\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\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 5.  Generalities---very general---on graded algebras and modules}

\bigskip\noindent
{\bf 5.1.~Definition.}  
\smallskip  
\item{$i$)} Suppose $k$ is a ring and $M$ is a $k$-module.  Let $X$ be a set.  An {\it $X$-grading of $M$ as a $k$-module\/} is 
a decomposition of $M$ as an internal direct sum of $k$-modules: 
$$M=\bigoplus_{\xi\in X}M_\xi\;.$$
The elements of $M_\xi$ are said to be {\it homogeneous of degree $\xi$}; we write $\deg(m) = \xi$ if $m\in M_\xi$.  Since $0$ belongs to all the $M_\xi$, 0 does not have a unique degree.  Every other element of $M$ does. 

\item{$ii$)} Let $\Gamma$ be a monoid.  A {\it $\Gamma$-graded $k$-algebra\/} is a $k$-algebra $A$ which is $\Gamma$-graded as a $k$-module and which in addition satisfies: $$A_\gamma A_{\lambda}\subseteq A_{\gamma+\lambda}\quad\hbox{and}\quad 1_A\in A_0.$$

\item{$iii$)} Let $A$ be a $\Gamma$-graded $k$-algebra.  A {\it $(A, \Gamma)$-graded-module\/} (also referred to as a {\it graded $A$-module\/}, when the grading is obvious from context) is an $A$-module that is $\Gamma$-graded as a $k$-module, {\it i.e.}, 
$$M=\bigoplus_{\gamma\in \Gamma}M_\gamma\;,$$
and which in addition satisfies $$A_\gamma M_\lambda \subseteq M_{\gamma+\lambda}\;.$$
Note that $A$ itself is an $(A, \Gamma)$-graded-module. 
\item{$iii'$)} {\it Remark.}  We can consider a more general concept.  Let $X$ be a $\Gamma$-set and let $A$ be a $\Gamma$-graded $k$-algebra.  An $A$-module $M$ that is $X$-graded as a $k$-module and that satisfies in addition the condition that $A_\gamma M_\xi\subseteq M_{\gamma\xi}$ is a construct that is occasionally useful.

\bigskip\noindent
{\bf Examples.} 
\smallskip 
\item{$i$)} Let $S:=k[x_1,\ldots,x_n]$.  Then $S$ is an $\N^n$-graded $k$-algebra.   

\item{$ii$)}  The monoid algebra $k[\Gamma]$ is $\Gamma$-graded.  ($k[\Gamma]_\gamma$ is the free $k$-module generated by $X^\gamma$.)

\item{$iii$)}  Suppose $A$ is a $\Gamma$-graded algebra and $\phi:\Gamma\to \Lambda$ is a monoid homomorphism.  Then $A$ can be given a $\Lambda$-grading by defining for $\lambda\in\Lambda$:
$$A_\lambda:=\bigoplus\{\,A_\gamma\mid \phi(\gamma)=\lambda\,\}\;.$$  
For example, $\N^n\to \N; \alpha\mapsto \alpha_1+\cdots+\alpha_n$ gives the usual grading by total degree on $k[x_1,\ldots,x_n]$.

\item{$iv$)}  Let $L$ and $M$ be graded $A$-modules.  Then, $L\oplus M$ is $A$-graded if we define
$$(L\oplus M)_\gamma := L_\gamma\oplus M_\gamma\;.$$   (We could also give this module a grading of the kind proposed in $iii'$) by taking $X=\Gamma\times\Gamma$, which is a $\Gamma$-set by $\gamma(\alpha, \beta)= (\gamma+\alpha, \gamma+\beta)$.)

\item{$v$)} Let $A$ be a $\Gamma$-graded $k$-algebra and let $B$ be a $\Lambda$-graded $k$-algebra.  Then $A\otimes_k B$ is a $k$-algebra with multiplication determined by $(a\oplus b)(c\oplus d) = (ac\oplus bd)$ and the distributive law.  $A\otimes_k B$ has a grading by $\Gamma\times \Lambda$:
$$(A\otimes_k B)_{(\gamma,\lambda)}:= A_\gamma\otimes_k B_\lambda\;.$$  
With respect to this grading, $A\otimes_k B$ is a $(\Gamma\times \Lambda)$-graded-$k$-algebra.  Note that the map $a\mapsto a\otimes 1:A\to A\otimes_k B$ preserves degree: $\deg(a)=\deg(a\otimes 1)$.  

\item{$vi$)}  Suppose that $L$ is an $(A,\Gamma)$-graded-module and $M$ is a $(B,\Lambda)$-graded-module.  Then $A\otimes_k B$ acts on $L\otimes_k M$ by $(a\otimes b)(\ell\otimes m) = a\ell\otimes bm$.  With respect to this action,  $L\otimes_k M$ is a $(A\oplus_k B, \Gamma\times \Lambda)$-graded-module.  It is also an $(A,\Gamma)$-graded-module, via the action $(a, \ell\otimes m)\mapsto a\ell\otimes  m$, which is the action induced by the map $a\mapsto a\otimes 1$.    

\item{$vii$)}  Suppose $L$ and $M$ are $(A,\Gamma)$-graded-modules.  Then $L\otimes_k M$ is $\Gamma \times \Gamma$-graded.  Via the monoid morphism $+:\Gamma\times\Gamma\to \Gamma$, it is also $\Gamma$-graded.  Now $H:=\langle\, am\otimes n - m\otimes an\mid a\in A\,,\;m\in M \,,\;n\in N\,\rangle$ is obviously a $\Gamma$-homogeneous submodule of $M\otimes_k N$.  Because $M\otimes_A N\cong \left(M\otimes_k N\right)/H$, $M\otimes_A N$ is a $(A,\Gamma)$-graded-module.  Note that $\left(M\otimes_A N\right)_\alpha$ is the sub-$k$-module generated by $\{\,m\otimes n\mid \deg(m)+\deg(n)=\alpha\,\}$.

\bigskip\noindent
{\bf 5.2.~Definition.}  Let $M$ be a $\Gamma$-graded $A$-module.  We can shift the grading by $\gamma\in\Gamma$.  It is customary to denote the result $M(-\gamma)$, where the graded pieces are defined by:
$$M(-\gamma)_{\delta}:=\bigoplus\{\,M_{\lambda}\mid \lambda+\gamma=\delta\,\}\;.$$
If $\Gamma$ is cancellative, this gives $M(-\gamma)_{\delta}:=M_{\delta-\gamma}$ if $\delta\geq \gamma$ and $M(-\gamma)_{\delta}:=0$ otherwise.  If $m\in M$ and the degree of $m$ in $M$ is 0, then the degree of $m$ in $M(-\gamma)$ is $\gamma$.

\proclaim Lemma. $M(-\gamma) = A(-\gamma)\otimes_A M$.

\noindent {\it Proof.} It suffices to show that the graded pieces on each side are the same.
$$\eqalign{\left(A(-\gamma)\otimes_A M\right)_\delta &= \langle\, a m\mid (\deg(a)+\gamma)+\deg(m) = \delta\,\rangle\cr
                                                     &= \langle\, m\mid \deg(m)+\gamma = \delta\,\rangle.\cr}$$

 
\bigskip\noindent
{\bf 5.3.~Definition.}  Let $A$ be a $\Gamma$-graded $k$-algebra.  A homomorphism $\phi:M\to N$ of $A$-modules is said to be {\it homogeneous of degree $\delta$\/} if for all $\gamma\in \Gamma$: $$\phi(M_\gamma)\subseteq N_\gamma+\delta\;.$$  The category of $(A,\Gamma)$-graded-modules and homogeneous homomorphisms of degree 0 is denoted ${\cal M}_{A,\Gamma}$ or just ${\cal M}_A$ if $\Gamma$ is understood.

\bigskip
\noindent  
Let $L$ be an (ungraded) $A$-submodule of an $(A,\Gamma)$-graded-module $M$.  $L$ is said to be a graded submodule of $M$ if $L$ is $(A,\Gamma)$-graded and the inclusion map is homogeneous, {\it i.e.}, $L_\gamma\subset M_\gamma$, {\it i.e.}, $L$ is generated by homogeneous elements. 

\bigskip\noindent If $\phi:M\to N$ is a morphism of ${\cal M}_{A,\Gamma}$, then $\ker \phi$ is a graded submodule of $M$.  If $L$ is a graded submodule of $M$, then $M/L = \bigoplus_{\gamma\in\Gamma} M_\gamma/L_\gamma$ is an $(A, \Gamma)$-graded-module. 

\bigskip
\noindent{\it Free modules}

\medskip
\noindent The $A$-module $\bigoplus_n A$ has generators $\epsilon^i := (0,\ldots,0,1_A,0,\ldots,0)$.  Given an $A$-module $M$ (not graded) and a set of elements $m_i\in M$, there is a unique $A$-module map $\phi: \bigoplus_n A \to M$ such that $\phi(\epsilon^i) = m_i$.  

\smallskip
\noindent Suppose, now, that $A$ is $\Gamma$-graded.  Then, so is $\bigoplus_n A$ (see Example $iv$).  The degree of each $\epsilon_i$ is $0$.   Unless $\deg(m_i)=0$ for all $i$, however, $\phi$ will not be a morphism of ${\cal M}_{A, \Gamma}$.  To remedy this, we shift the grading on $\bigoplus_n A$, using $\bigoplus_{i=1}^n A(-\deg(m_i))$ instead.     





\end