\left\{  \left(  r_{m}\right)  _{m\in M}\in R^{M}\ \mid\ \text{all but
finitely many }m\in M\text{ satisfy }r_{m}=0\right\}
\]
of $R^{M}$. If the set $M$ is finite, then $R^{\left(  M\right)  }=R^{M}$.

The $R$-module $R^{\left(  M\right)  }$ is free, and the \textbf{standard
basis} $\left(  e_{m}\right)  _{m\in M}$ of $R^{\left(  M\right)  }$ is
defined as follows: For each $m\in M$, the vector $e_{m}\in R^{\left(
M\right)  }$ is the family whose $m$-th entry is $1$ and whose all other
entries are $0$. (If $M=\left\{  1,2,\ldots,n\right\}  $ for some
$n\in\mathbb{N}$, then this recovers the classical linear-algebraic standard
basis: e.g., if $M=\left\{  1,2,3\right\}  $, then $e_{1}=\left(
1,0,0\right)  $ and $e_{2}=\left(  0,1,0\right)  $ and $e_{3}=\left(
0,0,1\right)  $.)

The standard basis $\left(  e_{m}\right)  _{m\in M}$ of $R^{\left(  M\right)
}$ is, of course, a basis of $R^{\left(  M\right)  }$.

We can now give a rigorous definition of the monoid algebra $R\left[
M\right]  $:

\begin{definition}
Let $M$ be a monoid, written multiplicatively (so that $\cdot$ denotes its
operation, and $1$ denotes its neutral element).

The \textbf{monoid algebra} of $M$ over $R$ (also known as the \textbf{monoid
ring} of $M$ over $R$) is the $R$-algebra $R\left[  M\right]  $ defined as follows:

As an $R$-module, it is the free $R$-module%
\[
R^{\left(  M\right)  }=\left\{  \left(  r_{m}\right)  _{m\in M}\in R^{M}%
\ \mid\ \text{all but finitely many }m\in M\text{ satisfy }r_{m}=0\right\}  .
\]
Its multiplication is defined to be the unique $R$-bilinear map $\mu
:R^{\left(  M\right)  }\times R^{\left(  M\right)  }\rightarrow R^{\left(
M\right)  }$ that satisfies%
\[
\mu\left(  e_{m},e_{n}\right)  =e_{mn}\ \ \ \ \ \ \ \ \ \ \text{for all
}m,n\in M.
\]
Here, $\left(  e_{m}\right)  _{m\in M}$ is the standard basis of $R^{\left(
M\right)  }$ (that is, $e_{m}\in R^{\left(  M\right)  }$ is the family whose
$m$-th entry is $1$ and whose all other entries are $0$). The unity of this
$R$-algebra is $e_{1}$.
\end{definition}

\begin{theorem}
This is indeed a well-defined $R$-algebra.
\end{theorem}

\begin{proof}
By linearity (and associativity of $M$). See Theorem 4.1.2 in the text for details.
\end{proof}

Recall that any group is a monoid. A group's monoid algebra is called a
\textquotedblleft group algebra\textquotedblright:

\begin{definition}
If $G$ is a group, then its monoid algebra $R\left[  G\right]  $ is called a
\textbf{group algebra} (or \textbf{group ring}).
\end{definition}

\subsubsection{Examples}

The above definition of a monoid algebra was rather abstract, so let us give
some examples.

\begin{example}
\label{exa.monalg.QC2}Consider the cyclic group $C_{2}$ of order $2$. We write
it multiplicatively as $C_{2}=\left\{  1,u\right\}  $ where $u^{2}=1$.

(This group is better known as $\mathbb{Z}/2$, but that would require writing
it additively, which we don't want.)

How does the group algebra (= monoid algebra) $\mathbb{Q}\left[  C_{2}\right]
$ look like?

As a $\mathbb{Q}$-module, it is%
\begin{align*}
\mathbb{Q}^{\left(  C_{2}\right)  }  &  =\mathbb{Q}^{\left(  \left\{
1,u\right\}  \right)  }=\mathbb{Q}^{\left\{  1,u\right\}  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }R^{\left(  I\right)  }=R^{I}\text{
when }I\text{ is finite}\right) \\
&  =\left\{  \left(  r_{m}\right)  _{m\in\left\{  1,u\right\}  }\ \mid
\ r_{m}\in\mathbb{Q}\text{ for all }m\in\left\{  1,u\right\}  \right\}  .
\end{align*}


A family of the form $\left(  r_{m}\right)  _{m\in\left\{  1,u\right\}  }$
contains just two entries: $r_{1}$ and $r_{u}$. By abuse of notation, we can
thus identify such a family with the pair $\left(  r_{1},r_{u}\right)  $
(although, formally speaking, there is merely a bijection between the former
families and the latter pairs). Thus, we can rewrite our above equality as%
\[
\mathbb{Q}^{\left(  C_{2}\right)  }=\left\{  \left(  r_{1},r_{u}\right)
\ \mid\ r_{1},r_{u}\in\mathbb{Q}\right\}  =\mathbb{Q}^{2}.
\]
The addition and the action of the group algebra $\mathbb{Q}\left[
C_{2}\right]  $ are entrywise. What about its multiplication?

Its standard basis is $\left(  e_{m}\right)  _{m\in\left\{  1,u\right\}
}=\left(  e_{1},e_{u}\right)  $, where the vector $e_{1}$ has $1$-th entry $1$
and $u$-th entry $0$, and where the vector $e_{u}$ has $1$-th entry $0$ and
$u$-th entry $1$. If we again encode each family $\left(  r_{m}\right)
_{m\in\left\{  1,u\right\}  }$ as a pair $\left(  r_{1},r_{u}\right)  $, then
we can restate this as%
\[
e_{1}=\left(  1,0\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ e_{u}=\left(  0,1\right)  .
\]


The multiplication of the group algebra $\mathbb{Q}\left[  C_{2}\right]  $ is
given by%
\[
\mu\left(  e_{m},e_{n}\right)  =e_{mn}\ \ \ \ \ \ \ \ \ \ \text{for all
}m,n\in C_{2}.
\]
In other words,%
\[
e_{m}e_{n}=e_{mn}\ \ \ \ \ \ \ \ \ \ \text{for all }m,n\in C_{2}.
\]
Thus,%
\begin{align*}
e_{1}e_{1}  &  =e_{1\cdot1}=e_{1},\ \ \ \ \ \ \ \ \ \text{\ }e_{1}e_{u}%
=e_{1u}=e_{u},\\
e_{u}e_{1}  &  =e_{u\cdot1}=e_{u},\ \ \ \ \ \ \ \ \ \ e_{u}e_{u}=e_{uu}%
=e_{1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }uu=u^{2}=1\right)  .
\end{align*}


Since $\left(  e_{1},e_{u}\right)  $ is a basis of $\mathbb{Q}\left[
C_{2}\right]  $, we can write each element of $\mathbb{Q}\left[  C_{2}\right]
$ uniquely as $ae_{1}+be_{u}$ for two numbers $a,b\in\mathbb{Q}$. How do we
multiply two such elements? Let's see:%
\begin{align}
&  \left(  ae_{1}+be_{u}\right)  \left(  ce_{1}+de_{u}\right) \nonumber\\
&  =ae_{1}\left(  ce_{1}+de_{u}\right)  +be_{u}\left(  ce_{1}+de_{u}\right)
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the multiplication }\mu\\
\text{is }\mathbb{Q}\text{-bilinear}%
\end{array}
\right) \nonumber\\
&  =a\left(  c\underbrace{e_{1}e_{1}}_{=e_{1}}+\,d\underbrace{e_{1}e_{u}%
}_{=e_{u}}\right)  +b\left(  c\underbrace{e_{u}e_{1}}_{=e_{u}}%
+\,d\underbrace{e_{u}e_{u}}_{=e_{1}}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since the
multiplication }\mu\text{ is }\mathbb{Q}\text{-bilinear}\right) \nonumber\\
&  =a\left(  ce_{1}+de_{u}\right)  +b\left(  ce_{u}+de_{1}\right) \nonumber\\
&  =\left(  ac+bd\right)  e_{1}+\left(  ad+bc\right)  e_{u}.
\label{eq.exa.monalg.QC2.mul1}%
\end{align}


If we again use the notation encode each family $\left(  r_{m}\right)
_{m\in\left\{  1,u\right\}  }$ as a pair $\left(  r_{1},r_{u}\right)  $, then
the element $ae_{1}+be_{u}$ simply becomes $\left(  a,b\right)  $ (since
$e_{1}=\left(  1,0\right)  $ and $e_{u}=\left(  0,1\right)  $), and thus the
multiplication rule (\ref{eq.exa.monalg.QC2.mul1}) rewrites as%
\begin{equation}
\left(  a,b\right)  \left(  c,d\right)  =\left(  ac+bd,\ ad+bc\right)  .
\label{eq.exa.monalg.QC2.mul2}%
\end{equation}
This is almost the rule for multiplying complex numbers! In fact, the latter
rule is%
\[
\left(  a,b\right)  \left(  c,d\right)  =\left(  ac-bd,\ ad+bc\right)  ,
\]
which differs from (\ref{eq.exa.monalg.QC2.mul2}) only in that it has a minus
instead of a plus. Thus, we can think of $\mathbb{Q}\left[  C_{2}\right]  $ as
a \textquotedblleft twin brother\textquotedblright\ of $\mathbb{C}$, except
that we are using rational numbers rather than real numbers as our entries
(but this is not a conceptual difference; we could play the same game with
$\mathbb{R}$ instead of $\mathbb{C}$).

However, it is a much less famous twin, and for a good reason: The ring
$\mathbb{C}$ is a field, but the ring $\mathbb{Q}\left[  C_{2}\right]  $ is
not. In fact, (\ref{eq.exa.monalg.QC2.mul2}) yields%
\[
\left(  1,1\right)  \left(  1,-1\right)  =\left(  1\cdot1+1\cdot\left(
-1\right)  ,\ 1\cdot\left(  -1\right)  +1\cdot1\right)  =\left(  0,0\right)
\]
in $\mathbb{Q}\left[  C_{2}\right]  $, so that $\mathbb{Q}\left[
C_{2}\right]  $ is not an integral domain, let alone a field.

Actually, we have%
\[
\mathbb{Q}\left[  C_{2}\right]  \cong\mathbb{Q}\times\mathbb{Q}%
\ \ \ \ \ \ \ \ \ \ \text{as rings and even as }\mathbb{Q}\text{-algebras.}%
\]
Next time, we will see why.
\end{example}


\end{document}