\magnification=1200

\input AMSSYM.DEF   % Either put a % befor this
%\def\Bbb{\bf}      % or before this
\def\Q{\mathord{\Bbb Q}}
\def\R{\mathord{\Bbb R}}
\def\N{\mathord{{\Bbb N}}}
\def\Z{\mathord{{\Bbb Z}}}

\def\monoid{*monoid}
\def\condition{$(\bf po)$}
\def\po{\mathrel{\leq}}
\def\supp{\mathop{{\rm supp}}}
\def\tmo{\mathrel{\preceq}}
\def\stmo{\mathrel{\prec}}

\def\graver{\mathop{\rm Gr}}
\def\mb{\mathop{\rm MinBal}}

\noindent {\bf Lecture 2. Algebraic constructions and categories (expanded version)}

\bigskip
\noindent We define some categories and display some constructions that
will play a role later.

\bigskip
\noindent{\it $M$-sets}

\bigskip
\noindent {\bf 2.1.~Definition.}  Let $M$ be a monoid.  A {\it left
$M$-set\/} is a set $X$ equipped with a function
$M\times X\to X; (m,x)\mapsto m\,x$ such that $e\,x=x$ and
$m\,(m'\,x) = (m\ast m')\,x$ for all $m,m'\in M$ and $x\in X$.  A {\it left $M$-set morphism\/} is a
function $f$ between left $M$-sets that satisfies $m\,f(x) = f(m\,x)$.   

Right $M$-sets and $M$-set morphisms are defined analogously.  From now on, when we refer to an $M$-set without qualification (left or right), we mean a left $M$-set.

\bigskip
\noindent  {\bf 2.2.~Examples.}

\smallskip
\item{a)} Any monoid $M$ is itself is a (left) $M$-set, where the action of $m\in M$ on $x\in M$ is defined by the operation of $M$: $m\,x =  m\ast x$.

\item{b)}A congruence $C$ on $M$ is both a left and right
$M$-set, with the actions defined as follows:
$$\eqalign{M\times C \to C; (m, (m', m''))&\mapsto (m*m', m*m'')\cr
C\times M\to C; ((m',m''), m)&\mapsto (m'*x,m''*m).\cr}$$

\bigskip
\noindent  {\bf 2.3.~Lemma.} An equivalence relation $E$ on $M$ that is both a right and left $M$-set under the action described in 2.2.b) is a congruence.

\smallskip
\noindent{\it Proof.} Exercise.\hfill/////

\bigskip
\noindent  {\bf 2.4.~Definition.}  A
{\it sub-$M$-set\/} of an $M$-set  $X$ is a subset of $X$ that is closed under
the action of $M$.  A sub-$M$-set of $M$ is called an {\it ideal}.  

\bigskip
\noindent  {\bf 2.5.~Discussion.}  We will describe the ``subobject classifier,'' $\Omega_M$.  Let $X$ be an $M$-set, let $U\subseteq X$ be sub-$M$-set and let $x\in X$.  We define the ideal $x\backslash U\subseteq M$ by
$$m\in x\backslash U \;\Leftrightarrow m\,x \in U.$$  
Note that 
$$(m_0\, x)\backslash U = m_0\backslash(x\backslash U) \quad\hbox{for all $m_0\in M$}, \eqno{(1)}$$
since 
$$m\in (m_0\, x)\backslash U \;\Leftrightarrow 
		m\,m_0\,x\in U\;\Leftrightarrow 
			m\, m_0 \in x\backslash U\;\Leftrightarrow 
			m\in m_0 \backslash (x\backslash U).$$
Let $\Omega_M$ denote the set of all (left) ideals of $M$.  The special case of $(1)$ where $X=M$ and $U$ is an ideal of $M$ shows that $\Omega_M$ is an $M$-set, with the action defined by $(m, I)\mapsto m\backslash I$.  The largest ideal of $M$---namely, $M$ itself---will be denoted $1$.  The smallest ideal---namely, the empty set---will be denoted $0$. 

Let $X$ be an $M$-set.  We will show that there is a bijection between the set ${\cal P}_M(X)$ of sub-$M$-sets of $X$ and the set $\hom_M(X, \Omega_M)$ of $M$-set morphisms from $M$ to $\Omega_M$.  Let $U$ be a sub-$M$-set of $X$.  Then the map $\chi_U:X\to \Omega_M$ defined by $\chi_U(x):= x\backslash U$ is an $M$-set morphism by $(1)$, and 
$$\chi_U^{-1}(1)= U.\eqno{(2)}$$  Conversely, if $\phi: X\to \Omega_M$ is any $M$-set morphism, then
$$\eqalign{m\in x\backslash\phi^{-1}(1)&\Longleftrightarrow m\,x \in \phi^{-1}(1)\cr
                                       &\Longleftrightarrow \phi(m\,x) = 1\cr
                                       &\Longleftrightarrow m\backslash \phi(x) = 1\cr
                                       &\Longleftrightarrow 0 \in m\backslash \phi(x)\cr
                                       &\Longleftrightarrow m \in \phi(x)\cr}$$
This shows that $$\phi = \chi_{\phi^{-1}(1)}.\eqno{(3)}$$  Now, $(2)$ and $(3)$ show that $U\mapsto \chi_U$ is a bijection with inverse $\phi\mapsto\phi^{-1}(1)$.\hfill/////


\bigskip
\noindent
{\bf 2.6.~Definition.} The {\it product  of an $I$-indexed family of
$M$-sets
$\{\,X_i\mid i\in I\,\}$\/} is the set-theoretic product, with the
component-wise action: $(m\,y)_i := m\,(y_i)$, for $y\in  \prod_{i\in
I}X_i$.  A {\it left $M$-set congruence on $X$\/} is an equivalence
relation on $X$ that is also sub-left-$M$-set of $X\times X$.  The set of
equivalence classes of a congruence $C$ on $S$ is naturally an $M$-set.
It is denoted $X/C$ and is called the {\it quotient of $X$ by $C$\/}.


\bigskip
\noindent  {\bf 2.7.~Facts.}  We describe {\it free $M$-sets}.  Let
$E$ be
a set.  Then $E$ is a left $M$-set under the trivial action $(m,
y)\mapsto
y$, $y\in E$.  Suppose that $X$ is a left $M$-set and $g :E\to X$ is any function.
Then (prove this!) there is a unique left $M$-set morphism $\overline g
:M\times E\to X$ such that $\overline g (m, y) = m\,g(y)$.  Because of
this universal mapping property, $M\times E$ is said to be the {\it free
left $M$-set on $E$}.

\bigskip
\noindent  {\bf 2.8.~Fact.}  Suppose that $\phi:M\to Q$ is a morphism of
monoids.  This induces a functor ${\bf M}_\phi$ from the category of left
$Q$-sets to the category of left $M$-sets.  If $Y$ is a $Q$-set, ${\bf
M}_\phi(Y)$ is simply $Y$ endowed with the $M$-action $(m, y)\mapsto
\phi(m)y$.  To avoid confusion, we will use the notation $m\cdot_\phi
y$ to denote this action.  Thus, for a $Q$-set $Y$ and $m\in M$,
$$m\cdot_\phi y := \phi(m)\,y.$$

\bigskip
\noindent  We can define a functor ${\bf Q}_\phi$ from left $M$-sets to left $Q$-sets as follows.  Let $X$ be an $M$-set.  The underlying set of $X$
is $|X|$ and the free $Q$-set over $|X|$ is $Q\times |X|$.  We set
$${\bf Q}_\phi(X):= (Q\times |X|)/E(\phi, X),$$ where $E(\phi, X)$ is the
smallest $Q$-set congruence on $Q\times |X|$ such that
$$ (\phi(m), x)\,\sim_E\,(e, m'\,x)\;{\rm whenever}\;
\phi(m)=\phi(m').$$

\bigskip
\noindent  {\bf 2.9.~Theorem.}  ${\bf Q}_\phi$ is a functor and is left adjoint to ${\bf M}_\phi$.


\smallskip
\noindent  {\it Proof.} To see this is a functor, suppose $f:X\to Y$ is an $M$-set morphism. 
Then there is a unique $Q$-set morphism 
$$f_0: Q\times|X|\to {\bf Q}_\phi(Y)\,;\; f_0(q, x)=\overline{(q, f(x))}.$$  
Observe that if $\phi(m) = \phi(m')$, then 
$$f_0(\phi(m), x) = \overline{(\phi(m), f(x))} = \overline{(e, m'\,f(x))} = \overline{(e,
f(m'\,x))}= f_0(e, m'\,x).$$
This shows that $E(\phi, X)$ is contained in the kernel of $f_0$, and
hence $f_0$ induces a $Q$-set morphism 
$${\bf Q}_\phi(f):{\bf Q}_\phi(X)\to{\bf Q}_\phi(Y).$$
Now we will show that ${\bf Q}_\phi$ is left adjoint to ${\bf M}_\phi$,
using the criterion of [S.~MacLane, {\it Categories for the Working Mathematician\/},  Springer 1971. Chapter IV, Theorem 2 part(i), page 81].  Let $X$ be an $M$-set.  Define
the morphism
$\eta_X:X\to {\bf M}_\phi{\bf Q}_\phi(X)$ by $\eta_X(x) = \overline{(e,
x)}$.  This is an
$M$-set morphism because $$\eta_X(m\,x) = \overline{(e, m\,x)}=
\overline{(\phi(m)\,e,x)}=m\,\overline{(e, x)}=m\,\eta_X(x).$$  Let $Y$
be any $Q$-set and let $f:X\to {\bf M}_\phi(Y)$ be any $M$-set
morphism.    We will show that there is a unique $Q$-set morphism
$\overline f:{\bf Q}_\phi(X)\to Y$ such that $f = {\bf M}_\phi(\overline
f)\circ\eta_X$.  There is a $Q$-set morphism $f_0: Q\times|X|\to Y$
that is uniquely determined by setting $f_0\left((e,x)\right)=
f(x)$.  Now, $E(\phi, X)\subseteq \ker f_0$, because if
$\phi(m)=\phi(m')$, then
$$f_0\left((\phi(m),x)\right) = \phi(m)\,f(x) = \phi(m')\,f(x) =
m'\cdot_\phi f(x)=f(m'x) = f_0\left((e, m'x)\right).$$  Hence there is a
unique $Q$-set morphism $\overline f$ such that
$f_0 =
\overline f\circ p$, where $p$ is the canonical morphism $p:Q\times|X|\to
{\bf Q}_\phi(x)$, and from the definition of $f_0$, it is clear that $f =
\overline f\circ
\eta_X$.\hfill/////





\bigskip
\noindent{\it $M$-modules}

\bigskip
\noindent {\bf Definition.}  Let $M$ be a monoid.  A {\it left
$M$-module\/} is an abelian group $A$ equipped with a function
$M\times A\to A; (m,a)\mapsto ma$ such that $ea = a$ and $m(m'a) =
(m*m')a$ and $m(a+b)=ma+mb$ for
all $m,m'\in M$ and $a, b\in A$.  Right $M$-module is defined
analogously.

\bigskip
\noindent   From now on, unless otherwise stated, ``$M$-module'' means
left $M$-module.  A function $\phi:A\to A'$ between $M$-modules is a {\it
morphism of $M$-modules\/} if $\phi(m_1 a_1 + m_2 a_2) = m_1 \phi(a_1) +
m_2 \phi(a_2)$ for all $m_1, m_2\in M$ and $a_1, a_2\in A$.  A {\it
sub-$M$-module\/} of an $M$-module $A$ is a subgroup with the property
that $mb\in B$ whenever $m\in M$ and $b\in B$.

\bigskip
\noindent  {\bf Exercise.} Suppose that $B\subseteq A$ is a
sub-$M$-module, and $a\in A$.  Show that the quotient group $A/B$ is
naturally an $M$-module.  State and prove the First Isomorphism Theorem
for $M$-modules.

\bigskip
\noindent  {\bf Definition.}  Let $X$ be a left $M$-set and let $A$ an
abelian group.  We define the left $M$-module $A^{(X)}$ to be the abelian
group of all finitely supported functions $\overline a:X\to A; x\mapsto
\overline a_x$ with $M$ action defined as follows:
$$ (m\overline a)_x = \sum_{m*y=x} \overline a_y.$$  If $\overline a\in
A^{(X)}$, we can represent $\overline a$ as a formal sum $$\overline a =
\sum_{x\in X} \overline a_x x.$$  Here, of course, only finitely many of
the $\overline a_x$ are non-zero.  In this notation, the action of $M$
can be expressed in the (possibly more transparent) form:
$$m\sum_{x\in X} \overline a_x x = \sum_{x\in X} \overline a_x mx.$$

\bigskip
\noindent  {\bf Exercise.}  Let $A$ be an abelian group.  Define
$\epsilon:A\to A^{(M)}$ by $\epsilon(a)_e:=a$ and $\epsilon(a)_m:=0$ if
$m\not=e$ ($e$ being the neutral element of $M$). Show that $A^{(M)}$ has
the following universal mapping property: if $\phi:A\to B$ is any group
homomorphism from $A$ to an $M$-module $B$, then there is a unique
$M$-module morphism $\overline\phi :A^{(M)}\to B$ such that $\phi =
\overline\phi\ \epsilon$.  Show that $\left(\Z^{(I)}\right)^{(M)}$ is the
free $M$-module on the set $I$.

\bigskip
\noindent{\it $M$-Objects in general}

\bigskip
\noindent  The two constructions just described are special cases of a
more general construction.  Let $\bf C$ be a category.  An $M$-object in
$\bf C$ is an object $C$ of $\bf C$ together with a monoid morphism from
$M$ to the monoid $\hom(C,C)$ of $\bf C$-morphisms from $C$ to $C$.  This
is the same thing as a functor from $M$ to $\bf C$.  A morphism of
$M$-objects is a $\bf C$ morphism between $M$-objects that commutes with
the given actions of $M$ on the domain and codomain.  This is the same
thing as a natural transform between functors.

\bigskip
\noindent{\it Monoid Rings}

\bigskip
\noindent{\bf Definition.}  Let $R$ be a ring.  The monoid ring $R[M]$ is
the $M$-module $R^{(M)}$ equipped with the multiplication
$$(rr')_m:=\sum_{k*\ell=m}r_k {r'}_\ell.$$

\bigskip
\noindent  {\bf Exercises.}

\item{$i$)}Show that this multiplication can also be expressed as
follows:

$$\left(\sum_{m\in M} r_m m\right)\left(\sum_{n\in M} s_n n\right) =
\sum_{m\in M}\sum_{n\in M} r_m s_n (m\ast n).$$

\item{$ii$)}Prove that $R[M]$ is a ring.

\item{$iii$)} Is $R[M]$ and $M$-object in the category of rings?  Why or
why not?

\item{$iv$)}Let $A$ be a left $M$-module.  Show that $A$ is a left
$\Z[M]$-module if we define:

$$\left(\sum_{m\in M} n_m m\right) a = \sum_{m\in M} n_m(ma).$$

\end