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\begin{document}
\section*{Math 332 Winter 2023, Lecture 24: Monoid algebras}

\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}{2}

\section{Monoid algebras and polynomials}

Recall: For this entire chapter, we fix a \textbf{commutative} ring $R$.

\subsection{Monoid algebras (cont'd)}

\setcounter{subsubsection}{1}

Recall from Lecture 23: If $M$ is any monoid (written multiplicatively), then
$R\left[  M\right]  $ denotes the \textbf{monoid algebra} of $M$ over $R$.
This is the $R$-algebra that consists of all formal $R$-linear combinations
$\sum_{m\in M}r_{m}e_{m}$ of elements $e_{m}$ with $m\in M$. So, as an
$R$-module, it is just the free $R$-module%
\[
R^{\left(  M\right)  }=\left\{  \left(  r_{m}\right)  _{m\in M}\in R^{M}%
\ \mid\ \text{all but finitely many }m\in M\text{ satisfy }r_{m}=0\right\}  ,
\]
and the elements $e_{m}$ are just the vectors that the standard basis of this
free $R$-module $R^{\left(  M\right)  }$ comprises\footnote{Specifically, for
each $m\in M$, the vector $e_{m}$ is the $m$-th standard basis vector, i.e.,
the family $\left(  \delta_{m,n}\right)  _{n\in M}$ that has a $1$ in its
$m$-th position and $0$s in all other positions. It is the natural
generalization of the standard basis vectors $\left(  0,0,\ldots
,0,1,0,0,\ldots,0\right)  $ of an $R$-module $R^{n}$ (but is more abstract due
to the fact that the elements of an arbitrary monoid $M$ do not come with a
given order, and $M$ can be infinite).}. Its multiplication is the unique
$R$-bilinear map from $R^{\left(  M\right)  }\times R^{\left(  M\right)  }$ to
$R^{\left(  M\right)  }$ that satisfies%
\[
e_{m}e_{n}=e_{mn}\ \ \ \ \ \ \ \ \ \ \text{for all }m,n\in M.
\]
Its unity is $e_{1}$, where $1$ is the neutral element of $M$.

\subsubsection{Examples (cont'd)}

Last time, we began to analyze the monoid algebra (aka group algebra)
$\mathbb{Q}\left[  C_{2}\right]  $, where $C_{2}$ is the cyclic group of order
$2$ (given as $C_{2}=\left\{  1,u\right\}  $ where $u^{2}=1$).

The elements of this $\mathbb{Q}$-algebra $\mathbb{Q}\left[  C_{2}\right]  $
have the form $ae_{1}+be_{u}$ where $a,b\in\mathbb{Q}$. They are multiplied
according to the rule%
\[
\left(  ae_{1}+be_{u}\right)  \left(  ce_{1}+de_{u}\right)  =\left(
ac+bd\right)  e_{1}+\left(  ad+bc\right)  e_{u}%
\]
(for $a,b,c,d\in\mathbb{Q}$). This makes it clear that the ring $\mathbb{Q}%
\left[  C_{2}\right]  $ is commutative.

I explained that $\mathbb{Q}\left[  C_{2}\right]  $ is not a field, and
instead has zero-divisors:%
\[
\left(  e_{1}+e_{u}\right)  \left(  e_{1}-e_{u}\right)  =\underbrace{e_{1}%
^{2}}_{\substack{=e_{1}e_{1}\\=e_{1\cdot1}=e_{1}}}-\underbrace{e_{u}^{2}%
}_{\substack{=e_{u}e_{u}\\=e_{uu}=e_{1}}}=e_{1}-e_{1}=0.
\]


Note that $e_{1}$ is the unity of the algebra $\mathbb{Q}\left[  C_{2}\right]
$, so we can write $1$ for it. Thus, $ae_{1}+be_{u}$ becomes $a+be_{u}$.

I teased you with the claim that%
\[
\mathbb{Q}\left[  C_{2}\right]  \cong\mathbb{Q}\times\mathbb{Q}%
\ \ \ \ \ \ \ \ \ \ \text{as }\mathbb{Q}\text{-algebras.}%
\]
Let me quickly explain why this is true:

Recall that any central idempotent element of a ring $R$ breaks this ring $R$
into a direct product (Exercise 1 \textbf{(d)} on homework set \#4). Thus, we
want to find a central idempotent element of $\mathbb{Q}\left[  C_{2}\right]
$.

Here is one: Set $z:=\dfrac{1}{2}+\dfrac{1}{2}e_{u}\in\mathbb{Q}\left[
C_{2}\right]  $. Then,%
\begin{align*}
z^{2}  &  =\left(  \dfrac{1}{2}+\dfrac{1}{2}e_{u}\right)  ^{2}=\dfrac{1}%
{4}+\dfrac{1}{2}\cdot\dfrac{1}{2}e_{u}+\dfrac{1}{2}e_{u}\cdot\dfrac{1}%
{2}+\dfrac{1}{4}\underbrace{e_{u}^{2}}_{=1}\\
&  =\dfrac{1}{4}+\dfrac{1}{4}e_{u}+\dfrac{1}{4}e_{u}+\dfrac{1}{4}=\dfrac{1}%
{2}+\dfrac{1}{2}e_{u}=z.
\end{align*}
Thus, $z$ is idempotent. Since $\mathbb{Q}\left[  C_{2}\right]  $ is
commutative, this $z$ is furthermore central. Hence, by the above-mentioned
exercise, we obtain%
\[
\mathbb{Q}\left[  C_{2}\right]  \cong z\mathbb{Q}\left[  C_{2}\right]
\times\left(  1-z\right)  \mathbb{Q}\left[  C_{2}\right]
\]
(as rings, but by the same logic as $\mathbb{Q}$-algebras as well\footnote{To
be more precise: If we fix a commutative ring $S$ and replace the word
\textquotedblleft ring\textquotedblright\ by \textquotedblleft$S$%
-algebra\textquotedblright\ throughout Exercise 1 on homework set \#4, then
the exercise remains true, and the solution does not get much harder (there
are some more straightforward axioms to be verified).}). Now, I claim that
both $\mathbb{Q}$-algebras $z\mathbb{Q}\left[  C_{2}\right]  $ and $\left(
1-z\right)  \mathbb{Q}\left[  C_{2}\right]  $ are isomorphic to $\mathbb{Q}$.
Indeed, for $z\mathbb{Q}\left[  C_{2}\right]  $, this follows easily from the
fact that%
\begin{align*}
z\cdot\left(  a+be_{u}\right)   &  =\left(  \dfrac{1}{2}+\dfrac{1}{2}%
e_{u}\right)  \cdot\left(  a+be_{u}\right) \\
&  =\left(  \dfrac{1}{2}a+\dfrac{1}{2}b\right)  +\left(  \dfrac{1}{2}%
b+\dfrac{1}{2}a\right)  e_{u}\\
&  =\dfrac{a+b}{2}+\dfrac{a+b}{2}e_{u}=\left(  a+b\right)
z\ \ \ \ \ \ \ \ \ \ \text{for all }a,b\in\mathbb{Q}.
\end{align*}
For $\left(  1-z\right)  \mathbb{Q}\left[  C_{2}\right]  $, this is similar
(notice that $1-z=\dfrac{1}{2}-\dfrac{1}{2}e_{u}$). With a bit of work, this
leads to the $\mathbb{Q}$-algebra isomorphism%
\begin{align*}
g:\mathbb{Q}\left[  C_{2}\right]   &  \rightarrow\mathbb{Q}^{2},\\
a+be_{u}  &  \mapsto\left(  a+b,\ a-b\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{for all }a,b\in\mathbb{Q}\right)
\end{align*}
(where $\mathbb{Q}^{2}$ means the direct product $\mathbb{Q}\times\mathbb{Q}$
of $\mathbb{Q}$-algebras).

A few more remarks:

\begin{itemize}
\item There is an obvious reason why $\mathbb{Q}\left[  C_{2}\right]
\cong\mathbb{Q}^{2}$ as $\mathbb{Q}$-modules: Indeed, $\mathbb{Q}\left[
C_{2}\right]  $ has a basis $\left(  e_{1},e_{u}\right)  $ that consists of
two vectors, so we can obtain a $\mathbb{Q}$-module isomorphism $f:\mathbb{Q}%
\left[  C_{2}\right]  \rightarrow\mathbb{Q}^{2}$ by mapping $e_{1}%
\mapsto\left(  1,0\right)  $ and $e_{u}\mapsto\left(  0,1\right)  $.
Explicitly, this isomorphism $f$ sends each $a+be_{u}$ to $\left(  a,b\right)
$. But this is \textbf{not} a $\mathbb{Q}$-algebra isomorphism (e.g., because
$e_{1}e_{u}$ is nonzero but $\left(  1,0\right)  \left(  0,1\right)  $ is
zero). Our above $\mathbb{Q}$-algebra isomorphism $g:\mathbb{Q}\left[
C_{2}\right]  \rightarrow\mathbb{Q}^{2}$ is more sophisticated.

\item We can easily repeat the above explorations of $\mathbb{Q}\left[
C_{2}\right]  $ using $\mathbb{R}$ or $\mathbb{C}$ instead of $\mathbb{Q}$.

However, things change if we try to repeat them using $\mathbb{Z}$. Indeed,
the idempotent element $z=\dfrac{1}{2}+\dfrac{1}{2}e_{u}$ does not exist over
$\mathbb{Z}$ because $\dfrac{1}{2}\notin\mathbb{Z}$. Thus, the group algebra
$\mathbb{Z}\left[  C_{2}\right]  $ has no reason to be isomorphic to
$\mathbb{Z}^{2}$. And indeed, it is not. (For a proof, see Example 4.1.5
\textbf{(b)} in the text.)
\end{itemize}

Here is another example of a monoid ring:

\begin{example}
Consider the cyclic group $C_{3}=\left\{  1,u,v\right\}  $ of order $3$ with
$u^{3}=1$ and $u^{2}=v$. Its group algebra $\mathbb{Q}\left[  C_{3}\right]  $
has a central idempotent
\[
z:=\dfrac{1+e_{u}+e_{v}}{3}.
\]
More generally, for any finite group $G$, the group algebra $\mathbb{Q}\left[
G\right]  $ has a central idempotent
\[
z:=\dfrac{\sum\limits_{g\in G}e_{g}}{\left\vert G\right\vert }.
\]
(Exercise: Prove this!) The principal ideal $z\mathbb{Q}\left[  G\right]  $ is
always $\cong\mathbb{Q}$ as a $\mathbb{Q}$-algebra. Thus, using Exercise 1
\textbf{(d)} on homework set \#4 (extended from rings to algebras), we obtain
\[
\mathbb{Q}\left[  G\right]  \cong\mathbb{Q}\times S\ \ \ \ \ \ \ \ \ \ \left(
\text{as }\mathbb{Q}\text{-algebras}\right)  ,
\]
where $S=\left(  1-z\right)  \mathbb{Q}\left[  G\right]  $. In the case
$G=C_{2}$, we found $S\cong\mathbb{Q}$, but in the general case $S$ can be
more complicated.
\end{example}

\subsubsection{General properties of monoid algebras}

Let us now discuss some general properties of and conventions on monoid algebras.

\begin{proposition}
Let $M$ be an \textbf{abelian} monoid. Then, the monoid ring $R\left[
M\right]  $ is commutative.
\end{proposition}

\begin{proof}
See the text (Proposition 4.1.9). This is again a proof \textquotedblleft by
linearity\textquotedblright.
\end{proof}

\begin{proposition}
\label{prop.monalg.const}Let $M$ be a monoid with neutral element $1$. Then,
the map%
\begin{align*}
R  &  \rightarrow R\left[  M\right]  ,\\
r  &  \mapsto r\cdot e_{1}%
\end{align*}
is an injective $R$-algebra morphism.
\end{proposition}

\begin{proof}
See the text (Proposition 4.1.10). Injectivity is clear; morphicity relies on
$e_{1}e_{1}=e_{1}$.
\end{proof}

\begin{convention}
If $M$ is a monoid, then we shall identify each $r\in R$ with $r\cdot e_{1}\in
R\left[  M\right]  $. This identification is harmless (i.e., does not lead to
false conclusions)\footnotemark, and turns $R$ into an $R$-subalgebra of
$R\left[  M\right]  $.

An element of $R\left[  M\right]  $ will be called \textbf{constant} if it
lies in this subalgebra (i.e., if it is $r\cdot e_{1}$ for some $r\in R$).
\end{convention}

\footnotetext{This follows from Proposition \ref{prop.monalg.const}%
.}\textbf{Warning:} We previously explained that $\mathbb{Q}\left[
C_{2}\right]  \cong\mathbb{Q}\times\mathbb{Q}$ as $\mathbb{Q}$-algebras. Now
we have identified $\mathbb{Q}$ with a subalgebra of $\mathbb{Q}\left[
C_{2}\right]  $. But this subalgebra is not one of the two $\mathbb{Q}$
factors in $\mathbb{Q}\left[  C_{2}\right]  \cong\mathbb{Q}\times\mathbb{Q}$;
in fact, none of those two $\mathbb{Q}$ factors is a $\mathbb{Q}$-subalgebra.
Our $\mathbb{Q}$-algebra isomorphism from $\mathbb{Q}\left[  C_{2}\right]  $
to $\mathbb{Q}\times\mathbb{Q}$ sends the unity of $\mathbb{Q}\left[
C_{2}\right]  $ to $\left(  1,1\right)  \in\mathbb{Q}\times\mathbb{Q}$, which
does not lie completely in either factor.

\begin{proposition}
\label{prop.monalg.M-into-RM}Let $M$ be a monoid. Then, the map%
\begin{align*}
M  &  \rightarrow R\left[  M\right]  ,\\
m  &  \mapsto e_{m}%
\end{align*}
is a monoid morphism from $M$ to $\left(  R\left[  M\right]  ,\cdot,1\right)
$.
\end{proposition}

\begin{proof}
This is just saying that $e_{mn}=e_{m}e_{n}$ and $e_{1}=1_{R\left[  M\right]
}$. Both hold by definition of $R\left[  M\right]  $.
\end{proof}

Our last two propositions tell us that the monoid algebra $R\left[  M\right]
$ \textquotedblleft includes\textquotedblright\ both $R$ and $M$ in an
appropriate way (at least when $R$ is nontrivial). Thus, we can view it as
what comes out if we \textquotedblleft throw\textquotedblright\ the elements
of $M$ into $R$.

This suggests another convention:

\begin{convention}
\label{conv.monalg.m-as-em}Let $M$ be a monoid. Then, the elements $e_{m}$ of
the standard basis $\left(  e_{m}\right)  _{m\in M}$ of $R\left[  M\right]  $
will just be written as $m$ if there is no confusion to worry about.

For example, if $M$ is the cyclic group $C_{3}$ as above, then the element
$ae_{1}+be_{u}+ce_{v}$ will just be written as $a1+bu+cv=a+bu+cv$.

For another example, if $M$ is the cyclic group $C_{2}$ as above, then our
multiplication rule%
\[
\left(  ae_{1}+be_{u}\right)  \left(  ce_{1}+de_{u}\right)  =\left(
ac+bd\right)  e_{1}+\left(  ad+bc\right)  e_{u}%
\]
(for $a,b,c,d\in R$) rewrites as%
\[
\left(  a+bu\right)  \left(  c+du\right)  =\left(  ac+bd\right)  +\left(
ad+bc\right)  u.
\]

\end{convention}

\subsection{Polynomial rings}

\subsubsection{Univariate polynomials}

We can now effortlessly define univariate polynomials: They are just elements
of certain monoid algebras. Which ones?

Recall that $R$ denotes a commutative ring. Recall also that $\mathbb{N}%
=\left\{  0,1,2,\ldots\right\}  $.

\begin{definition}
Let $C$ be the free monoid with a single generator $x$. This is the monoid
whose elements are countably many distinct symbols called%
\[
x^{0},\ \ x^{1},\ \ x^{2},\ \ x^{3},\ \ \ldots,
\]
and whose operation is defined by%
\[
x^{i}\cdot x^{j}=x^{i+j}\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\mathbb{N}.
\]
Of course, this monoid is just the well-known additive monoid $\left(
\mathbb{N},+,0\right)  $ in a multiplicative disguise (with each nonnegative
integer $i$ renamed as $x^{i}$ in order to avoid overloading the notation
$i\cdot j$).

The neutral element of this monoid $C$ is $x^{0}$. We set $x:=x^{1}$.

The elements of $C$ are called \textbf{monomials} in the variable $x$. The
specific element $x$ is called the \textbf{indeterminate}.

Now, the \textbf{univariate polynomial ring} $R\left[  x\right]  $ over $R$ is
defined to be the monoid algebra $R\left[  C\right]  $. Following Convention
\ref{conv.monalg.m-as-em}, we simply write $m$ for the standard basis vector
$e_{m}$ when $m\in C$. That is, we write $x^{i}$ for the basis vector
$e_{x^{i}}$. Thus, $R\left[  x\right]  $ is a free $R$-module with basis%
\[
\left(  x^{0},x^{1},x^{2},x^{3},\ldots\right)  =\left(  1,x,x^{2},x^{3}%
,\ldots\right)  .
\]
Hence, any $p\in R\left[  x\right]  $ can be written as a finite $R$-linear
combination of powers of $x$. That is, $p$ can be written as%
\[
p=a_{0}x^{0}+a_{1}x^{1}+a_{2}x^{2}+\cdots+a_{n}x^{n}=a_{0}+a_{1}x+a_{2}%
x^{2}+\cdots+a_{n}x^{n}%
\]
for some $n\in\mathbb{N}$ and some $a_{0},a_{1},\ldots,a_{n}\in R$. This
representation is unique up to trailing zeroes (i.e., up to adding extra terms
of the form $0x^{n+1}$ and $0x^{n+2}$ and so on).

Elements of $R\left[  x\right]  $ are called \textbf{polynomials} in $x$ over
$R$.
\end{definition}

Thus, up to notation, the univariate polynomial ring $R\left[  x\right]  $ is
just the monoid ring $R\left[  \mathbb{N}\right]  $ of the abelian monoid
$\mathbb{N}=\left(  \mathbb{N},+,0\right)  $. Hence, it is commutative (since
$\mathbb{N}$ is abelian).

\begin{example}
\textbf{(a)} The sum%
\[
1+3x^{2}+6x^{3}=1e_{x^{0}}+3e_{x^{2}}+6e_{x^{3}}%
\]
belongs to $R\left[  x\right]  $, i.e., is a polynomial in $x$.

\textbf{(b)} The infinite sum%
\[
1+x+x^{2}+x^{3}+\cdots
\]
is \textbf{not} a polynomial, because it is not a \textbf{finite} $R$-linear
combination of powers of $x$ (unless $R$ is trivial). We defined the
polynomial ring $R\left[  x\right]  $ to be $R\left[  C\right]  $, so that it
is $R^{\left(  C\right)  }$ as an $R$-module (not $R^{C}$). Infinite sums like
$1+x+x^{2}+\cdots$ would make sense in $R^{C}$ and are known as \textbf{formal
power series}, but they are not polynomials; they are a subject of their own.
\end{example}

So we have defined \textbf{univariate} polynomial rings (i.e., polynomial
rings in a single variable). Likewise, we can define \textbf{multivariate}
polynomial rings (i.e., polynomial rings in several variables). For
simplicity, let me restrict myself to finitely many variables.

\subsubsection{Multivariate polynomials}

\begin{definition}
Let $n\in\mathbb{N}$. Let $C^{\left(  n\right)  }$ be the free abelian monoid
with $n$ generators $x_{1},x_{2},\ldots,x_{n}$. This is the monoid whose
elements are the distinct symbols%
\[
x_{1}^{i_{1}}x_{2}^{i_{2}}\cdots x_{n}^{i_{n}}\ \ \ \ \ \ \ \ \ \ \text{with
}i_{1},i_{2},\ldots,i_{n}\in\mathbb{N},
\]
and whose operation is given by%
\[
\left(  x_{1}^{i_{1}}x_{2}^{i_{2}}\cdots x_{n}^{i_{n}}\right)  \left(
x_{1}^{j_{1}}x_{2}^{j_{2}}\cdots x_{n}^{j_{n}}\right)  =x_{1}^{i_{1}+j_{1}%
}x_{2}^{i_{2}+j_{2}}\cdots x_{n}^{i_{n}+j_{n}}.
\]
If we wrote this monoid additively, it would just be $\mathbb{N}^{n}$ (the
additive monoid of $n$-tuples of nonnegative integers, with entrywise
addition), but we write it multiplicatively.

The elements of $C^{\left(  n\right)  }$ are called \textbf{monomials}.

For each $i\in\left\{  1,2,\ldots,n\right\}  $, the monomial%
\[
x_{1}^{0}x_{2}^{0}\cdots x_{i-1}^{0}x_{i}^{1}x_{i+1}^{0}x_{i+2}^{0}\cdots
x_{n}^{0}%
\]
will be denoted by $x_{i}$. These specific monomials $x_{1},x_{2},\ldots
,x_{n}$ are called the \textbf{indeterminates}.

Now, the $R$-algebra $R\left[  x_{1},x_{2},\ldots,x_{n}\right]  $ is defined
to be the monoid algebra $R\left[  C^{\left(  n\right)  }\right]  $. It is
called the \textbf{polynomial ring in }$n$ \textbf{variables }$x_{1}%
,x_{2},\ldots,x_{n}$ \textbf{over }$R$. Any element of this $R$-algebra can be
written as an $R$-linear combination%
\[
\sum_{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\mathbb{N}^{n}}%
r_{i_{1},i_{2},\ldots,i_{n}}x_{1}^{i_{1}}x_{2}^{i_{2}}\cdots x_{n}^{i_{n}}%
\]
with $r_{i_{1},i_{2},\ldots,i_{n}}\in R$ (such that all but finitely many of
these coefficients $r_{i_{1},i_{2},\ldots,i_{n}}$ are $0$).

Elements of $R\left[  x_{1},x_{2},\ldots,x_{n}\right]  $ are called
\textbf{polynomials} in $x_{1},x_{2},\ldots,x_{n}$.
\end{definition}

For example, for $R=\mathbb{Z}$, the sum%
\[
4x_{1}^{2}+x_{2}x_{3}+7x_{3}^{5}-3
\]
is a polynomial in $x_{1},x_{2},\ldots,x_{n}$ whenever $n\geq3$. For
$R=\mathbb{R}$, the sums%
\[
3x_{1}+14x_{1}x_{2}^{3}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \sqrt
{2}x_{1}^{7}-\dfrac{3}{2}x_{1}^{3}x_{2}+\pi
\]
are polynomials in $x_{1},x_{2},\ldots,x_{n}$ whenever $n\geq2$. You can
easily construct examples like this ad infinitum.

Note that the univariate polynomial ring $R\left[  x\right]  $ is just the
particular case of the polynomial ring $R\left[  x_{1},x_{2},\ldots
,x_{n}\right]  $ in $n$ variables $x_{1},x_{2},\ldots,x_{n}$ obtained by
setting $n=1$ and renaming the indeterminate $x_{1}$ as $x$.
\end{document}