\magnification=1200

\input AMSSYM.DEF
\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}

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

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


\bigskip
\noindent {\bf Definition.}  Let $M$ be a monoid.  A {\it left
$M$-set\/} is a set $X$ equipped with a function 
$M\times X\to X; (m,x)\mapsto mx$ such that $ex=x$ and $m(m'x) = (m\ast m')x$ for
all $m,m'\in M$ and $x\in X$.  A right $M$-set is a set $X$ equipped
with a
function $X\times M\to X; (x, m)\mapsto xm$ such that $xe = x$ and $(xm)m' = x(m\ast m')$
for all $m,m'\in M$ and $x\in X$.

\bigskip
\noindent  {\bf 2.1.~Facts.}
\smallskip
\item{1.}A congruence $C$ on $M$ is both a left and right
$M$-set, with the actions defined as follows:
$$\eqalign{M\times C \to C; (m, (m', m''))&\mapsto (m*m', m*m'')\cr C\times M\to C;
((m',m''), m)&\mapsto (m'*x,m''*m).\cr}$$  \item{2.}An equivalence relation $E$ on $M$
that is both a right and left $M$-set under the action described above is a
congruence.

\bigskip
\noindent  {\bf 2.2.~Definition.}  A left $M$-set morphism is a function $\phi$ between left $M$-sets that satisfies $m\phi(x) = \phi(mx)$. A sub-left-$M$-set of an $M$-set is a subset that is closed under the action of $M$.  The 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: $(my)_i := m(y_i)$, for $y\in  \prod_{i\in I}X_i$.  A 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 quotient of $X$ by $C$.


\bigskip
\noindent  {\bf 2.3.~Exercise.}  Let $I$ be a set.  Show that $I$ is a left $M$-set under the
trivial action $(m, i)\mapsto i$.  Suppose that $X$ is a left $M$-set and
$\phi:I\to X$ is any function.  Show that there is a unique left $M$-set
morphism $\overline\phi:M\times I\to X$ such that $\overline\phi(m, i)
= m\phi(i)$.  (In other words, $M\times I$ is the free left $M$-set on $I$.)

\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 2.4.~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 2.5.~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 2.6.~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