Killing of supports on graded algebras
Roberto Mart�nez Villa
Instituto de Matem�ticas
UNAM
AP 61-3
58089 Morelia
Michoac�n
MEXICO
Manuel Saor�n
Departamento de Matem�ticas
Universidad de Murcia
Aptdo. 4021
30100 Espinardo
Murcia
SPAIN
{Killing of supports along subsets $\mathcal{U}$ of a group $G$ and regradings along certain maps of groups $\varphi :G'\longrightarrow G$ are studied, in the context of group-graded algebras. We show that, under precise conditions on $\mathcal{U}$ and $\varphi$, the module theories over the initial and the final algebras are functorially well-connected. Special attention is paid to $G=\mathbf{Z}$, in which case the results can be applied to $n$-Koszul algebras }
math0505076.tex
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\begin{document}

\title{Killing of supports  on graded  algebras}





\author{Roberto Mart�nez Villa\\
Instituto de Matem�ticas, UNAM, AP 61-3 \\
58089 Morelia, Michoac�n\\ MEXICO \\ {\it mvilla@matmor.unam.mx} \vspace*{0.5cm} \and Manuel Saor�n\\ Departamento de Matem�ticas\\
Universidad de Murcia, Aptdo. 4021\\
30100 Espinardo, Murcia\\
SPAIN\\ {\it msaorinc@um.es}}




\date{}



\thanks{
The first named author thanks CONACYT for funding the research
project. The second one thanks the D.G.I. of the Spanish Ministry
of Education and Science and the Fundaci�n "S�neca" of Murcia for
their financial support}









\maketitle



\begin{abstract}

{\bf Killing of supports along subsets $\mathcal{U}$ of a group
$G$ and regradings along certain maps of groups $\varphi
:G'\longrightarrow G$ are studied, in the context of group-graded
algebras. We show that, under precise conditions on $\mathcal{U}$
and $\varphi$, the module theories over the initial and the final
algebras are functorially well-connected. Special attention is
paid to  $G=\mathbf{Z}$, in which case the results can be applied
to $n$-Koszul algebras
 }

\end{abstract}




\section{Introduction}

In a recent paper (\cite{GMMVZ}, see also \cite{BM}),   the
authors showed that if $\Lambda$ is a $n$-Koszul graded algebra
and $\Lambda^!$ is its $n$-homogeneous dual, then the Yoneda
algebra $E=E(\Lambda )$ of $\Lambda$ is obtained as follows. One
takes $\mathcal{U}=n\mathbf{Z}\cup (n\mathbf{Z}+1)$,  kills the
part of the support of $\Lambda^!$ in
$\mathbf{Z}\setminus\mathcal{U}$ and then obtains a new
$\mathbf{Z}$-graded algebra $\Lambda^!_\mathcal{U}$ by keeping the
same multiplication of homogeneous elements,   when its degree
falls into $\mathcal{U}$, and making it zero otherwise.   Then $E$
is obtained from $\Lambda^!_\mathcal{U}$ by regrading along
certain function $\delta :\mathbf{Z}\longrightarrow\mathbf{Z}$
which has image $Im(\delta )=\mathcal{U}$.

Motivated by the above fact, we consider an arbitrary subset
$\mathcal{U}$ of a group $G$ and define a multiplication of
nonzero  homogeneous elements  on
$A_\mathcal{U}=\oplus_{u\in\mathcal{U}}A_u$ by the rule $a\cdot
b=ab$, in case $deg(a)deg(b)\in\mathcal{U}$, and $a\cdot b=0$
otherwise. We then extend by linearity and  call this operation
the {\bf support-restricted multiplication} on $A_\mathcal{U}$. In
this situation, we try to answer the following:

\begin{ques} \label{pregunta1}
When is it true that, for every $G$-graded algebra $A=\oplus_{g\in
G}A_g$, the subspace $A_\mathcal{U}=\oplus_{u\in\mathcal{U}}A_u$
inherites a structure of $G$-graded algebra with the
support-restricted multiplication?. For such a $\mathcal{U}$, how
are connected the graded module theories of $A$ and
$A_\mathcal{U}$?
\end{ques}

Also, concerning regradings, we try to answer the following:

\begin{ques} \label{pregunta2}
Let $G',G$ be groups and $\varphi :G'\longrightarrow G$ be a map.
When is it true that, for every $G$-graded algebra $B=\oplus_{g\in
G}B_g$ with support contained in $Im(\varphi )$, taking
$\tilde{B}_\sigma =B_{\varphi (\sigma )}$, for every $\sigma\in
G'$, gives rise to a $G'$-graded algebra $\tilde{B}$  which
coincides with $B$ as ungraded algebra?. In that case, how are
connected the graded module theories of $B$ and the (regraded)
$\tilde{B}$?
\end{ques}



  The structure of the paper goes as follows. In Section 2, we
  show that the
  answer to Question \ref{pregunta1} is: 'when $\mathcal{U}$ is a ring-supporting
  subset (see definition below) of $G$' (cf. Proposition
  \ref{caracterizacion-de-ring-supporting}).
We then introduce the concept of (right) modular pair of subsets
(see Definition \ref{definicion}) of $G$. If
$(\mathcal{S},\mathcal{U})$ is such a pair, in which case
$\mathcal{U}$ is ring-supporting, then the assigment
$M\rightsquigarrow M_\mathcal{S}$ gives an exact functor
$(-)_\mathcal{S}:Gr_A\longrightarrow Gr_{A_\mathcal{U}}$, which is
fully faithful when restricted to a well-described subcategory
$\mathcal{G}(\mathcal{S},\mathcal{U})$
 (Theorem \ref{fullyfaithful-killing-supports}). The essential
image of $\mathcal{G}(\mathcal{S},\mathcal{U})$ by
$(-)_\mathcal{S}$, denoted $\mathcal{L}(\mathcal{S},\mathcal{U})$,
contains all the  graded $A_\mathcal{U}$-modules presented in
degrees belonging to $(\mathcal{S}:\mathcal{U})=\{g\in G:$
$g\mathcal{U}=\mathcal{S}\}$. In Section 3, we show that the
answer to Question \ref{pregunta2} is: 'when $\varphi$ is a
pseudomorphism of groups'. Then we show that there is an
equivalence  between the category $\Sigma_{Im(\varphi )}$ of
$G$-graded $B$-modules $X$ with support $Supp(X)\subseteq
Im(\varphi )$ and the category $\tilde{\Sigma}_{Im(\varphi )}$ of
$G'$-graded $\tilde{B}$-modules $V$ such that
$V_\sigma\tilde{B}_\tau =0$ whenever $\varphi (\sigma )\varphi
(\tau )\not\in Im(\varphi )$ (cf. Proposition
\ref{caracterizacion-de-pseudomorphisms}). We then mix both
questions, namely, study the case when $\mathcal{U}$ is a
ring-supporting subset of $G$ which is the image of a
pseudomorphism $\varphi :G'\longrightarrow G$. Then, passing from
$A$ to $B=A_\mathcal{U}$ and from $B$ to $\tilde{B}$, we obtain an
equivalence of categories between the subcategory
$\mathcal{G}(\mathcal{S},\mathcal{U})$ of $Gr_A$ and a  full
subcategory $\mathcal{L}_{\tilde{B}}$ of $Gr_{\tilde{B}}$
(Proposition \ref{ring-supporting=image of pseudomorphism}).

In Section 4, we study the ring-supporting subsets
$\mathcal{U}\subseteq\mathbf{Z}$  of the integers, giving
essentially their structure (Theorem \ref{ring-supporting}). Then
precise criteria are given for a graded $A_\mathcal{U}$-module
 to
belong to $\mathcal{L}(\mathcal{S},\mathcal{U})$, criteria which
are specially simple when
 $\mathcal{U}$ is the translation of an inverval (cf. Corollary
 \ref{liftable-interval}). In addition, in this last case,
 $\mathcal{U}$ is the image of a precise pseudomorphism
 $\mathbf{Z}\longrightarrow\mathbf{Z}$. Then, for a
 positively  $\mathbf{Z}$-graded algebra $A=\oplus_{n\geq
 0}A_n$, taking $B=A_\mathcal{U}$ and then the regraded
 $\tilde{B}$, we obtain a precise description of the subcategories
 $\mathcal{G}(\mathcal{S},\mathcal{U})$ and
 $\mathcal{L}_{\tilde{B}}$ mentioned above. In case $\Lambda$ is a
 $n$-Koszul algebra and $A=\Lambda^!$, the results can be applied
 to show equivalences between appropriate full subcategories of
 $Gr_{\Lambda^!}$ and $Gr_E$ (cf. Remark \ref{n-Koszul}).




All algebras in the paper are associative with unit and modules
are right modules, unless explicitly said otherwise. Our basic
reference for graded rings and modules will be
 \cite{NVO}, while the terminology concerning torsion theories in Grothendieck categories can be found in \cite{St}.
 The few definitions and properties that we need about
  $n$-Koszul algebras can be found in \cite{B} and
 \cite{GMMVZ}. For the non-defined terms concerning algebras  and modules, we refer to
 \cite{ARS}.

\section{Killing of supports and induced equivalences}

In this and next section, we assume that $R$ is a commutative ring, $G$ is a group and $A=%
\underset{g\in G}{\oplus }A_{g}$ is a $G-$graded $R$-algebra.
 Let us fix a subset $\mathcal{S}$ of $G$
  and, for every graded $A$-module $M=\oplus_{g\in G}M_{g}$,  put
  $M_{\mathcal{S}}$ =$\oplus_{
s\in \mathcal{S}}M_{s}$,  letting $M_\emptyset=0$. We take
$\mathcal{T}=:\mathcal{T}_{\mathcal{S}}=\{ M\in Gr_A:$
$Supp(M)\subseteq G\setminus\mathcal{S} \}$, where $Supp(M)=\{g\in
G:$ $M_g\neq 0\}$ is the support of $M$. It is a hereditary
torsion class closed for products in $Gr_A$. We can view $R$ as a
$G$-graded ring, with $R_1=R$ and $R_g=0$ for $g\neq 1$, and then
an object of $Gr_R$ is just a family $(M_g)_{g\in G}$ of
$R$-modules. We view $M_\mathcal{S}$ as an object of $Gr_R$,  with
$(M_\mathcal{S})_g=0$ when $g\in G\setminus\mathcal{S}$.

\begin{prop} \label{faithful-killing-supports}
The assignment $M\longrightarrow M_{\mathcal{S}}$ gives an exact
functor, $(-)_{\mathcal{S}}:Gr_{A}\rightarrow Gr_R$ with kernel $
\mathcal{T}$.  Moreover, the induced functor
$Gr_{A}/\mathcal{T}\rightarrow Gr_R$ is faithful and identifies
$Gr_{A}/\mathcal{T}$ with a (not necessarily full) subcategory of
$Gr_R$.
\end{prop}

\begin{proof}
As a functor $Gr_{A}\longrightarrow Mod_R$ , $M\rightsquigarrow
M_{s}$,  is exact, for every $s\in \mathcal{S}$ . Consequently,
the assignment $M\rightsquigarrow M_{\mathcal{S}}$ yields an exact
functor: $Gr_{A}\rightarrow Gr_{R}$.

The kernel of $(-)_{\mathcal{S}}$ is $\{ M\in Gr_{A}:$
$M_{\mathcal{S}}=0\} =\mathcal{T} $.  Then the universal property
of the quotient category (cf \cite{G}) yields a unique $R$-linear
functor $F:Gr_{A}/\mathcal{T}\longrightarrow Gr_R$ such that the
diagram:

\vspace*{0.5cm}

\setlength{\unitlength}{1mm}
\begin{picture}(140,30)


\put(25,24){$Gr_A$} \put(32,25){\vector(1,0){53}}
\put(87,24){$Gr_R$} \put(55,2){$Gr_A/\mathcal{T}$}
\put(28,22){\vector(3,-2){26}} \put(62,5){\vector(3,2){25}}
\put(57,27){$(-)_\mathcal{S}$} \put(38,12){$p$} \put(79,13){$F$}
\end{picture}



is commutative, where $p$ is the canonical projection. All we need
to check is that $F$ is faithful. By \cite{G},  we have
$Hom_{Gr_{A}/ \mathcal{T}}(M,N)=\underset{N^{\prime
}\text{,}M/M'\in
\mathcal{T}}{\underrightarrow{\lim }}%
Hom_{Gr_{A}}(M^{\prime },N/N^{\prime }).$
 Here the order in the
index set is: $(M^{\prime },N^{\prime })\leq (M^{\prime \prime
},N^{\prime \prime })$ iff $M^{\prime \prime }\subseteq M^{\prime
}$ and $N^{\prime }\subseteq N^{\prime \prime }.$ In our situation
we have a unique maximum in that set. Indeed, since $\mathcal{T}$
is closed for products and subobjects in $Gr_A$, we have that
$\underset{M^{\prime }\leq M,M/M^{\prime }\in \mathcal{T}}{\cap
}M^{\prime }=\overset{\sim }{M}$ is a graded submodule of $M$ such
that $M/\overset{\sim }{M}\in \mathcal{T}.$ On the other hand,
$N^{\prime }\subseteq t(N),$ where $t(N)$ is by definition the
largest submodule of $N$ belonging to $\mathcal{T}.$

Therefore $(M^{\prime },N^{\prime })\leq (\overset{\sim
}{M},t(N))$ for all $(M^{\prime },N^{\prime })$ in the index
set.\newline
Then Hom$_{Gr_{A}/\mathcal{T}}($M$,$N$)$=Hom$%
_{Gr_{A}}$($\overset{\sim }{\text{M}}$,N/t(N))$.$

In order to prove the faithful condition of $F$ we need a more
precise
description of $\overset{\sim }{M}$ and $t(N).$ Notice that if \ $%
M^{\prime }\leq M$ is a graded submodule, then  $M/M^{\prime }\in
\mathcal{T}$ iff $M_{s}=M_{s}^{\prime }$ for every $s\in
\mathcal{S}$ iff  $M_{\mathcal{S}}\subseteq M^{\prime
}.$ As a consequence, $\overset{\sim }{M}=M_{\mathcal{S}}A$ is the graded $%
A- $ submodule of $M$ generated by $M_{\mathcal{S}}.$ On the other hand, in case $%
N^{\prime }\leq N$, we have that  $N^{\prime }\in \mathcal{T}$ iff
 $N_{s}^{\prime }=0$ for all $s\in \mathcal{S}.$ As a
consequence, $t(N)$ is the largest graded submodule of $N$
contained in $N_{G\smallsetminus \mathcal{S}}.$ In particular,
this implies that $N_{\mathcal{S}}=(N/$ $t(N)_{\mathcal{S}}$.

Now the canonical map
$Hom_{Gr_{A}}(M_{\mathcal{S}}A,N/t(N))\rightarrow
Hom_{Gr_R}(M_{\mathcal{S}},N_{\mathcal{S}})$
maps $f\in Hom_{Gr_{A}}(M_{\mathcal{S}}A,N/t(N))$ onto $%
f_{\mid M_{\mathcal{S}}}.$ If $F(f)=0,$ then $f_{\mid
M_{\mathcal{S}}}=0$ and, since $M_{\mathcal{S}}A$ is generated by
$M_{\mathcal{S}}$ as A- module, $f=0$. That proves the faithful
condition of $F$.

\end{proof}


An arbitrary killing of supports won't lead to a reasonably good
module theory over a new graded algebra. We need to impose some
restrictions.

\begin{defi} \label{definicion}
Let $G$ be a group and $(\mathcal{S},\mathcal{U})$ a pair of
nonempty subsets of $G$ such that $1\in\mathcal{U}$. We shall say
that $(\mathcal{S},\mathcal{U})$ is a  {\bf right premodular pair}
if whenever
$(s,u,v)\in\mathcal{S}\times\mathcal{U}\times\mathcal{U}$ is a
triple, with $suv\in\mathcal{S}$, one has that $su\in\mathcal{S}$
if, and only if, $uv\in\mathcal{U}$. By symmetry we get the notion
of left premodular pair.

A subset $\mathcal{U}\subseteq G$ containing $1$ will be called
{\bf ring-supporting} in case $(\mathcal{U},\mathcal{U})$ is a
(left or right) premodular pair. A  {\bf right} (resp. {\bf left})
{\bf modular pair} is a right (resp. left) premodular pair  such
that $\mathcal{U}$ is ring-supporting.
\end{defi}

The terms above are justified by the following results.

\begin{prop} \label{caracterizacion-de-ring-supporting}
Let $\mathcal{U}\subseteq G$ be a  subset of the group $G$. The
following assertions are equivalent:

\begin{enumerate}
\item $\mathcal{U}$ is ring-supporting \item For every $G$-graded
algebra $A=\oplus_{g\in G}A_g$, the sub-bimodule
$A_\mathcal{U}=\oplus_{u\in\mathcal{U}}A_u$ has a  structure of
$G$-graded $R$-algebra by putting $(A_\mathcal{U})_g=0$, when
$g\in G\setminus\mathcal{U}$, and considering the
support-restricted multiplication.

\end{enumerate}
\end{prop}

\begin{proof}
$1)\Longrightarrow 2)$ The ring-supporting condition of
$\mathcal{U}$ guarantees that the given multiplication in
$A_\mathcal{U}$ is associative and, from that, it follows easily
that $A_\mathcal{U}$ becomes a $G$-graded algebra.

$2)\Longrightarrow 1)$ Take the group algebra $A=RG$. Then
$A_\mathcal{U}=R\mathcal{U}=\oplus_{u\in\mathcal{U}}Ru$. Denote by
$\cdot$ the support-restricted multiplication on $R\mathcal{U}$.
Since $R\mathcal{U}$ has an identity, necessarily
$1\in\mathcal{U}$. On the other hand, in case
$u,v,w\in\mathcal{U}$ and $uvw\in\mathcal{U}$, one has $(u\cdot
v)\cdot w=uvw$ exactly when $uv\in\mathcal{U}$, and $uvw=u\cdot
(v\cdot w)$ exactly when $vw\in\mathcal{U}$. The associative
property of $\cdot$ then gives that $uv\in\mathcal{U}$ if, and
only if, $vw\in\mathcal{U}$.
\end{proof}


\begin{prop} \label{caracterizacion-de-right-modular}

Let $(\mathcal{S},\mathcal{U})$ be a pair of subsets of the group
$G$, where $\mathcal{U}$ is ring-supporting. The following
assertions are equivalent:

\begin{enumerate}
\item $(\mathcal{S},\mathcal{U})$ is right modular \item  For
every $G$-graded algebra and every $M\in Gr_A$, the sub-bimodule $M_{\mathcal{S}%
}$ =$\underset{s\in \mathcal{S}}{\oplus }M_{s}$ has a  structure
of graded right $A_{\mathcal{U}}-$ module by putting
$(M_{\mathcal{S}})_g=0$, when $g\in G\setminus\mathcal{S}$, and
defining the exterior multiplication of nonzero homogeneous
elements by: $x\cdot a=\left\{
\begin{array}{cc}
xa & \text{if }\deg (x)\text{deg( }a\text{)}\in \mathcal{S} \\
0 & \text{otherwise}%
\end{array}%
\right. .$

\end{enumerate}



In the above situation, the functor $(-)_\mathcal{S}$ of
Proposition \ref{faithful-killing-supports} may  be viewed as an
$R$-linear functor $Gr_A\longrightarrow Gr_{A_\mathcal{U}}$.

\end{prop}



\begin{proof}
  Along the lines of the previous proposition.
\end{proof}

We now give some properties of the just introduced subsets of a
group. In case $(\mathcal{S},\mathcal{U})$ is a  pair of subsets
of the group $G$, we shall put $(\mathcal{S}:\mathcal{U})=\{g\in
G:$ $g\mathcal{U}=\mathcal{S}\}$. If $1\in\mathcal{U}$ then
$(\mathcal{S}:\mathcal{U})$ is a (possibly empty) subset of
$\mathcal{S}$.

\begin{prop} \label{ejemplos-modular-pairs}
Let $G$ be a group and $\mathcal{S},\mathcal{U}$ be nonempty
subsets of $G$. The following assertions hold:
\begin{enumerate}
\item If $\mathcal{U}$ is  ring-supporting, then
$\mathcal{U}^{-1}$ is also  ring-supporting  \item The pair
$(\mathcal{S},\mathcal{U})$ is right modular if, and only if,
$(\mathcal{U}^{-1},\mathcal{S}^{-1})$ is left modular. \item
$(\mathcal{U},\mathcal{S})$ is  left modular  if, and only if,
$(\mathcal{S}^{-1},\mathcal{U})$ is  right modular  \item If
$\mathcal{U}$ is  ring-supporting  then
$(\mathcal{U},\mathcal{U}^{-1})$ is  left and right modular \item
If $1\in \mathcal{U}$  then $(\mathcal{U}:\mathcal{U})$ is a
subgroup of $G$ contained in $\mathcal{U}$ which, in case
$\mathcal{U}$ is ring-supporting, coincides with
$\mathcal{U}\cap\mathcal{U}^{-1}$ \item If
$(\mathcal{S},\mathcal{U})$ is a right modular pair and $g\in
(\mathcal{S}:\mathcal{U})$, then $\mathcal{S}=g\mathcal{U}$ and
$(\mathcal{S}:\mathcal{U})=g(\mathcal{U}:\mathcal{U})$
\end{enumerate}
\end{prop}


\begin{proof}
We leave to the reader the easy verification of 1 and 2. In 3, by
symmetry it will  be enough to check one of the implications.
Suppose then that $(\mathcal{U},\mathcal{S})$ is a left modular
pair.  We consider elements $s\in\mathcal{S}$ and
$u,v\in\mathcal{U}$ such that $s^{-1}uv\in\mathcal{S}^{-1}$ or,
equivalently, $v^{-1}u^{-1}s\in\mathcal{S}$. We want to prove that
$uv\in\mathcal{U}$ if, and only if, $s^{-1}u\in\mathcal{S}^{-1}$.
That is, we want to prove that $uv\in\mathcal{U}$ if, and only if,
$u^{-1}s\in\mathcal{S}$. But we have elements  $u$, $v$ in
$\mathcal{U}$ and $v^{-1}u^{-1}s\in\mathcal{S}$ with product
$uvv^{-1}u^{-1}s=s\in\mathcal{S}$. The left modularity of
$(\mathcal{U},\mathcal{S})$ gives that $uv\in\mathcal{U}$ if, and
only if, $vv^{-1}u^{-1}s=u^{-1}s\in\mathcal{S}$, as desired.

Assertion 4 follows from 2 and 3.

We prove now assertion 5. Clearly, $(\mathcal{U}:\mathcal{U})$ is
a subgroup of $G$ contained in $\mathcal{U}$. Then
$(\mathcal{U}:\mathcal{U})\subseteq\mathcal{U}\cap\mathcal{U}^{-1}$.
For the reverse inclusion in case $\mathcal{U}$ is
ring-supporting, take $u\in\mathcal{U}\cap \mathcal{U}^{-1}$ and
suppose that $u\not\in (\mathcal{U}:\mathcal{U})$. Then we have
two possibilities: i) there exists a $v\in\mathcal{U}$ such that
$uv\not\in\mathcal{U}$; ii) there exists $w\in\mathcal{U}$ such
that $w\not\in u\mathcal{U}$. In case i)  $u^{-1},u,v$ are
elements of $\mathcal{U}$ with product $u^{-1}uv=v\in\mathcal{U}$
such that $u^{-1}u=1\in\mathcal{U}$, but $uv\not\in\mathcal{U}$.
That contradicts the ring-supporting condition of $\mathcal{U}$.
In case ii), we have that $u^{-1}w\not\in\mathcal{U}$ and again we
get elements $u,u^{-1},w\in\mathcal{U}$ with product
$uu^{-1}w=w\in\mathcal{U}$ such that $uu^{-1}=1\in\mathcal{U}$,
but $u^{-1}w\not\in\mathcal{U}$. That is a contradiction. Then $u$
cannot exist, so that $\mathcal{U}\cap \mathcal{U}^{-1}\subseteq
(\mathcal{U}:\mathcal{U})$ as desired.

The proof of assertion 6 is left to the reader.
\end{proof}

\begin{lema} \label{torsionfree-cogeneration}
Let $A=\oplus_{g\in G}A_g$ be a $G$-graded algebra,
$\mathcal{S}\subseteq G$ be a subset and put
$\mathcal{T}=\mathcal{T}_\mathcal{S}$. The following assertions
are equivalent for a graded $A$-module $M$:

\begin{enumerate}
\item $M$ is $\mathcal{T}$-torsionfree (i.e. $Hom_{Gr_A}(T,M)=0$,
for all $T\in\mathcal{T}$) \item For every nonzero graded
submodule $N$ of $M$, one has $N_\mathcal{S}\neq 0$
\end{enumerate}
\end{lema}

\begin{proof}
If $T\in\mathcal{T}$ then $Supp(T)\subseteq
G\setminus\mathcal{S}$. Hence $M$ is $\mathcal{T}$-torsionfree if,
and only if, for every nonzero graded submodule $N$ of $M$, one
has $Supp(N)\not\subseteq G\setminus\mathcal{S}$.
\end{proof}


If $M\in Gr_A$ satisfies the conditions of the above lemma, we
shall indistinctly say that $M$ is $\mathcal{T}$-torsionfree or
that $M$ is {\bf cogenerated in degrees belonging to}
$\mathcal{S}$. In case $(\mathcal{S}, \mathcal{U})$ is a right
modular pair of subsets of $G$, the induced functor $F:Gr_{A}/
\mathcal{T}\longrightarrow Gr_{A_\mathcal{U}}$ is not full. We
shall see that it is full when restricted to an appropriate full
subcategory of $Gr_{A}/\mathcal{T}$.

\begin{lema} \label{posterior}
Let $(\mathcal{S},\mathcal{U})$ be a right modular pair of subsets
of the group $G$ and $A=\oplus_{\sigma\in G}A_\sigma$ be a
$G$-graded algebra. Suppose that $g\in (\mathcal{S}:\mathcal{U})$.
Then we have a commutative diagram of functors:

 \vspace*{0.5cm}

\setlength{\unitlength}{1mm}
\begin{picture}(140,30)


\put(42,25){$Gr_A$}\put(50,26){\vector(1,0){15}}
\put(68,25){$Gr_{A}$}



\put(42,4){$Gr_{A_\mathcal{U}}$}\put(50,5){\vector(1,0){15}}
\put(69,4){$Gr_{A_\mathcal{U}}$}

  \put(45,23){\vector(0,-1){14}}
\put(72,23){\vector(0,-1){14}}

\put(36,16){$(-)_{\mathcal{U}}$}  \put(73,16){$(-)_{\mathcal{S}}$}

\put(58,27){$\cong$}   \put(58,2){$\cong$}

\end{picture}

where the horizontal arrows are the shifting equivalences
$?[g^{-1}]:  V\rightsquigarrow V[g^{-1}]$
\end{lema}

\begin{proof}
We just need to prove that
$M[g^{-1}]_\mathcal{S}=M_\mathcal{U}[g^{-1}]$, for every $M\in
Gr_A$. Indeed $Supp(M_\mathcal{U}[g^{-1}])=\{h\in G:$ $g^{-1}h\in
Supp(M_\mathcal{U})\}$, which is clearly contained in
$g\mathcal{U}=\mathcal{S}$. Now if $s\in\mathcal{S}$ then
$M[g^{-1}]_s=M_{g^{-1}s}$ while
$M_\mathcal{U}[g^{-1}]_s=(M_\mathcal{U})_{g^{-1}s}$. But
$(M_\mathcal{U})_{g^{-1}s}=M_{g^{-1}s}$ since
$g^{-1}s\in\mathcal{U}$.

\end{proof}

 In order to give our next result we
consider the full subcategory
$\mathcal{G}(\mathcal{S},\mathcal{U})$ of $Gr_A$ with objects the
graded $A$-modules generated in degrees belonging to
$(\mathcal{S}:\mathcal{U})$ and cogenerated in degrees belonging
to $\mathcal{S}$ (i.e. $\mathcal{T}_\mathcal{S}$-torsionfree). On
the other hand, the objects of the essential image of the functor
$(-)_{\mathcal{S}}:Gr_A\longrightarrow Gr_{A_{\mathcal{U}}}$ will
be called {\bf liftable} graded $A_{\mathcal{U}}$-modules with
respect to $(-)_\mathcal{S}$. We shall denote by
$\mathcal{L}(\mathcal{S},\mathcal{U})$ the full subcategory of
$Gr_{A_\mathcal{U}}$ with objects the liftable
$A_\mathcal{U}$-modules generated in degrees belonging to
$(\mathcal{S}:\mathcal{U})$. A graded module (over any $G$-graded
algebra) $M$ will be called {\bf presented in degrees belonging
to} $\mathcal{X}\subseteq G$, when there is an exact sequence of
graded modules $P\longrightarrow Q\twoheadrightarrow M\rightarrow
0$, with $P,Q$ gr-projective and generated in degrees belonging to
$\mathcal{X}$.  We are now ready to prove:






\begin{teor} \label{fullyfaithful-killing-supports}
Let $(\mathcal{S},\mathcal{U})$ be a right modular pair of subsets
of $G$ and $\mathcal{T}=\mathcal{T}_\mathcal{S}$. The following
assertions hold:

\begin{enumerate}
\item The composition
$\mathcal{G}(\mathcal{S},\mathcal{U})\hookrightarrow
Gr_A\stackrel{p}{\longrightarrow}Gr_A/\mathcal{T}$ is fully
faithful \item The restriction of $(-)_\mathcal{S}$ to
$\mathcal{G}(\mathcal{S},\mathcal{U})$ is a fully faithful functor
which induces an equivalence of categories
$\mathcal{G}(\mathcal{S},\mathcal{U})\stackrel{\cong}{\longrightarrow}
\mathcal{L}(\mathcal{S},\mathcal{U})$ \item For every $g\in
(\mathcal{S}:\mathcal{U})$, there is a commutative diagram of
equivalences of categories:

 \vspace*{0.5cm}

\setlength{\unitlength}{1mm}
\begin{picture}(140,30)


\put(38,25){$\mathcal{G}(\mathcal{U},\mathcal{U})$}\put(50,26){\vector(1,0){15}}
\put(68,25){$\mathcal{G}(\mathcal{S},\mathcal{U})$}



\put(38,4){$\mathcal{L}(\mathcal{U},\mathcal{U})$}\put(50,5){\vector(1,0){15}}
\put(69,4){$\mathcal{L}(\mathcal{S},\mathcal{U})$}

  \put(45,23){\vector(0,-1){14}}
\put(72,23){\vector(0,-1){14}}

\put(36,16){$(-)_{\mathcal{U}}$} \put(46,16){$\cong$}
  \put(73,16){$(-)_{\mathcal{S}}$}

\put(58,27){$\cong$} \put(69,16){$\cong$}  \put(58,2){$\cong$}

\end{picture}

where the horizontal arrows are the shifting equivalences
$?[g^{-1}]$
\end{enumerate}

Moreover, $\mathcal{L}(\mathcal{S},\mathcal{U})$ contains all the
graded $A_\mathcal{U}$-modules presented in degrees belonging to
$(\mathcal{S}:\mathcal{U})$
\end{teor}

\begin{proof} Let us take $M,N\in\mathcal{G}(\mathcal{S},\mathcal{U})$. By the proof
of Proposition \ref{faithful-killing-supports},  we know that
$Hom_{Gr_A/\mathcal{T}}(M,N)=Hom_{Gr_A}(\tilde{M},N/t(N))$, where
$\tilde{M}$ is the $A$-submodule of $M$ generated by
$M_\mathcal{S}$. Since
$(\mathcal{S}:\mathcal{U})\subseteq\mathcal{S}$ and $M$ is
generated in degrees belonging to $(\mathcal{S}:\mathcal{U})$,  we
conclude that $\tilde{M}=M$. On the other hand, since $N$ is
torsionfree we get $t(N)=0$ and, hence,
$Hom_{Gr_A/\mathcal{T}}(M,N)=Hom_{Gr_A}(M,N)$, which proves
assertion 1.

By assertion 1, we can view $\mathcal{G}(\mathcal{S},\mathcal{U})$
as a full subcategory of $Gr_A/\mathcal{T}$, and then  the
restriction of $(-)_\mathcal{S}$ to
$\mathcal{G}(\mathcal{S},\mathcal{U})$, which we denote by $H$ in
the sequel,  can be identified with the restriction of
$F:Gr_A/\mathcal{T}\longrightarrow Gr_{A_\mathcal{U}}$ to
$\mathcal{G}(\mathcal{S},\mathcal{U})$. Since $F$ is faithful (cf.
Proposition \ref{faithful-killing-supports}) $H$ is faithful. We
next prove that it is full. Let
$M,N\in\mathcal{G}(\mathcal{S},\mathcal{U})$ and
$M_{\mathcal{S}}\overset{g}{\rightarrow }N_{\mathcal{S}}$ be a
morphism in $Gr_{A_{\mathcal{U}}}.$
Notice that  we have epimorphisms $\underset{i\in I}{%
\oplus }A\left[ s_{i}^{-1}\right] $ $\overset{\varepsilon _{M}}{\rightarrow }%
M$ and $\underset{j\in J}{\oplus }A\left[ t_{j}^{-1}\right] \overset{%
\varepsilon _{N}}{\rightarrow }N$ in $Gr_{A},$ where
$(s_{i})_{i\in I}$ and $(t_{j})_{j\in J}$ are families in
$(\mathcal{S}:\mathcal{U}).$ From the proof of Lemma
\ref{posterior} we get that
$A[g^{-1}]_\mathcal{S}=A_\mathcal{U}[g^{-1}]$, for every $g\in
(\mathcal{S}:\mathcal{U})$. Then, by applying $(-)_{\mathcal{S}}$,  we get epimorphisms $\underset{%
i\in I}{\oplus }A_{\mathcal{U}}\left[ s_{i}^{-1}\right] $ $\overset{\overset{%
\sim }{\varepsilon }_{M}}{\rightarrow }M_{\mathcal{S}}$ and $\underset{j\in J%
}{\oplus }A_{\mathcal{U}}\left[ t_{j}^{-1}\right] \overset{\overset{\sim }{%
\varepsilon }_{N}}{\rightarrow }N_{\mathcal{S}}$ in
$Gr_{A_\mathcal{U}}$
 By the projective condition of $\underset{i\in I}{\oplus }A_{\mathcal{U}}\left[ s_{i}^{-1}%
\right] $ in $Gr_{A_\mathcal{U}}$ we get a commutative diagram in
this latter category:

\begin{center}
$%
\begin{array}{ccc}
\underset{i\in I}{\oplus }A_{\mathcal{U}}\left[ s_{i}^{-1}\right] & \overset{%
\overset{\sim }{\varepsilon }_{M}}{\rightarrow } & M_{\mathcal{S}} \\
\downarrow \phi &  & \downarrow g \\
\underset{j\in J}{\oplus }A_{\mathcal{U}}\left[ t_{j}^{-1}\right] & \overset{%
\overset{\sim }{\varepsilon }_{N}}{\rightarrow } & N_{\mathcal{S}}%
\end{array}%
$
\end{center}

Recall that $\underset{i\in I}{\oplus }A_{\mathcal{U}}\left[ s_{i}^{-1}%
\right] =A_{\mathcal{U}}^{(I)}$ as ungraded right
$A_{\mathcal{U}}-$ module,
with the grading obtained by assigning degree $s_{i}$ to the i-th element $%
e_{i}$ of the canonical basis of $A_{\mathcal{U}}^{(I)}$ as
$A_{\mathcal{U}}
$ - module. Then $\phi (e_{i})$ is an element of $\underset{j\in J}{\oplus }%
A_{\mathcal{U}}\left[ t_{j}^{-1}\right] $ of degree $s_{i}.$ i.e.
$\phi
(e_{i})$ $\in $ $\underset{j\in J}{\oplus }A_{\mathcal{U}}\left[ t_{j}^{-1}%
\right]_{s_i} =\underset{j\in J}{\oplus }A_{t_{j}^{-1}s_{i}}.$ We define now $%
\overset{\_}{\phi }:\underset{i\in I}{\oplus }A\left[ s_{i}^{-1}\right] $ $%
\rightarrow $ $\underset{j\in J}{\oplus }A\left[ t_{j}^{-1}\right]
$ as the
only homomorphism of (graded) right $A$-modules such that $\overset{\_}{\phi }%
(e_{i})=\phi (e_{i})$ for all $i.$

This yields a diagram in $Gr_{A}:$

\begin{center}
$%
\begin{array}{ccc}
\underset{i\in I}{\oplus }A\left[ s_{i}^{-1}\right] &
\overset{\varepsilon
_{M}}{\rightarrow } & M \\
\downarrow \overset{\_}{\phi } &  &  \\
\underset{j\in J}{\oplus }A\left[ t_{j}^{-1}\right] & \overset{\varepsilon _{N}}%
{\rightarrow } & N%
\end{array}%
$
\end{center}

We claim that  $\varepsilon _{N}\overset{\_}{\phi }$ vanishes on $%
Ker\varepsilon _{M},$ which implies the existence of a unique morphism $%
f:M\rightarrow N$ in $Gr_{A}$ such that $\varepsilon _{N}%
\overset{\_}{\phi }$ = $f\varepsilon _{M}.$ From that one easily
gets that $f_\mathcal{S}=g$ and the full condition for $H$ will
follow. In order to prove our claim, let $x\in Ker\epsilon _{M}$
be a homogeneous element. If $\deg (x)\in \mathcal{S}$, then $x\in
Ker(\overset{\sim}{\varepsilon}_M)$, so that
$\varepsilon_N\bar{\phi}(x)=\overset{\sim}{\varepsilon}_N\phi
(x)=g\overset{\sim}{\varepsilon}_M(x)=0$. In case $\deg (x)\notin
\mathcal{S}$,
we have $\deg ((\varepsilon _{N}\overset{\_}{\phi }%
)(x))=\deg (x)\notin \mathcal{S}$. In either case,  every element
in
$(\varepsilon _{N}\overset{%
\_}{\phi })(Ker\varepsilon _{M})$ has support in $G\smallsetminus
\mathcal{S} $ . This implies $(\varepsilon _{N}\overset{\_}{\phi
})(Ker\varepsilon _{M})\subseteq t(N)=0$  as desired.

In order to end the proof of assertion 2, it only remains to prove
that if $X\in\mathcal{L}(\mathcal{S},\mathcal{U})$  then there is
a $M\in Gr_A$ such that $M_\mathcal{S}=X$ and $M$ is generated in
degrees belonging to $(\mathcal{S}:\mathcal{U})$. In that case, we
could always assume that $M$ is $\mathcal{T}$-torsionfree, so that
$M\in\mathcal{G}(\mathcal{S},\mathcal{U})$. Since $X$ is liftable,
we can  fix a $N\in Gr_A$ such that $N_\mathcal{S}=X$. We take a
family $(x_i)_{i\in I}$ of homogeneous generators of $X$, where
$deg(x_i)=:s_i\in (\mathcal{S}:\mathcal{U})$ for all $i\in I$.
Then $x_i\in X_{s_i}=N_{s_i}$ and we can consider $M=\sum_{i\in
I}x_iA$, i.e., the graded $A$-submodule of $N$ generated by the
$x_i$. Then $X=\sum_{i\in I}x_iA_\mathcal{U}\subseteq
M_\mathcal{S}$, due to the fact that $s_i\in
(\mathcal{S}:\mathcal{U})$ for all $i\in I$. From that we get that
$X\subseteq M_\mathcal{S}\subseteq N_\mathcal{S}=X$ and, hence,
equalities. Then $M_\mathcal{S}=X$ as desired.

Assertion 3 follows directly from Lemma \ref{posterior}. For the
final statement, notice that every gr-projective
$A_\mathcal{U}$-module $P$ generated in degrees belonging to
$(\mathcal{S}:\mathcal{U})$ is a direct summand of $\oplus_{i\in
I}A_\mathcal{U}[g_i^{-1}]$, where $(g_i)_{i\in I}$ is a family of
elements of $(\mathcal{S}:\mathcal{U})$. The proof of Lemma
\ref{posterior} implies that $P$ is liftable and, hence, belongs
to $\mathcal{L}(\mathcal{S},\mathcal{U})$. But, due to assertion 2
and the exactness of $(-)_\mathcal{S}$, the class
$\mathcal{L}(\mathcal{S},\mathcal{U})$ is closed for cokernels.
\end{proof}













\begin{ejem}
Let $G$ be a group, $A=\oplus_{g\in G}A_g$ be a $G$-graded algebra
and $H<G$ be a subgroup. We put $\mathcal{S}=\mathcal{U}=H$. Then
$A_H=\oplus_{h\in H}A_h$ is in this case a subalgebra of $A$.
Moreover, $(\mathcal{S}:\mathcal{U})=(H:H)=\{g\in G:$ $gH=H\}=H$.
In this case, $\mathcal{G}(H,H)$ is the full subcategory of $Gr_A$
with objects those graded $A$-modules $M$ such that $M=M_HA$ and
$M$ contains no nonzero graded submodule with support in
$G\setminus H$.  Our Theorem \ref{fullyfaithful-killing-supports}
says that the functor $(-)_H:Gr_A\longrightarrow Gr_{A_H}$
establishes equivalences of categories
$\mathcal{G}(H,H)\stackrel{\cong}{\longrightarrow}
Gr_A/\mathcal{T}\stackrel{\cong}{\longrightarrow}Gr_{A_H}$,
understanding $Gr_{A_H}$ as the category of $G$-graded
$A_H$-modules with support in $H$. The reason is that every graded
$A_H$-module $X$ is liftable, because $M=:X\otimes_{A_H}A$ is a
graded $A$-module generated in degrees belonging to $H$ such that
$M_H\cong X$.
\end{ejem}


\section{Pseudomorphisms of groups and regradings}


In this section we answer Question \ref{pregunta2} (see the
introduction).

\begin{defi}
Let $G,G'$ be groups. A {\bf pseudomorphism} $\varphi
:G'\longrightarrow G$ is an injective map satisfying the following
two properties:

\begin{enumerate}
\item $\varphi (1)=1$ \item If $\sigma ,\tau\in G'$ and
 $\varphi (\sigma )\varphi (\tau )\in Im(\varphi )$, then
 $\varphi (\sigma\tau )=\varphi (\sigma )\varphi (\tau )$
 \end{enumerate}
\end{defi}

\begin{prop} \label{caracterizacion-de-pseudomorphisms}
Let $G',G$ be groups and $\varphi :G'\longrightarrow G$ be a map. The following
assertions are equivalent:

\begin{enumerate}
\item[a)] $\varphi$ is a pseudomorphism of groups \item[b)]  $1\in
Im(\varphi )$ and, for every $G$-graded algebra $B=\oplus_{g\in
G}B_g$ with support contained in $Im(\varphi )$, putting
$\tilde{B}=B$ and $\tilde{B}_\sigma =B_{\varphi (\sigma )}$, for
every $\sigma\in G'$, one obtains a $G'$-graded algebra
$\tilde{B}$ coinciding with $B$ as an ungraded algebra
\end{enumerate}

If $B$ and $\tilde{B}$ are $G$-graded and $G'$-graded algebras as
in b), then, for every element $g\in G$, there is an equivalence
of (abelian) categories between:

\begin{enumerate}
\item The full subcategory $\Sigma_{gIm(\varphi )}$ of $Gr_B$ with
objects the graded $B$-modules $X$ such that $Supp(X)\subseteq
gIm(\varphi )$ \item  The full subcategory
$\tilde{\Sigma}_{Im(\varphi )}$ of $Gr_{\tilde{B}}$ with objects
the graded $\tilde{B}$-modules $V$ such that
$V_\sigma\tilde{B}_\tau=0$ whenever $\varphi (\sigma )\varphi
(\tau )\not\in Im(\varphi )$
\end{enumerate}
\end{prop}

\begin{proof}
$a)\Longrightarrow b)$ Let $B=\oplus_{g\in G}B_g$ be any
$G$-graded algebra with $Supp(B)\subseteq Im(\varphi )$, and put
$\tilde{B}=B$ and $\tilde{B}_\sigma =B_{\varphi (\sigma )}$, for
all $\sigma\in G'$. Then
$\tilde{B}_\sigma\tilde{B}_\tau=B_{\varphi (\sigma )}B_{\varphi
(\tau )}$ is  zero,  in case $\varphi (\sigma )\varphi (\tau
)\not\in Im(\varphi )$, and is an $R$-submodule of $B_{\varphi
(\sigma\tau )}=\tilde{B}_{\sigma\tau}$, in case $\varphi (\sigma
)\varphi (\tau )\in Im(\varphi )$. Hence we always have
$\tilde{B}_\sigma\tilde{B}_\tau\subseteq\tilde{B}_{\sigma\tau}$,
so that $\tilde{B}$ is a $G'$-graded algebra.

$b)\Longrightarrow a)$ Since $1\in Im(\varphi )$, by taking
$G$-graded algebras concentrated in degree $1$, we can ensure that
there exist graded $G$-algebras with support contained in
$Im(\varphi )$. Moreover, for any such algebra $B=\oplus_{g\in
G}B_g$, the hypothesis says that putting $\tilde{B}=B$ and
$\tilde{B}_\sigma =B_{\varphi (\sigma )}$, for every $\sigma\in
G'$, we get a $G'$-graded algebra $\tilde{B}$ coinciding with $B$
as an ungraded algebra. Since the identity of $\tilde{B}=B$ is
always a homogeneous element of degree $1$, we conclude that
$1\in\tilde{B}_1=B_{\varphi (1)}$ and $1\in B_1$. But, by
definition of $G$-grading, the sum $\oplus_{g\in G}B_g$ is direct.
It follows that $\varphi (1)=1$.

On the other hand, if $1\neq g\in Im(\varphi )$
  then we can always find a $G$-graded algebra $B$, with $Supp(B)\subseteq Im(\varphi
)$, such that $B_g\neq 0$. Indeed, take the algebra $B=R[x]/(x^2)$
with $G$-grading given by putting $B_1=R$ and $deg(\bar{x})=g$.
Suppose now that $\sigma ,\tau\in G'$ are elements such that
$\varphi (\sigma )=\varphi (\tau)$. Then we consider any
$G$-graded algebra $B$, with $Supp(B)\subseteq Im(\varphi )$, such
that $B_{\varphi (\sigma )}\neq 0$. Then we have that
$\tilde{B}_\sigma =B_{\varphi (\sigma )}=B_{\varphi (\tau
)}=\tilde{B}_\tau$ is nonzero.  Then, by definition of
$G'$-grading,  we necessarily have $\sigma =\tau$. That proves
that $\varphi$ is injective.

Finally,  if $\sigma ,\tau\in G'$ and $\varphi (\sigma )\varphi
(\tau )\in Im(\varphi )$, then we have
$\tilde{B}_\sigma\tilde{B}_\tau=B_{\varphi (\sigma )}B_{\varphi
(\tau )}\subseteq B_{\varphi (\sigma )\varphi (\tau )}$ and, since
$\tilde{B}$ is a $G'$-graded algebra, we also have
$\tilde{B}_\sigma\tilde{B}_\tau\subseteq\tilde{B}_{\sigma\tau}=B_{\varphi
(\sigma\tau )}$. It follows that $\varphi (\sigma\tau )=\varphi
(\sigma )\varphi (\tau )$ unless $B_{\varphi (\sigma )}B_{\varphi
(\tau )}=0$, for every graded $G$-algebra $B$ with
$Supp(B)\subseteq Im(\varphi )$. We should then discard this last
possibility, for which we prove that if $g,h\in G\setminus{1}$ are
elements such that $g,h,gh\in Im(\varphi )$, then there exists a
$G$-graded algebra $B$, with $Supp(B)\subseteq Im(\varphi )$, such
that $B_gB_h\neq 0$. We consider all the possibilities:

\begin{enumerate}
\item[i)] $g=h$ and $gh=1$:  Take the group algebra $B=RC$, where
$C=<g>$ is the subgroup of $G$ generated by $g$ with the obvious
$G$-grading  \item[ii)] $g=h$ and $gh\neq 1$: Take $B=R[x]/(x^3)$,
$B_1=R$ and $deg(\bar{x})=g$  \item[iii)] $g\neq h$ and $gh=1$:
Take $B=R<x,y>/(x^2,y^2,yx)$, with $B_1=R+R\bar{x}\bar{y}$,
$deg(\bar{x})=g$ and $deg(\bar{y})=h$ \item[iv)] $g\neq h$ and
$gh\neq 1$: Take $B=R<x,y>/(x^2,y^2,yx)$, with $B_1=R$,
$deg(\bar{x})=g$ and $deg(\bar{y})=h$
\end{enumerate}

For the final part, given a subset $\mathcal{X}\subseteq G$, we
denote by $\Sigma_\mathcal{X}$ the full subcategory of $Gr_B$ with
objects those $M\in Gr_B$ such that $Supp(M)\subseteq\mathcal{X}$.
It is clear that, for every $g\in G$, the shifting equivalence
$?[g]:Gr_B\stackrel{\cong}{\longrightarrow}Gr_B$ induces by
restriction an equivalence $\Sigma_{gIm(\varphi
)}\cong\Sigma_{Im(\varphi )}$. So it is not restrictive to assume,
something that we do in the rest of the proof, that $g=1$. Let
then  $X=\oplus_{g\in G}X_g$ be a graded $B$-module with
$Supp(X)\subseteq Im(\varphi )$. Then we define $\tilde{X}=X$ as
ungraded $\tilde{B}$- (or $B$-)module and give it a structure of
$G'$-graded $\tilde{B}$-module by putting $\tilde{X}_\sigma
=X_{\varphi (\sigma )}$, for all $\sigma\in G'$. It is a mere
routine to check that the assignment $X\rightsquigarrow\tilde{X}$
defines a fully faithful exact functor $F:\Sigma_{Im(\varphi
)}\longrightarrow Gr_{\tilde{B}}$. Moreover, if $\sigma ,\tau\in
G'$ are elements such that $\varphi (\sigma )\varphi (\tau
)\not\in Im(\varphi )$, in which case $X_{\varphi (\sigma )\varphi
(\tau )}=0$, then we have $\tilde{X}_\sigma\tilde{B}_\tau
=X_{\varphi (\sigma )}B_{\varphi (\tau )}\subseteq X_{\varphi
(\sigma )\varphi (\tau )}=0$, thus proving that
$Im(F)\subseteq\tilde{\Sigma}_{Im(\varphi )}$. Conversely, if
$V\in\tilde{\Sigma}_{Im(\varphi )}$ then we can form a $G$-graded
$B$-module $X$ as follows. We put $X=V$ as ungraded $B-$ (or
$\tilde{B}-$)module and define $X_g=0$, when $g\not\in Im(\varphi
)$, and $X_g=V_\sigma$ provided $g=\varphi (\sigma )$. The
injective condition of $\varphi$ implies that the given
$G$-grading on $X$ is well-defined. We leave to the reader the
routinary task of checking that $X\in\Sigma_{Im(\varphi )}$ and
$\tilde{X}=V$. From that it follows that $F$ induces the desired
equivalence $\Sigma_{Im(\varphi
)}\stackrel{\cong}{\longrightarrow}\tilde{\Sigma}_{Im(\varphi )}$.


\end{proof}




It turns out that some of the most interesting ring-supporting
subsets of $\mathbf{Z}$ are the image of a pseudomorphism (see
next section). It is then of interest to consider the case when
$\mathcal{U}\subseteq G$ is a ring-supporting subset such that,
for some pseudomorphism $\varphi :G'\longrightarrow G$, one has
$Im(\varphi )=\mathcal{U}$. In that situation, starting with any
$G$-graded algebra $A$, we can take $B=A_\mathcal{U}$ in the above
proposition, so that we get a $G'$-graded algebra
$\tilde{B}=\tilde{A}_\mathcal{U}$. If now
$(\mathcal{S},\mathcal{U})$ is a right modular pair of subsets of
$G$ such that $(\mathcal{S}:\mathcal{U})\neq\emptyset$ then,
according to Proposition \ref{ejemplos-modular-pairs}(6), we have
$\mathcal{S}=g\mathcal{U}=gIm(\varphi )$, where $g\in
(\mathcal{S}:\mathcal{U})$. Moreover, since the objects in the
image of the functor $(-)_\mathcal{S}:Gr_A\longrightarrow
Gr_{A_\mathcal{U}}$ are graded $A_\mathcal{U}$-modules with
support in $\mathcal{S}$, we get a well-defined $R$-lineal exact
functor $\Phi :Gr_A\longrightarrow Gr_{\tilde{B}}$, which is the
following composition
$Gr_A\stackrel{(-)_\mathcal{S}}{\longrightarrow}\Sigma_{\mathcal{S}}
\stackrel{\cong}{\longrightarrow}\tilde{\Sigma}_{\mathcal{U}}\hookrightarrow
Gr_{\tilde{B}}$, where the central equivalence is that of
Proposition \ref{caracterizacion-de-pseudomorphisms}. We now have:

\begin{prop} \label{ring-supporting=image of pseudomorphism}
In the above situation, the following assertions hold:

\begin{enumerate}
\item $H'=\varphi^{-1}[(\mathcal{U}:\mathcal{U})]$ is a subgroup
of $G'$ \item  The functor $\Phi$ induces an equivalence of
categories between $\mathcal{G}(\mathcal{S},\mathcal{U})$ and the
full subcategory $\mathcal{L}_{\tilde{B}}$ of $Gr_{\tilde{B}}$
consisting of those $V\in Gr_{\tilde{B}}$ such that
$V\cong\tilde{X}$, for some
$X\in\mathcal{L}(\mathcal{S},\mathcal{U})$.  \item All
$V\in\mathcal{L}_{\tilde{B}}$ are graded $\tilde{B}$-modules
generated in degrees belonging to $H'$ and
$\mathcal{L}_{\tilde{B}}$ contains all the graded
$\tilde{B}$-modules presented in degrees belonging to $H'$

\end{enumerate}
\end{prop}

\begin{proof}
1) We prove, more generally, that if $\varphi :G'\longrightarrow
G$ is a pseudomorphism and $H<G$ is a subgroup contained in
$Im(\varphi )$, then $H'=\varphi^{-1}(H)$ is a subgroup of $G'$.

If $\sigma ,\tau\in H'$ then $\varphi (\sigma ), \varphi (\tau
)\in H$. Since $H$ is a subgroup of $G$, we have that $\varphi
(\sigma )\varphi (\tau )\in H\subseteq Im(\varphi )$. The fact
that $\varphi$ is a pseudomorphism implies that $\varphi
(\sigma\tau )=\varphi (\sigma )\varphi (\tau )$ and, hence,
$\sigma\tau\in\varphi^{-1}(H)=H'$.

On the other hand, if $\sigma\in H'$ then $\varphi (\sigma
)^{-1}\in H\subseteq Im(\varphi )$ due to the fact that $H$ is a
subgroup of $G$. Then we can choose $\tau\in G'$ such that
$\varphi (\sigma )^{-1}=\varphi (\tau )$. But since $\varphi
(\sigma )\varphi (\tau )=1\in Im(\varphi )$, we get that $\varphi
(1)=1=\varphi (\sigma )\varphi (\tau )=\varphi (\sigma\tau )$. The
fact that $\varphi$ is injective implies that $1=\sigma\tau$. Then
$\tau =\sigma^{-1}$ and hence $\sigma^{-1}\in H'=\varphi^{-1}(H)$.
That proves that $H'$ is a subgroup of $G'$.

2) According to the definition of $\Phi$ and Theorem 2.7, we get
equivalences
$\mathcal{G}(\mathcal{S},\mathcal{U})\stackrel{\cong}{\longrightarrow}
\mathcal{L}(\mathcal{S},\mathcal{U})\stackrel{\cong}{\longrightarrow}\mathcal{L}_{\tilde{B}}$.

3) Put $B=A_\mathcal{U}$. Then, as shown in the proof of
Proposition \ref{caracterizacion-de-pseudomorphisms},  the
equivalence $\Sigma_\mathcal{S}=\Sigma_{gIm(\varphi
)}\stackrel{\cong}{\longrightarrow}\tilde{\Sigma}_{Im(\varphi)}=\tilde{\Sigma}_\mathcal{U}$
is a composition
$\Sigma_{\mathcal{S}}\stackrel{[g]}{\longrightarrow}\Sigma_{\mathcal{U}}
\stackrel{\cong}{\longrightarrow}\tilde{\Sigma}_\mathcal{U}$,
where the second equivalence takes graded $B$-modules generated
(resp. presented) in degrees belonging to
$(\mathcal{U}:\mathcal{U})$ to graded $\tilde{B}$-modules
generated (resp. presented) in degrees belonging to
$H'=\varphi^{-1}[(\mathcal{U}:\mathcal{U})]$. On the other hand,
the first equivalence
$[g]:\Sigma_{\mathcal{S}}\stackrel{\cong}{\longrightarrow}\Sigma_{\mathcal{U}}$
takes graded $B$-modules generated (resp. presented) in degrees
belonging to $(\mathcal{S}:\mathcal{U})$ to graded $B$-modules
generated (resp. presented) in degrees belonging to
$(\mathcal{U}:\mathcal{U})$. Now, from the chain of inclusions
 $\{X\in Gr_B:$ $X\text{ presented in
degrees belonging to }(\mathcal{S}:\mathcal{U})\}
\subseteq\mathcal{L}(\mathcal{S}:\mathcal{U})\subseteq\{X\in
Gr_B:$ $X\text{ generated in degrees belonging to
}(\mathcal{S}:\mathcal{U})\}\subseteq\Sigma_\mathcal{S}$ (cf.
Theorem \ref{fullyfaithful-killing-supports}),  assertion 3
follows.






\end{proof}

\section{The case $G=\mathbf{Z}$}

In this section we particularize the previous arguments to
$G=\mathbf{Z}$, the additive group of integers.  We thereby shift
from multiplicative to additive notation, and our algebra will be
a $\mathbf{Z}$-graded algebra $A=\oplus_{i\in\mathbf{Z}}A_i$. Our
first goal is to get as much information as possible  on
ring-supporting subsets $\mathcal{U}\subseteq\mathbf{Z}$ and
hence, by Proposition \ref{ejemplos-modular-pairs},    on modular
pairs $(\mathcal{S},\mathcal{U})$ such that $\emptyset\neq
(\mathcal{S}:\mathcal{U})=:\{m\in\mathbf{Z}:$
$m+\mathcal{U}=\mathcal{S}\}$. Those modular pairs will be of the
form $(m+\mathcal{U},\mathcal{U})$, for some integer $m$.

\begin{lema} \label{reduction-modulo-(U:U)}
Let $\mathcal{U}\subseteq\mathbf{Z}$ be a subset containing $0$
such that $(\mathcal{U}:\mathcal{U})=n\mathbf{Z}$, and let
$p:\mathbf{Z}\longrightarrow\mathbf{Z}_n$ be the canonical
projection. The following assertions are equivalent:

\begin{enumerate}
\item  $\mathcal{U}$ is a ring-supporting subset of $\mathbf{Z}$
\item  $\bar{\mathcal{U}}=p(\mathcal{U})$ is a ring-supporting
subset of $\mathbf{Z}_n$ such that
$(\bar{\mathcal{U}}:\bar{\mathcal{U}})=\{\bar{0}\}$
\end{enumerate}

In that case
$\mathcal{U}=\bigcup_{i\in\bar{\mathcal{U}}}(n\mathbf{Z}+i)$,
viewing $\bar{\mathcal{U}}$ as a subset of $[0,n)$ in case $n>0$.
\end{lema}

\begin{proof}

 We put $\bar{m}=p(m)$, for every $m\in\mathbf{Z}$. Let us first notice
 that  $p^{-1}(\bar{m}+\bar{\mathcal{U}}))=m+\mathcal{U}$. Indeed,
if $v\in p^{-1}(\bar{m}+\bar{\mathcal{U}}))$ then
$\bar{v}=p(v)=\bar{m}+\bar{u}$, for some $u\in\mathcal{U}$, which
implies that $v-m-u\in n\mathbf{Z}=(\mathcal{U}:\mathcal{U})$.
Then $v=m+u+(v-m-u)\in m+(v-m-u)+\mathcal{U}$. But
$(v-m-u)+\mathcal{U}=\mathcal{U}$, and then $v\in m+\mathcal{U}$
as desired. In particular $m+\mathcal{U}=\mathcal{U}$ iff
$\bar{m}+\bar{\mathcal{U}}=\bar{\mathcal{U}}$. In other words,
$m\in (\mathcal{U}:\mathcal{U})$ iff $\bar{m}\in
(\bar{\mathcal{U}}:\bar{\mathcal{U}})$. That gives that
$(\bar{\mathcal{U}}:\bar{\mathcal{U}})=\{\bar{0}\}$.

$1)\Longrightarrow 2)$ If  $u,v,w\in\mathcal{U}$ are elements such
that $\bar{u}+\bar{v}+\bar{w}\in\bar{\mathcal{U}}$ or,
equivalently, $u+v+w\in\mathcal{U}$ then
$\bar{u}+\bar{v}=\overline{u+v}\in\bar{\mathcal{U}}$ iff
$u+v\in\mathcal{U}$. Since $\mathcal{U}$ is ring-supporting, the
latter happens iff $v+w\in\mathcal{U}$ iff
$\overline{v+w}=\bar{v}+\bar{w}\in\bar{\mathcal{U}}$. Therefore
$\bar{\mathcal{U}}$ is a ring-supporting subset of $\mathbf{Z}_n$.

$2)\Longrightarrow 1)$ Go backward in the other implication.

Since $p^{-1}(\bar{\mathcal{U}})=\mathcal{U}$, we immediately get
$\mathcal{U}=\bigcup_{i\in\bar{\mathcal{U}}}(n\mathbf{Z}+i)$, thus
ending the proof.



\end{proof}



\begin{teor} \label{ring-supporting}
Let $\mathcal{U}$ be a  ring-supporting subset of $\mathbf{Z}$
such that $(\mathcal{U}:\mathcal{U})=n\mathbf{Z}$. Then
$\mathcal{U}=\bigcup_{j\in J}(n\mathbf{Z}+j)$, where $J$ is a
ring-supporting subset of $\mathbf{Z}_n$ such that $(J:J)=J\cap
(-J)=\{\bar{0}\}$. Moreover, the
 following
assertions hold:

\begin{enumerate}
\item If $\mathcal{U}$ contains an infinite interval then exactly
one of the following conditions hold:

\begin{enumerate}
\item $\mathcal{U}=\mathbf{Z}=(\mathcal{U}:\mathcal{U})$ \item
$\mathcal{U}$ is an additive submonoid of $\mathbf{N}$ \item
$\mathcal{U}$ is an additive submonoid of $-\mathbf{N}$
\end{enumerate}
 \item
If $\mathcal{U}$ does not contain an infinite interval, then
$\mathcal{U}$ is a union $\mathcal{U}=\bigcup_{i\in I}[a_i,b_i]$
of finite intervals such that $a_i\leq b_i<a_{i+1}-1$ for all
$i\in I$, where the index set $I$ is an interval of $\mathbf{Z}$.
 Moreover, the interval $[a_{i_0},b_{i_0}]$ containing $0$ is either $[0,r]$ or
$[-r,0]$, for some natural number $r$ which, in case $n>1$,
satisfies that $2r<n$
\end{enumerate}


\end{teor}

\begin{proof}

The first part of the theorem follows directly from Lemma
\ref{reduction-modulo-(U:U)}. To prove assertion 1, notice that,
since $(\mathcal{U}:\mathcal{U})$ is a subgroup of $\mathbf{Z}$
contained in $\mathcal{U}$, one has that
$(\mathcal{U}:\mathcal{U})=\mathbf{Z}$ if and only if
$\mathcal{U}=\mathbf{Z}$. So in the rest of the proof we assume
that $\mathcal{U}$ is a proper subset of $\mathbf{Z}$. Suppose
that $\mathcal{U}$  contains an infinite interval $[m,+\propto )$.
  If $m<0$ then the subgroup
$(\mathcal{U}:\mathcal{U})=\mathcal{U}\cap (-\mathcal{U})$
contains the interval $[m,-m]$, which contains $1$. That implies
that $(\mathcal{U}:\mathcal{U})=\mathbf{Z}=\mathcal{U}$, which is
a contradiction. Therefore $m\geq 0$ and we choose $m$ minimal
with the property that $[m,+\propto )\subseteq\mathcal{U}$. Notice
that if $m>0$ then $m>1$, because $0\in\mathcal{U}$.  We also have
$(\mathcal{U}:\mathcal{U})=0$, for otherwise the subgroup
$(\mathcal{U}:\mathcal{U})$ would be lower and upper unbounded and
then $\mathbf{Z}=(\mathcal{U}:\mathcal{U})+[m,+\propto )\subseteq
(\mathcal{U}:\mathcal{U})+\mathcal{U}=\mathcal{U}$, which
contradicts the fact that $\mathcal{U}$ is proper. Then
$\mathcal{U}\cap(-\mathcal{U})=(\mathcal{U}:\mathcal{U})=0$ and,
in particular, $(-\propto ,-m]\cap\mathcal{U}=\emptyset$ unless
$m=0$, in which case that intersection is  $\{0\}$. Then
$\mathcal{U}\subseteq [-m,+\propto )$. In order to prove that
$\mathcal{U}\subset\mathbf{N}$ it will be enough to consider the
case $m>1$ and prove that $(-m ,0)\cap\mathcal{U}=\emptyset$.
Indeed, if  that intersection is nonempty, then we can take $k>0$
such that  $-k=Inf[(-m ,0)\cap\mathcal{U}]$. We choose now
$u=v=-k$ and $w=m+2k$. Then $u,v,w$ and $u+v+w=m$ belong to
$\mathcal{U}$. However $v+w=m+k\in\mathcal{U}$ while
$u+v=-2k\not\in\mathcal{U}$. That contradicts the ring-supporting
condition of $\mathcal{U}$. Therefore
$\mathcal{U}\subseteq\mathbf{N}$. We finally check that
$\mathcal{U}$ is an additive submonoid of $\mathbf{N}$. Suppose
that it is not the case, so that there are $u,v\in\mathcal{U}$
such that $u+v\not\in\mathcal{U}$. Then, taking $w=m$, we get that
$u,v,w\in\mathcal{U}$ and $u+v+w\geq m$, so that also
$u+v+w\in\mathcal{U}$. But now $v+w=m+v\in
[m,+\propto)\subseteq\mathcal{U}$ and $u+v\not\in\mathcal{U}$.
That contradicts the ring-supporting condition of $\mathcal{U}$.

Since $-\mathcal{U}$ is a ring-supporting subset whenever so is
$\mathcal{U}$ (cf. Proposition \ref{ejemplos-modular-pairs}), the
above paragraph also gives that if $(-\propto
,-m]\subseteq\mathcal{U}$, for some $m\in\mathbf{Z}$, then
$\mathcal{U}$ is an additive submonoid of $-\mathbf{N}$.






 Suppose now that  $\mathcal{U}$ does not
contain infinite intervals. We can always express it as a union
$\mathcal{U}=\bigcup_{i\in I}[a_i,b_i]$  of finite intervals such
that $a_i\leq b_i<a_{i+1}-1$ for all $i\in I$, where the index set
$I$ is an interval of the integers which we convene that contains
$0$ and, in addition $0\in [a_0,b_0]$.
 Suppose now that $a_0<0$ and
$b_0>0$. We then take $u=a_0$, $v=-1$, $w=1$. Then $u,v,w\in
[a_0,b_0]\subset\mathcal{U}$, and also $u+v+w=u\in\mathcal{U}$.
Now $v+w=0\in\mathcal{U}$ but $u+v=a_0-1\notin\mathcal{U}$. That
contradicts the ring-supporting condition of $\mathcal{U}$.
Therefore either $a_0=0$ or $b_0=0$, so that we can assume that
$[a_0,b_0]=[0,r]$ or $[a_0,b_0]=[-r,0]$, for some natural number
$r$. It only remains to prove that, in case $n>1$, we have $2r<n$.
We do it when $[a_0,b_0]=[0,r]$, leaving the dualization when
$[a_0,b_0]=[-r,0]$ for the reader. Since $\bar{\mathcal{U}}$ is a
proper subset of $\mathbf{Z}_n\cong [0,n)$, we necessarily have
$r<n-1$. Suppose now that $2r\geq n$. Then we  take $u=n-r$, $v=r$
and $w=1$. Since $0<n-r\leq 2r-r=r$, and $u+v+w=n+1$, we conclude
that $u,v,w$ and $u+v+w$ belong to $\mathcal{U}$. Now
$u+v=n\in\mathcal{U}$, but $v+w=r+1\notin\mathcal{U}$ because
$r=b_0<a_1-1$. That contradicts the fact that $\mathcal{U}$ is
ring-supporting and, hence, we necessarily have $2r<n$.


\end{proof}

\begin{rem}
From the above result we deduce that, in order to classify all the
ring-supporting subsets of $\mathbf{Z}$, one should classify all
the ring-supporting subsets $\mathcal{U}$ such that
$(\mathcal{U}:\mathcal{U})=0$ and, for every $n>0$, all the
ring-supporting subsets $J\subset\mathbf{Z}_n$ such that
$(J:J)=\{\bar{0}\}$.  It seems to be a very difficult task, and
the only reasonable expectation is to tackle the case when $n>0$
is not large. The following is a sample. We leave the verification
to the reader.
\end{rem}

\begin{ejems}
The following are the ring-supporting subsets
$\mathcal{U}\subseteq\mathbf{Z}$ such that
$(\mathcal{U}:\mathcal{U})=n\mathbf{Z}$, with $0<n\leq 5$

\begin{enumerate}
\item If $n=1$ then $\mathcal{U}=\mathbf{Z}$ \item If $n=2$ then
$\mathcal{U}=2\mathbf{Z}=\bigcup_{k\in\mathbf{Z}}[2k,2k]$ \item If
$n=3$ then $\mathcal{U}$ is one of the subsets
$3\mathbf{Z}=\bigcup_{k\in\mathbf{Z}}[3k,3k]$, $3\mathbf{Z}\cup
(3\mathbf{Z}+1)=\bigcup_{k\in\mathbf{Z}}[3k,3k+1]$ or
$3\mathbf{Z}\cup
(3\mathbf{Z}+2)=\bigcup_{k\in\mathbf{Z}}[3k-1,3k]$ \item If $n=4$
then $\mathcal{U}$ is one of the subsets
$4\mathbf{Z}=\bigcup_{k\in\mathbf{Z}}[4k,4k]$, $4\mathbf{Z}\cup
(4\mathbf{Z}+1)=\bigcup_{k\in\mathbf{Z}}[4k,4k+1]$ or
$4\mathbf{Z}\cup
(4\mathbf{Z}+3)=\bigcup_{k\in\mathbf{Z}}[4k-1,4k]$ \item If $n=5$
then $\mathcal{U}$ is one of the subsets of the following list:

\begin{enumerate}
\item[a)]$5\mathbf{Z}=\bigcup_{k\in\mathbf{Z}}[5k,5k]$

\item[b)] $5\mathbf{Z}\cup
(5\mathbf{Z}+1)=\bigcup_{k\in\mathbf{Z}}[5k,5k+1]$ \item[b*)]
$5\mathbf{Z}\cup
(5\mathbf{Z}+4)=\bigcup_{k\in\mathbf{Z}}[5k-1,5k]$ \item[c)]
 $5\mathbf{Z}\cup
(5\mathbf{Z}+1)\cup
(5\mathbf{Z}+2)=\bigcup_{k\in\mathbf{Z}}[5k,5k+2]$ \item[c*)]
$5\mathbf{Z}\cup (5\mathbf{Z}+3)\cup
(5\mathbf{Z}+4)=\bigcup_{k\in\mathbf{Z}}[5k-2,5k]$ \item[d)]
$5\mathbf{Z}\cup
(5\mathbf{Z}+2)=\bigcup_{k\in\mathbf{Z}}([5k,5k]\cup [5k+2,5k+2])$
\item [d*)] $5\mathbf{Z}\cup
(5\mathbf{Z}+3)=\bigcup_{k\in\mathbf{Z}}([5k,5k]\cup [5k+3,5k+3])$
\item[e)] $5\mathbf{Z}\cup (5\mathbf{Z}+1)\cup
(5\mathbf{Z}+3)=\bigcup_{k\in\mathbf{Z}}([5k,5k+1]\cup
[5k+3,5k+3])$ \item[e*)] $5\mathbf{Z}\cup (5\mathbf{Z}+2)\cup
(5\mathbf{Z}+4)=\bigcup_{k\in\mathbf{Z}}([5k-1,5k]\cup
[5k+2,5k+2])$
\end{enumerate}

\end{enumerate}
\end{ejems}



The last part of  Theorem \ref{ring-supporting} gives the
following  peculiar ring-supporting subset, which will be called
{\bf translation of an interval} in the sequel:

\begin{cor} \label{translation-of-interval}
Let $\mathcal{U}$ be a  subset of $\mathbf{Z}$ containing $0$ such
that $(\mathcal{U}:\mathcal{U})=n\mathbf{Z}$, with $n\neq 0,1$.
The following assertions are equivalent:

\begin{enumerate}
\item $\mathcal{U}$ is ring-supporting and
$\mathcal{U}=I+n\mathbf{Z}$, for some finite interval
$I\subset\mathbf{Z}$ \item
$\mathcal{U}=\bigcup_{k\in\mathbf{Z}}[nk,nk+r]$ or
$\mathcal{U}=\bigcup_{k\in\mathbf{Z}}[nk-r,nk]$, for some natural
number $r$ such that $2r<n$
\end{enumerate}
\end{cor}

\begin{proof}
Theorem \ref{ring-supporting} gives the implication
$1)\Longrightarrow 2)$. For the converse, it will enough to check
that $\mathcal{U}=\bigcup_{k\in\mathbf{Z}}[nk,nk+r]$ is
ring-supporting and, for that, we will check that
$\bar{\mathcal{U}}$ is a ring-supporting subset of $\mathbf{Z}_n$.
Indeed, if $u,v,w$ are natural numbers such that $u,v,w\leq r$ and
we assume that $\bar{u}+\bar{v}+\bar{w}\in\bar{\mathcal{U}}$,
then, since $u+v\leq 2r<n$, we have that either $u+v+w\leq r$ or
$n\leq u+v+w\leq n+r$. In the first case, both $u+v\leq r$ and
$v+w\leq r$. In the second case, both $u+v>r$ and $v+w>r$ for if,
say, $u+v\leq r$ then $u+v+w\leq r+r=2r<n$, against the assumption
that $u+v+w\in [n,n+r]$. Therefore
$\bar{u}+\bar{v}\in\bar{\mathcal{U}}$ if, and only if,
$\bar{v}+\bar{w}\in\bar{\mathcal{U}}$, as desired.
\end{proof}



In the rest of the section, we assume that our algebra is a
positively graded algebra $A=\oplus_{i\geq 0}A_i$ over a field
$K$,  which is generated in degrees $0,1$ and has the property
that $A_0$ is semisimple. The symbol $\otimes$ will  always mean
tensor product over $A_0$.





In case $(\mathcal{S},\mathcal{U})$ is a right modular pair of
subsets of $\mathbf{Z}$ and $X\in Gr_{A_\mathcal{U}}$ has
$Supp(X)\subseteq\mathcal{S}$, we denote by $\mu_{s,u}:X_s\otimes
A_u\longrightarrow X_{s+u}$ the multiplication map, whenever
$(s,u)\in\mathcal{S}\times\mathcal{U}$ and $s+u\in\mathcal{S}$.
Since $X\otimes A$ is a graded right $A$-module,
$Ker(\mu_{s,u})A_i$ is a well-defined $A_0$-submodule of
$X_s\otimes A_{u+i}$, for all $i\geq 0$. With this terminology in
mind, we can give criteria for $X$ to be liftable.

\begin{prop} \label{liftable}
Let $A=\oplus_{i\geq 0}A_i$ be a positively graded algebra
generated in degrees $0,1$, with $A_0$ semisimple,
$(\mathcal{S},\mathcal{U})$ be a right modular pair of subsets of
$\mathbf{Z}$  and consider an $X\in Gr_{A_\mathcal{U}}$ generated
in degrees belonging to $(\mathcal{S}:\mathcal{U})$.
 The following assertions
are equivalent:

\begin{enumerate}
\item $X$ is liftable with respect to the functor
$(-)_\mathcal{S}:Gr_A\longrightarrow Gr_{A_\mathcal{U}}$ \item
$Ker(\mu_{m,u})A_{v-u}\subseteq Ker(\mu_{m,v})$ whenever
$(m,u,v)\in
(\mathcal{S}:\mathcal{U})\times\mathcal{U}\times\mathcal{U}$ is a
triple such that $0\leq u<v$ and $v-u\not\in\mathcal{U}$
\end{enumerate}
\end{prop}

\begin{proof}
$1)\Longrightarrow 2)$ Take $M\in Gr_A$ such that
$M_\mathcal{S}=X$. Then, for all $(m,u)\in
(\mathcal{S}:\mathcal{U})\times\mathcal{U}$,  the multiplication
maps $\mu_{m,u}:X_m\otimes A_u=M_m\otimes A_u\longrightarrow
M_{m+u}=X_{m+u}$ are those of the $A$-module $M$. Now condition 2
follows from the associative property of the multiplication for
the $A$-module $M$

$2)\Longrightarrow 1)$ When $(\mathcal{S}:\mathcal{U})=\emptyset$
there is nothing to prove. Hence, we assume that
$(\mathcal{S}:\mathcal{U})\neq\emptyset$. Since $X$ is generated
in degrees belonging to $(\mathcal{S}:\mathcal{U})$, the
multiplication map $\mu :X_{(\mathcal{S}:\mathcal{U})}\otimes
A_\mathcal{U}\longrightarrow X$ is surjective and
$Supp(X)\subseteq\mathcal{S}$. We then consider the graded
$A$-module $M=\frac{X_{(\mathcal{S}:\mathcal{U})}\otimes A}{(Ker
\mu )A}$. Then
$M_\mathcal{S}=\frac{(X_{(\mathcal{S}:\mathcal{U})}\otimes
A)_\mathcal{S}}{[(Ker \mu
)A]_\mathcal{S}}=\frac{X_{(\mathcal{S}:\mathcal{U})}\otimes
A_\mathcal{U}}{[(Ker \mu )A]_\mathcal{S}}$. We shall prove that
$[(Ker \mu)A]_\mathcal{S}=Ker\mu$ and the result will follow, for
$\frac{X_{(\mathcal{S}:\mathcal{U})}\otimes A_\mathcal{U}}{Ker \mu
}$ is isomorphic to $X$ in $Gr_{A_\mathcal{U}}$. Let us fix
$t\in\mathcal{S}$. Then the $t$-homogeneous component of $(Ker\mu
)A$ is $\sum_{s\in\mathcal{S}, s\leq t}(Ker\mu )_sA_{t-s}$. The
task is whence reduced to prove that if $s\leq t$ are elements of
$\mathcal{S}$ then $(Ker\mu )_sA_{t-s}\subseteq (Ker\mu )_t$. That
inclusion is clear in case $t-s\in\mathcal{U}$ due to the
associative property of the multiplication in the
$A_\mathcal{U}$-module  $X$. So we assume that
$t-s\not\in\mathcal{U}$. Let $\sum_{1\leq i\leq r}x_i\otimes a_i$
belong to $(Ker\mu )_s$, where $deg(x_i)=m_i\in
(\mathcal{S}:\mathcal{U})$ and $a_i\in A_{s-m_i}$ for $i=1,...,r$.
We take $p=max\{m\in (\mathcal{S}:\mathcal{U}):$ $m\leq s\}$. The
hypothesis on $A$ implies that $A_{s-m_i}=A_{p-m_i}\cdot A_{s-p}$,
so that $a_i=\sum_jb_{ij}c_{ij}$, where $b_{ij}\in A_{p-m_i}$ and
$c_{ij}\in A_{s-p}$, for all $i,j$. Then
$0=\sum_ix_ia_i=\sum_{i,j}x_i(b_{ij}c_{ij})=\sum_{i,j}(x_ib_{ij})c_{ij}$,
using the associative property of $X$ again (notice that
$deg(b_{ij})=p-m_i$ and $deg(c_{ij})=s-p$ belong to
$\mathcal{U}$). Then $\sum_{i,j}(x_ib_{ij})\otimes c_{ij}$ belongs
to the kernel of the multiplication map $\mu_{p,s-p}:X_p\otimes
A_{s-p}\longrightarrow A_s$. Since $p\in
(\mathcal{S}:\mathcal{U})$, we have that $u=s-p$ and $v=t-p$ are
elements of $\mathcal{U}$ such that $0\leq u<v$ and
$v-u=t-s\not\in\mathcal{U}$. Our hypothesis says that
$Ker(\mu_{p,s-p})A_{t-s}\subseteq Ker(\mu_{p,t-p})$. Then, if
$d\in A_{t-s}$ the element $\sum_{i,j}(x_ib_{ij})\otimes
(c_{ij}d)$ belongs to $Ker(\mu_{p,t-p})$. That gives
$0=\sum_{i,j}(x_ib_{ij})(c_{ij}d)=\sum_{i,j}x_{i}[b_{ij}(c_{ij}d)]$,
using the associative property of the multiplication in $X$. But
the associative property of $A$ gives that
$b_{ij}(c_{ij}d)=(b_{ij}c_{ij})d$. Hence we have
$0=\sum_{i,j}x_i[(b_{ij}c_{ij})d]=\sum_ix_i(a_id)$, which implies
that the element $(\sum_ix_i\otimes a_i)d$ belongs to
$(Ker\mu)_t$. That proves that $(Ker\mu_s)A_{t-s}\subseteq (Ker\mu
)_t$, as desired.

\end{proof}



 The conditions get much simpler in the case that we
are more interested in, namely, the translation of an interval
(cf. Corollary \ref{translation-of-interval}).





\begin{cor} \label{liftable-interval}
In the situation of the proposition, adopting the notations of
Corollary \ref{translation-of-interval}, suppose that
$\mathcal{U}$ is the translation of an interval. Then the
following assertions hold for a graded $A_\mathcal{U}$-module $X$
generated in degrees belonging to $(\mathcal{S}:\mathcal{U})$:

\begin{enumerate}
\item If $\mathcal{U}=\bigcup_{k\in\mathbf{Z}}[kn,kn+r]$ ($0\leq
2r<n$), then $X$ is liftable if, and only if,
$Ker(\mu_{m,r})A_{n-r}\subseteq Ker(\mu_{m,n})$, for all $m\in
(\mathcal{S}:\mathcal{U})$ \item If
$\mathcal{U}=\bigcup_{k\in\mathbf{Z}}[kn-r,kn]$ ($0\leq 2r<n$),
then $X$ is liftable if, and only if,
$Ker(\mu_{m,u})A_{v-u}\subseteq Ker(\mu_{m,v})$, for all $m\in
(\mathcal{S}:\mathcal{U})$ and all $n-r\leq u<v\leq n$
\end{enumerate}
\end{cor}

\begin{proof}
We prove 1, for which we only need to check the 'if' part. The
proof of assertion 2 follows a similar pattern. We first prove by
induction on $k\geq 0$ that $Ker(\mu_{m,kn+r})A_{n-r}\subseteq
Ker(\mu_{m,(k+1)n})$ (*), for all $m\in
(\mathcal{S}:\mathcal{U})$. The case $k=0$ is just the hypothesis.
Suppose  now  that $k>0$ and $\sum_ix_i\otimes a_i\in
Ker(\mu_{m,kn+r})$, where $x_i\in X_m$ and $a_i\in A_{kn+r}$ for
all indices $i$. Since $A_{kn+r}=A_n\cdot A_{(k-1)n+r}$, we can
decompose $a_i=\sum_jb_{ij}c_{ij}$, where $b_{ij}\in A_n$ and
$c_{ij}\in A_{(k-1)n+r}$. With an argument similar to that in the
proof of Proposition \ref{liftable}, we see that
$\sum_{i,j}(x_ib_{ij})\otimes c_{ij}$ belongs to
$Ker(\mu_{m+n,(k-1)n+r})$. If now $d\in A_{n-r}$ then the
induction hypothesis says that $(\sum_{i,j}(x_ib_{ij})\otimes
c_{ij})d=\sum_{i,j}(x_ib_{ij})\otimes (c_{ij}d)$ belongs to
$Ker(\mu_{m+n,kn})$ and, as in the proof of Proposition
\ref{liftable} again, one sees that $(\sum_ix_i\otimes
a_i)d=\sum_i x_i\otimes (a_id)\in Ker(\mu_{m,(k+1)n})$.

Let $(m,u,v)\in
(\mathcal{S}:\mathcal{U})\times\mathcal{U}\times\mathcal{U}$ be a
triple such that $0\leq u<v$ and $v-u\not\in\mathcal{U}$. Then we
have $u\in [a_i,b_i]=[in,in+r]$ and $v\in [a_j,b_j]=[jn,jn+r]$,
for some indices $0\leq i<j$. Since $A$ is generated in degrees
$0,1$, we have that $A_{v-u}=A_{b_i-u}\cdot A_{a_{i+1}-b_i}\cdot
...\cdot A_{a_j-b_{j-1}}\cdot A_{v-a_j}$. Each of the factors of
this product is of the form $A_{v'-u'}$, where either $u',v'\in
[a_k,b_k]=[kn,kn+r]$ or $u'=b_k=kn+r$, $v'=a_{k+1}=(k+1)n$, for
some index $k$. In the first case $0\leq v'-u'\leq r$ and, hence,
$v'-u'\in\mathcal{U}$. In the second case, $v'-u'=n-r$. One then
obtains that $Ker(\mu_{m,u})A_{v-u}\subseteq Ker(\mu_{m,v})$ by
recursive use of the associative property of the
$A_\mathcal{U}$-module $X$ and the above condition (*).
\end{proof}

Using Theorem \ref{fullyfaithful-killing-supports}, we have the
following straightforward consequence:

\begin{cor} \label{equivalence-for-translation-of-intervals}
Let $A=\oplus_{i\geq 0}A_i$ be a positively graded algebra
generated in degrees $0,1$ such that $A_0$ is semisimple. We
consider  $\mathcal{U}=\bigcup_{k\in\mathbf{Z}}[nk,nk+r]$ and
$\mathcal{S}=m+\mathcal{U}$, where $n>1$,  $0\leq 2r<n$ and
$m\in\mathbf{Z}$. The functor $(-)_\mathcal{S}:Gr_A\longrightarrow
Gr_{A_\mathcal{U}}$ establishes an equivalence  between the
following categories:

\begin{enumerate}
\item The full subcategory $\mathcal{G}(\mathcal{S},\mathcal{U})$
of $Gr_A$ whose objects are the graded modules generated in
degrees belonging to $m+n\mathbf{Z}$ and cogenerated in degrees
belonging to $\mathcal{S}$ \item The full subcategory
$\mathcal{L}(\mathcal{S},\mathcal{U})$ of $Gr_{A_\mathcal{U}}$
whose objects are those $X$ generated in degrees belonging to
$m+n\mathbf{Z}$ and satisfying that
$(Ker\mu_{m+kn,r})A_{n-r}\subseteq Ker\mu_{m+kn,n}$  for all
$k\in\mathbf{Z}$
\end{enumerate}
\end{cor}



By the induction principle, given a subset
$\mathcal{S}\subseteq\mathbf{Z}$, which is neiter upper nor lower
bounded,  and a fixed element $s_0\in\mathcal{S}$, there is a
unique strictly increasing map $\delta
=\delta_{(\mathcal{S},s_0)}:\mathbf{Z}\longrightarrow\mathbf{Z}$
such that $\delta (0)=s_0$ and $Im(\delta )=\mathcal{S}$.

Let $(\mathcal{S},\mathcal{U})$ be a right modular pair of subsets
  of $\mathbf{Z}$ such that $0\neq
(\mathcal{U}:\mathcal{U})=n\mathbf{Z}$ and $m\in
(\mathcal{S}:\mathcal{U})$. We want to identify
$\delta_{(\mathcal{S},m)}$. As in Lemma
\ref{reduction-modulo-(U:U)}, we can think of $\bar{\mathcal{U}}$
as a subset of $[0,n)$, in which case we can write
$\bar{\mathcal{U}}=\{i_0,i_1,...,i_{t-1}\}$, with
$t=|\bar{\mathcal{U}}|$ and $0=i_0<i_1<...<i_{t-1}<n$.

\begin{lema} \label{funcion asociada a un ring-supporting}
In the above situation, the following assertions hold:

\begin{enumerate}

\item $\delta_{(\mathcal{S},m)}=m+\delta_{(\mathcal{U},0)}$ \item
$\delta_{(\mathcal{S},m)}:\mathbf{Z}\longrightarrow\mathbf{Z}$
maps $tj+k\rightsquigarrow m+nj+i_k$, for all $j\in\mathbf{Z}$ and
all $k\in [0,t)$ \item If $\mathcal{U}$ is the translation of an
interval, then  $\delta_{(\mathcal{U},0)}$ is a pseudomorphism of
groups.
\end{enumerate}
\end{lema}

\begin{proof}

1) Clearly, $m+\delta_{(\mathcal{U},0)}:j\rightsquigarrow
m+\delta_{(\mathcal{U},0)}(j)$ is strictly increasing and maps
$0\rightsquigarrow m+\delta_{(\mathcal{U},0)}(0)=m$. Moreover,
$Im(m+\delta_{(\mathcal{U},0)})=m+Im(\delta_{(\mathcal{U},0)})=m+\mathcal{U}=\mathcal{S}$.
Then $\delta_{(\mathcal{S},m)}=m+\delta_{(\mathcal{U},0)}$.

2) Use an argument as in 1.

3)  We then prove that $\delta=:\delta_{(\mathcal{U},0)}$ is a
pseudomorphism in case
$\mathcal{U}=\bigcup_{k\in\mathbf{Z}}[nk,nk+r]$ or
$\mathcal{U}=\bigcup_{k\in\mathbf{Z}}[nk-r,nk]$, where $r$ and $n$
are  natural numbers such that $0\leq 2r<n$. We consider the
second possibility, leaving the first one as an exercise. Notice
that $t=|\bar{\mathcal{U}}|=r+1$ in this case. Then,  according to
assertion 2, we have $\delta [(r+1)j-i]=nj-i$, for all
$j\in\mathbf{Z}$ and $i\in [0,r]$. If $m=(r+1)j-i$ and
$m'=(r+1)j'-i'$, where $i,i'\in [0,r]$, then $\delta (m)+\delta
(m')=n(j+j')-i-i'$ belongs to $\mathcal{U}=Im(\delta )$ if, and
only if, $i+i'\in [0,r]$. In that case, $\delta (m+m')=\delta
[(r+1)(j+j')-(i+i')]=n(j+j')-(i+i')=\delta (m)+\delta (m')$. Then
$\delta$ is a pseudomorphism of groups.
\end{proof}

We can now combine the general results of Sections 2 and 3 with
those of this section. If
$\mathcal{U}=\bigcup_{k\in\mathbf{Z}}[nk,nk+r]$, with $0<2r<n$,
then we consider the graded algebra $\tilde{B}$ obtained from
$B=:A_\mathcal{U}$ by regrading along the pseudomorphism $\delta
=\delta_{(\mathcal{U},0)}$. If now $V\in Gr_{\tilde{B}}$, then we
have a multiplication maps
$\tilde{\mu}_{(r+1)k,r}:V_{(r+1)k}\otimes\tilde{B}_r=V_{(r+1)k}\otimes
A_r\longrightarrow V_{(r+1)(k+1)-1}$ and
$\tilde{\mu}_{(r+1)k,r+1}:V_{(r+1)k}\otimes\tilde{B}_{r+1}=V_{(r+1)k}\otimes
A_n\longrightarrow V_{(r+1)(k+1)}$, so that, working in the right
$A$-module $V_{(r+1)\mathbf{Z}}\otimes A$, it makes sense to
consider $Ker(\tilde{\mu})_{(r+1)k,r}A_{n-r}$, which will be
$A_0-A_0-$sub-bimodule of
$V_{(r+1)k}\otimes\tilde{B}_{r+1}=V_{(r+1)k}\otimes A_n$.  We then
get the following:

\begin{cor}
Let $A=\oplus_{n\geq 0}A_n$ be a positively graded algebra
generated in degrees $0,1$ such that $A_0$ is semisimple,
$(\mathcal{S}, \mathcal{U})=(m+\mathcal{U},\mathcal{U})$ be a
right modular pair, where
$\mathcal{U}=\bigcup_{k\in\mathbf{Z}}[nk,nk+r]$, with $0\leq
2r<n$, and $\tilde{B}$ be the graded algebra obtained from
$B=A_\mathcal{U}$ by regrading along the pseudomorphism $\delta
=\delta_{(\mathcal{U},0)}:\mathbf{Z}\longrightarrow\mathbf{Z}$.
The functor $\Phi :Gr_A\longrightarrow Gr_{\tilde{B}}$ of
Proposition \ref{ring-supporting=image of pseudomorphism} gives an
equivalence of categories between:

\begin{enumerate}
\item The full subcategory $\mathcal{G}(\mathcal{S},\mathcal{U})$
of $Gr_A$ whose objects are the graded $A$-modules generated in
degrees belonging to $(\mathcal{S}:\mathcal{U})=m+n\mathbf{Z}$ and
cogenerated in degrees belonging to $\mathcal{S}$ \item The full
subcategory $\mathcal{L}_{\tilde{B}}$ of $Gr_{\tilde{B}}$
consisting of those $V$ generated in degrees belonging to
$(r+1)\mathbf{Z}$ such that
$Ker(\tilde{\mu}_{(r+1)k,r})A_{n-r}\subseteq
Ker(\tilde{\mu}_{(r+1)k,r+1})$, for all $k\in\mathbf{Z}$.
\end{enumerate}

Moreover, $\mathcal{L}_{\tilde{B}}$ contains all the graded
$\tilde{B}$-modules presented in degrees belonging to
$(r+1)\mathbf{Z}$.
\end{cor}

\begin{proof}
It is a direct consequence of Proposition
\ref{ring-supporting=image of pseudomorphism} and Corollary
\ref{equivalence-for-translation-of-intervals}, after checking
that the equivalence $\Sigma_\mathcal{S}=\Sigma_{m+Im(\delta
)}\stackrel{\cong}{\longrightarrow}\tilde{\Sigma}_{Im(\delta
)}=\tilde{\Sigma}_\mathcal{U}$ of Proposition
\ref{caracterizacion-de-pseudomorphisms} restricts to an
equivalence between $\mathcal{L}(\mathcal{S},\mathcal{U})$ and
$\mathcal{L}_{\tilde{B}}$.
\end{proof}

\begin{rem} \label{n-Koszul}
If $\Lambda$ is a $n$-Koszul algebra, with $n>2$, then we can
apply the above corollary to $A=\Lambda^!$, with $r=1$, so that
$\tilde{B}\cong E=E(\Lambda )$ is the Yoneda algebra of $\Lambda$
(cf. \cite{GMMVZ}[Theorem 9.1], see also \cite{BM}[Proposition
3.1]) and $(\mathcal{S},\mathcal{U})=(m+\mathcal{U},\mathcal{U})$,
with $\mathcal{U}=n\mathbf{Z}\cup (n\mathbf{Z}+1)$. Hence, if
$V\in Gr_E$, we have multiplication maps
$\tilde{\mu}_{2k,1}:V_{2k}\otimes
E_1=V_{2k}\otimes\Lambda^!_1\longrightarrow V_{2k+1}$ and
$\tilde{\mu}_{2k,2}:V_{2k}\otimes
E_2=V_{2k}\otimes\Lambda^!_n\longrightarrow V_{2k+2}$ and we can
take $Ker(\tilde{\mu}_{2k,1})\Lambda^!_{n-1}$, which is a
$\Lambda_0-\Lambda_0-$sub-bimodule of
$V_{2k}\otimes\Lambda^!_n=V_{2k}\otimes E_2$. The last corollary
gives then an equivalence of categories between:

\begin{enumerate}
\item The full subcategory $\mathcal{G}(\mathcal{S},\mathcal{U})$
of $Gr_{\Lambda^!}$ consisting of those $M$ such that
$M=M_{m+n\mathbf{Z}}\Lambda^!$ and, for every graded submodule
$0\neq N<M$, one has $N_\mathcal{S}\neq 0$ \item The full
subcategory $\mathcal{L}_E$ of $Gr_E$ consisting of those $V\in
Gr_E$ which are generated in even degrees and satisfy that
$Ker(\tilde{\mu}_{2k,1})\Lambda^!_{n-1}\subseteq
Ker(\tilde{\mu}_{2k,2})$, for all $k\in\mathbf{Z}$.
\end{enumerate}

Moreover, $\mathcal{L}_E$ contains all the graded $E$-modules
presented in even degrees.
\end{rem}

\begin{thebibliography}{9999}



\bibitem{ARS}{\sc AUSLANDER, M.; REITEN, I.; SMALO, S.O.}:
"Representation theory of Artin algebras". Cambridge Univ. Press
{\bf 36}, in Cambridge Studies in Advanced Mathematics. Cambridge
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\bibitem{B} {\sc BERGER, R.}: Koszulity for nonquadratic algebras.
J. Algebra {\bf 239} (2001), 705-734.



\bibitem{BM} {\sc BERGER, R.; MARCONNET, N.}: Koszul and
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math.QA/0310070 v2



\bibitem{G}{\sc GABRIEL, P.}: Des cat�gories abeliennes. Bull.
Soc. Math. France, {\bf 90} (1962), 323-448


\bibitem{GMMVZ} {\sc GREEN, E.L.; MARCOS, E.N.; MARTINEZ-VILLA, R.; ZHANG,
P.}: D-Koszul algebras. J. Pure and Appl. Algebra {\bf 193}
(2004), 141-162

\bibitem{NVO} {\sc NASTASESCU, C.; VAN OYSTAEYEN, F.}: "Graded Ring Theory". North-Holland (1982)


\bibitem{St} {\sc STENSTR�M, B.}: "Rings os quotients".
Springer-Verlag (1975)

\end{thebibliography}

\end{document}

math0505076.extracted_bib
\begin{thebibliography}{9999}



\bibitem{ARS}{\sc AUSLANDER, M.; REITEN, I.; SMALO, S.O.}:
"Representation theory of Artin algebras". Cambridge Univ. Press
{\bf 36}, in Cambridge Studies in Advanced Mathematics. Cambridge
(1995).

\bibitem{B} {\sc BERGER, R.}: Koszulity for nonquadratic algebras.
J. Algebra {\bf 239} (2001), 705-734.



\bibitem{BM} {\sc BERGER, R.; MARCONNET, N.}: Koszul and
Gorenstein properties for homogeneous algebras. Preprint,   arXiv:
math.QA/0310070 v2



\bibitem{G}{\sc GABRIEL, P.}: Des cat�gories abeliennes. Bull.
Soc. Math. France, {\bf 90} (1962), 323-448


\bibitem{GMMVZ} {\sc GREEN, E.L.; MARCOS, E.N.; MARTINEZ-VILLA, R.; ZHANG,
P.}: D-Koszul algebras. J. Pure and Appl. Algebra {\bf 193}
(2004), 141-162

\bibitem{NVO} {\sc NASTASESCU, C.; VAN OYSTAEYEN, F.}: "Graded Ring Theory". North-Holland (1982)


\bibitem{St} {\sc STENSTR�M, B.}: "Rings os quotients".
Springer-Verlag (1975)

\end{thebibliography}