Graded analogues of theorems in commutative algebra












9














Many theorems in commutative algebra hold true in a ($mathbb{Z}$-)graded context. More precisely, we can take any theorem in commutative algebra and replace every occurrence of the word




  • commutative ring by commutative graded ring (without the sign for commutativity)

  • module by graded module

  • element by homogeneous element

  • ideal by homogeneous ideal (i.e. ideal generated by homogeneous elements)


This results in further substitutions, e.g. a $ast$local ring is a graded ring with a unique maximal homogeneous ideal, we get a notion of graded depth etc. After all these substitutions we can ask whether the theorem is still true.



One book that does some steps in this direction is Cohen-Macaulay rings by Bruns and Herzog, especially Section 1.5. For example, they have as Exercise 1.5.24 the following graded analogue of the Nakayama lemma:




Let $(R,mathfrak{m})$ be a $ast$local ring, $M$ be a finitely generated graded $R$-module and $N$ a graded submodule. Assume $M = N + mathfrak{m}M$. Then $M = N$.




Moreover, a student of mine recently showed that the graded analogue of Lazard's theorem (a module is flat if and only if it is a filtered colimit of free modules) is also true.



Usually one proves this kind of theorems essentially by a combination of two techniques:




  1. Copy the ungraded proof and just substitute ungraded for graded concepts in the manner sketched above.

  2. If one is annoyed by the length of the resulting argument, use some shortcuts by some translations between ungraded and graded. (E.g. a noetherian graded ring that is graded Cohen-Macaulay is also ungraded Cohen-Macaulay.)


Sometimes one can also be lucky and the statement is suitably algebro-geometry that one can argue geometrically with the stack $[Spec R/mathbb{G}_m]$ for a graded ring $R$, using that a $mathbb{Z}$-grading corresponds to a $mathbb{G}_m$-action.



In any case, my question is the following:




Is there any class of statements, where one knows automatically that the graded analogue is true if the original statement in ungraded commutative algebra is true, without going through the whole proof?




I am not sure whether one can hope here for a model-theoretic approach as I know almost nothing about model theory, but any such statement could save a lot of work in proving graded analogues of known theorems.










share|cite|improve this question



























    9














    Many theorems in commutative algebra hold true in a ($mathbb{Z}$-)graded context. More precisely, we can take any theorem in commutative algebra and replace every occurrence of the word




    • commutative ring by commutative graded ring (without the sign for commutativity)

    • module by graded module

    • element by homogeneous element

    • ideal by homogeneous ideal (i.e. ideal generated by homogeneous elements)


    This results in further substitutions, e.g. a $ast$local ring is a graded ring with a unique maximal homogeneous ideal, we get a notion of graded depth etc. After all these substitutions we can ask whether the theorem is still true.



    One book that does some steps in this direction is Cohen-Macaulay rings by Bruns and Herzog, especially Section 1.5. For example, they have as Exercise 1.5.24 the following graded analogue of the Nakayama lemma:




    Let $(R,mathfrak{m})$ be a $ast$local ring, $M$ be a finitely generated graded $R$-module and $N$ a graded submodule. Assume $M = N + mathfrak{m}M$. Then $M = N$.




    Moreover, a student of mine recently showed that the graded analogue of Lazard's theorem (a module is flat if and only if it is a filtered colimit of free modules) is also true.



    Usually one proves this kind of theorems essentially by a combination of two techniques:




    1. Copy the ungraded proof and just substitute ungraded for graded concepts in the manner sketched above.

    2. If one is annoyed by the length of the resulting argument, use some shortcuts by some translations between ungraded and graded. (E.g. a noetherian graded ring that is graded Cohen-Macaulay is also ungraded Cohen-Macaulay.)


    Sometimes one can also be lucky and the statement is suitably algebro-geometry that one can argue geometrically with the stack $[Spec R/mathbb{G}_m]$ for a graded ring $R$, using that a $mathbb{Z}$-grading corresponds to a $mathbb{G}_m$-action.



    In any case, my question is the following:




    Is there any class of statements, where one knows automatically that the graded analogue is true if the original statement in ungraded commutative algebra is true, without going through the whole proof?




    I am not sure whether one can hope here for a model-theoretic approach as I know almost nothing about model theory, but any such statement could save a lot of work in proving graded analogues of known theorems.










    share|cite|improve this question

























      9












      9








      9


      4





      Many theorems in commutative algebra hold true in a ($mathbb{Z}$-)graded context. More precisely, we can take any theorem in commutative algebra and replace every occurrence of the word




      • commutative ring by commutative graded ring (without the sign for commutativity)

      • module by graded module

      • element by homogeneous element

      • ideal by homogeneous ideal (i.e. ideal generated by homogeneous elements)


      This results in further substitutions, e.g. a $ast$local ring is a graded ring with a unique maximal homogeneous ideal, we get a notion of graded depth etc. After all these substitutions we can ask whether the theorem is still true.



      One book that does some steps in this direction is Cohen-Macaulay rings by Bruns and Herzog, especially Section 1.5. For example, they have as Exercise 1.5.24 the following graded analogue of the Nakayama lemma:




      Let $(R,mathfrak{m})$ be a $ast$local ring, $M$ be a finitely generated graded $R$-module and $N$ a graded submodule. Assume $M = N + mathfrak{m}M$. Then $M = N$.




      Moreover, a student of mine recently showed that the graded analogue of Lazard's theorem (a module is flat if and only if it is a filtered colimit of free modules) is also true.



      Usually one proves this kind of theorems essentially by a combination of two techniques:




      1. Copy the ungraded proof and just substitute ungraded for graded concepts in the manner sketched above.

      2. If one is annoyed by the length of the resulting argument, use some shortcuts by some translations between ungraded and graded. (E.g. a noetherian graded ring that is graded Cohen-Macaulay is also ungraded Cohen-Macaulay.)


      Sometimes one can also be lucky and the statement is suitably algebro-geometry that one can argue geometrically with the stack $[Spec R/mathbb{G}_m]$ for a graded ring $R$, using that a $mathbb{Z}$-grading corresponds to a $mathbb{G}_m$-action.



      In any case, my question is the following:




      Is there any class of statements, where one knows automatically that the graded analogue is true if the original statement in ungraded commutative algebra is true, without going through the whole proof?




      I am not sure whether one can hope here for a model-theoretic approach as I know almost nothing about model theory, but any such statement could save a lot of work in proving graded analogues of known theorems.










      share|cite|improve this question













      Many theorems in commutative algebra hold true in a ($mathbb{Z}$-)graded context. More precisely, we can take any theorem in commutative algebra and replace every occurrence of the word




      • commutative ring by commutative graded ring (without the sign for commutativity)

      • module by graded module

      • element by homogeneous element

      • ideal by homogeneous ideal (i.e. ideal generated by homogeneous elements)


      This results in further substitutions, e.g. a $ast$local ring is a graded ring with a unique maximal homogeneous ideal, we get a notion of graded depth etc. After all these substitutions we can ask whether the theorem is still true.



      One book that does some steps in this direction is Cohen-Macaulay rings by Bruns and Herzog, especially Section 1.5. For example, they have as Exercise 1.5.24 the following graded analogue of the Nakayama lemma:




      Let $(R,mathfrak{m})$ be a $ast$local ring, $M$ be a finitely generated graded $R$-module and $N$ a graded submodule. Assume $M = N + mathfrak{m}M$. Then $M = N$.




      Moreover, a student of mine recently showed that the graded analogue of Lazard's theorem (a module is flat if and only if it is a filtered colimit of free modules) is also true.



      Usually one proves this kind of theorems essentially by a combination of two techniques:




      1. Copy the ungraded proof and just substitute ungraded for graded concepts in the manner sketched above.

      2. If one is annoyed by the length of the resulting argument, use some shortcuts by some translations between ungraded and graded. (E.g. a noetherian graded ring that is graded Cohen-Macaulay is also ungraded Cohen-Macaulay.)


      Sometimes one can also be lucky and the statement is suitably algebro-geometry that one can argue geometrically with the stack $[Spec R/mathbb{G}_m]$ for a graded ring $R$, using that a $mathbb{Z}$-grading corresponds to a $mathbb{G}_m$-action.



      In any case, my question is the following:




      Is there any class of statements, where one knows automatically that the graded analogue is true if the original statement in ungraded commutative algebra is true, without going through the whole proof?




      I am not sure whether one can hope here for a model-theoretic approach as I know almost nothing about model theory, but any such statement could save a lot of work in proving graded analogues of known theorems.







      ac.commutative-algebra graded-rings-modules






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      asked 6 hours ago









      Lennart Meier

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          Let $mathbb P(A,M)$ be a property of a finitely generated module $M$ over a
          Noetherian local ring $A$.
          Let $R$ be a $mathbb Z^n$-graded ring, $N$ a finitely generated graded $R$-module, and
          $P$ a prime ideal of $R$.
          Let $P^*$ be the largest homogeneous ideal contained in $P$ (it is a prime ideal).
          For many important $mathbb P$, we have that $mathbb P(R_P,N_P)$ is equivalent to
          $mathbb P(R_{P^*},N_{P^*})$.



          For example, the property $M=0$ satisfies the equivalence, and the graded Nakayama
          follows from this.



          Indeed, if $mathcal C$ and $mathcal D$ are classes of Noetherian local rings and
          $mathbb P(A,M)$ is a property of a finitely generated module over a Noetherian
          local ring, and assume that



          (1) If $Ainmathcal C$, $M$ a finite $A$-module, $mathbb P(A,M)$, and $Arightarrow B$ is a regular local
          homomorphism essentially of finite type, then $Binmathcal D$ and $mathbb P(B,B
          otimes_AM)$
          holds.



          (2) If $A$ is a Noetherian local ring and $M$ a finite $A$-module, $Arightarrow B$ is a regular local homomorphism essentially of finite type, $Binmathcal D$ and $mathbb P(B,Botimes_AM)$ holds, then $Ainmathcal D$ and $mathbb P(A,M)$ holds.



          Then for any $mathbb Z^n$-graded Noetherian ring $R$ and a finitely generated graded $R$-module
          $N$, if $R_{P^*}inmathcal C$ and $mathbb P(R_{P^*},N_{P^*})$ holds, then $R_Pinmathcal D$ and $mathbb P(R_P,N_P)$.



          For example, letting $mathcal C=mathcal D=text{Cohen-Macaulay}$ and $mathbb P$ be
          anything, we have that $R_{P^*}$ CM implies $R_P$ CM.
          So if $(R,mathfrak m)$ is a ${}^*$local ring and $R$ is graded CM, then
          $R_{mathfrak m}$ is CM, and so $R_Q$ is CM for any graded prime ideal $Q$.
          So $R_P$ is CM for any prime, since $R_{P^*}$ is CM.
          We have proved that $R$ is CM.



          Similar argument is applicable to Gorenstein, local complete intersection, and regular.
          It is also applicable to some $F$-singularities.



          On the other hand, letting $mathcal C=mathcal D=text{Noetherian local}$, and
          letting $mathbb P(A,M)$ to be finite projective dimension,'finite injective
          dimension,' torsionless,'reflexive' and so on, the condition is satisfied.



          Maybe this does not produce any graded "analogue," but I hope this is useful.
          Please see section 7 of M. Hashimoto and M. Miyazaki, Comm. Algebra 2013 for details.






          share|cite|improve this answer








          New contributor




          P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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            0














            Maybe not really what you are after, but anyway: This holds for all those statements that depend only on the structure of abelian category on the category of graded modules. Namely, if $G$ is any commutative group, then the category of $G$-graded $R$-modules, where $R$ is a $G$-graded ring, is abelian and fulfills AB5 and AB4, hence has a projective generator and an injective cogenerator.



            However, one has to be careful with this approach, since there are still many differences between these categories for different groups $G$. For example, suppose that $R$ is a noetherian $G$-graded ring. If $G$ is finite, then the category of $G$-graded $R$-modules has a noetherian generator, but if $G$ is infinite then it does not.






            share|cite|improve this answer





















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              Let $mathbb P(A,M)$ be a property of a finitely generated module $M$ over a
              Noetherian local ring $A$.
              Let $R$ be a $mathbb Z^n$-graded ring, $N$ a finitely generated graded $R$-module, and
              $P$ a prime ideal of $R$.
              Let $P^*$ be the largest homogeneous ideal contained in $P$ (it is a prime ideal).
              For many important $mathbb P$, we have that $mathbb P(R_P,N_P)$ is equivalent to
              $mathbb P(R_{P^*},N_{P^*})$.



              For example, the property $M=0$ satisfies the equivalence, and the graded Nakayama
              follows from this.



              Indeed, if $mathcal C$ and $mathcal D$ are classes of Noetherian local rings and
              $mathbb P(A,M)$ is a property of a finitely generated module over a Noetherian
              local ring, and assume that



              (1) If $Ainmathcal C$, $M$ a finite $A$-module, $mathbb P(A,M)$, and $Arightarrow B$ is a regular local
              homomorphism essentially of finite type, then $Binmathcal D$ and $mathbb P(B,B
              otimes_AM)$
              holds.



              (2) If $A$ is a Noetherian local ring and $M$ a finite $A$-module, $Arightarrow B$ is a regular local homomorphism essentially of finite type, $Binmathcal D$ and $mathbb P(B,Botimes_AM)$ holds, then $Ainmathcal D$ and $mathbb P(A,M)$ holds.



              Then for any $mathbb Z^n$-graded Noetherian ring $R$ and a finitely generated graded $R$-module
              $N$, if $R_{P^*}inmathcal C$ and $mathbb P(R_{P^*},N_{P^*})$ holds, then $R_Pinmathcal D$ and $mathbb P(R_P,N_P)$.



              For example, letting $mathcal C=mathcal D=text{Cohen-Macaulay}$ and $mathbb P$ be
              anything, we have that $R_{P^*}$ CM implies $R_P$ CM.
              So if $(R,mathfrak m)$ is a ${}^*$local ring and $R$ is graded CM, then
              $R_{mathfrak m}$ is CM, and so $R_Q$ is CM for any graded prime ideal $Q$.
              So $R_P$ is CM for any prime, since $R_{P^*}$ is CM.
              We have proved that $R$ is CM.



              Similar argument is applicable to Gorenstein, local complete intersection, and regular.
              It is also applicable to some $F$-singularities.



              On the other hand, letting $mathcal C=mathcal D=text{Noetherian local}$, and
              letting $mathbb P(A,M)$ to be finite projective dimension,'finite injective
              dimension,' torsionless,'reflexive' and so on, the condition is satisfied.



              Maybe this does not produce any graded "analogue," but I hope this is useful.
              Please see section 7 of M. Hashimoto and M. Miyazaki, Comm. Algebra 2013 for details.






              share|cite|improve this answer








              New contributor




              P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.























                2














                Let $mathbb P(A,M)$ be a property of a finitely generated module $M$ over a
                Noetherian local ring $A$.
                Let $R$ be a $mathbb Z^n$-graded ring, $N$ a finitely generated graded $R$-module, and
                $P$ a prime ideal of $R$.
                Let $P^*$ be the largest homogeneous ideal contained in $P$ (it is a prime ideal).
                For many important $mathbb P$, we have that $mathbb P(R_P,N_P)$ is equivalent to
                $mathbb P(R_{P^*},N_{P^*})$.



                For example, the property $M=0$ satisfies the equivalence, and the graded Nakayama
                follows from this.



                Indeed, if $mathcal C$ and $mathcal D$ are classes of Noetherian local rings and
                $mathbb P(A,M)$ is a property of a finitely generated module over a Noetherian
                local ring, and assume that



                (1) If $Ainmathcal C$, $M$ a finite $A$-module, $mathbb P(A,M)$, and $Arightarrow B$ is a regular local
                homomorphism essentially of finite type, then $Binmathcal D$ and $mathbb P(B,B
                otimes_AM)$
                holds.



                (2) If $A$ is a Noetherian local ring and $M$ a finite $A$-module, $Arightarrow B$ is a regular local homomorphism essentially of finite type, $Binmathcal D$ and $mathbb P(B,Botimes_AM)$ holds, then $Ainmathcal D$ and $mathbb P(A,M)$ holds.



                Then for any $mathbb Z^n$-graded Noetherian ring $R$ and a finitely generated graded $R$-module
                $N$, if $R_{P^*}inmathcal C$ and $mathbb P(R_{P^*},N_{P^*})$ holds, then $R_Pinmathcal D$ and $mathbb P(R_P,N_P)$.



                For example, letting $mathcal C=mathcal D=text{Cohen-Macaulay}$ and $mathbb P$ be
                anything, we have that $R_{P^*}$ CM implies $R_P$ CM.
                So if $(R,mathfrak m)$ is a ${}^*$local ring and $R$ is graded CM, then
                $R_{mathfrak m}$ is CM, and so $R_Q$ is CM for any graded prime ideal $Q$.
                So $R_P$ is CM for any prime, since $R_{P^*}$ is CM.
                We have proved that $R$ is CM.



                Similar argument is applicable to Gorenstein, local complete intersection, and regular.
                It is also applicable to some $F$-singularities.



                On the other hand, letting $mathcal C=mathcal D=text{Noetherian local}$, and
                letting $mathbb P(A,M)$ to be finite projective dimension,'finite injective
                dimension,' torsionless,'reflexive' and so on, the condition is satisfied.



                Maybe this does not produce any graded "analogue," but I hope this is useful.
                Please see section 7 of M. Hashimoto and M. Miyazaki, Comm. Algebra 2013 for details.






                share|cite|improve this answer








                New contributor




                P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.





















                  2












                  2








                  2






                  Let $mathbb P(A,M)$ be a property of a finitely generated module $M$ over a
                  Noetherian local ring $A$.
                  Let $R$ be a $mathbb Z^n$-graded ring, $N$ a finitely generated graded $R$-module, and
                  $P$ a prime ideal of $R$.
                  Let $P^*$ be the largest homogeneous ideal contained in $P$ (it is a prime ideal).
                  For many important $mathbb P$, we have that $mathbb P(R_P,N_P)$ is equivalent to
                  $mathbb P(R_{P^*},N_{P^*})$.



                  For example, the property $M=0$ satisfies the equivalence, and the graded Nakayama
                  follows from this.



                  Indeed, if $mathcal C$ and $mathcal D$ are classes of Noetherian local rings and
                  $mathbb P(A,M)$ is a property of a finitely generated module over a Noetherian
                  local ring, and assume that



                  (1) If $Ainmathcal C$, $M$ a finite $A$-module, $mathbb P(A,M)$, and $Arightarrow B$ is a regular local
                  homomorphism essentially of finite type, then $Binmathcal D$ and $mathbb P(B,B
                  otimes_AM)$
                  holds.



                  (2) If $A$ is a Noetherian local ring and $M$ a finite $A$-module, $Arightarrow B$ is a regular local homomorphism essentially of finite type, $Binmathcal D$ and $mathbb P(B,Botimes_AM)$ holds, then $Ainmathcal D$ and $mathbb P(A,M)$ holds.



                  Then for any $mathbb Z^n$-graded Noetherian ring $R$ and a finitely generated graded $R$-module
                  $N$, if $R_{P^*}inmathcal C$ and $mathbb P(R_{P^*},N_{P^*})$ holds, then $R_Pinmathcal D$ and $mathbb P(R_P,N_P)$.



                  For example, letting $mathcal C=mathcal D=text{Cohen-Macaulay}$ and $mathbb P$ be
                  anything, we have that $R_{P^*}$ CM implies $R_P$ CM.
                  So if $(R,mathfrak m)$ is a ${}^*$local ring and $R$ is graded CM, then
                  $R_{mathfrak m}$ is CM, and so $R_Q$ is CM for any graded prime ideal $Q$.
                  So $R_P$ is CM for any prime, since $R_{P^*}$ is CM.
                  We have proved that $R$ is CM.



                  Similar argument is applicable to Gorenstein, local complete intersection, and regular.
                  It is also applicable to some $F$-singularities.



                  On the other hand, letting $mathcal C=mathcal D=text{Noetherian local}$, and
                  letting $mathbb P(A,M)$ to be finite projective dimension,'finite injective
                  dimension,' torsionless,'reflexive' and so on, the condition is satisfied.



                  Maybe this does not produce any graded "analogue," but I hope this is useful.
                  Please see section 7 of M. Hashimoto and M. Miyazaki, Comm. Algebra 2013 for details.






                  share|cite|improve this answer








                  New contributor




                  P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  Let $mathbb P(A,M)$ be a property of a finitely generated module $M$ over a
                  Noetherian local ring $A$.
                  Let $R$ be a $mathbb Z^n$-graded ring, $N$ a finitely generated graded $R$-module, and
                  $P$ a prime ideal of $R$.
                  Let $P^*$ be the largest homogeneous ideal contained in $P$ (it is a prime ideal).
                  For many important $mathbb P$, we have that $mathbb P(R_P,N_P)$ is equivalent to
                  $mathbb P(R_{P^*},N_{P^*})$.



                  For example, the property $M=0$ satisfies the equivalence, and the graded Nakayama
                  follows from this.



                  Indeed, if $mathcal C$ and $mathcal D$ are classes of Noetherian local rings and
                  $mathbb P(A,M)$ is a property of a finitely generated module over a Noetherian
                  local ring, and assume that



                  (1) If $Ainmathcal C$, $M$ a finite $A$-module, $mathbb P(A,M)$, and $Arightarrow B$ is a regular local
                  homomorphism essentially of finite type, then $Binmathcal D$ and $mathbb P(B,B
                  otimes_AM)$
                  holds.



                  (2) If $A$ is a Noetherian local ring and $M$ a finite $A$-module, $Arightarrow B$ is a regular local homomorphism essentially of finite type, $Binmathcal D$ and $mathbb P(B,Botimes_AM)$ holds, then $Ainmathcal D$ and $mathbb P(A,M)$ holds.



                  Then for any $mathbb Z^n$-graded Noetherian ring $R$ and a finitely generated graded $R$-module
                  $N$, if $R_{P^*}inmathcal C$ and $mathbb P(R_{P^*},N_{P^*})$ holds, then $R_Pinmathcal D$ and $mathbb P(R_P,N_P)$.



                  For example, letting $mathcal C=mathcal D=text{Cohen-Macaulay}$ and $mathbb P$ be
                  anything, we have that $R_{P^*}$ CM implies $R_P$ CM.
                  So if $(R,mathfrak m)$ is a ${}^*$local ring and $R$ is graded CM, then
                  $R_{mathfrak m}$ is CM, and so $R_Q$ is CM for any graded prime ideal $Q$.
                  So $R_P$ is CM for any prime, since $R_{P^*}$ is CM.
                  We have proved that $R$ is CM.



                  Similar argument is applicable to Gorenstein, local complete intersection, and regular.
                  It is also applicable to some $F$-singularities.



                  On the other hand, letting $mathcal C=mathcal D=text{Noetherian local}$, and
                  letting $mathbb P(A,M)$ to be finite projective dimension,'finite injective
                  dimension,' torsionless,'reflexive' and so on, the condition is satisfied.



                  Maybe this does not produce any graded "analogue," but I hope this is useful.
                  Please see section 7 of M. Hashimoto and M. Miyazaki, Comm. Algebra 2013 for details.







                  share|cite|improve this answer








                  New contributor




                  P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  share|cite|improve this answer



                  share|cite|improve this answer






                  New contributor




                  P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  answered 41 mins ago









                  P. Grape

                  714




                  714




                  New contributor




                  P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                  New contributor





                  P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  P. Grape is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.























                      0














                      Maybe not really what you are after, but anyway: This holds for all those statements that depend only on the structure of abelian category on the category of graded modules. Namely, if $G$ is any commutative group, then the category of $G$-graded $R$-modules, where $R$ is a $G$-graded ring, is abelian and fulfills AB5 and AB4, hence has a projective generator and an injective cogenerator.



                      However, one has to be careful with this approach, since there are still many differences between these categories for different groups $G$. For example, suppose that $R$ is a noetherian $G$-graded ring. If $G$ is finite, then the category of $G$-graded $R$-modules has a noetherian generator, but if $G$ is infinite then it does not.






                      share|cite|improve this answer


























                        0














                        Maybe not really what you are after, but anyway: This holds for all those statements that depend only on the structure of abelian category on the category of graded modules. Namely, if $G$ is any commutative group, then the category of $G$-graded $R$-modules, where $R$ is a $G$-graded ring, is abelian and fulfills AB5 and AB4, hence has a projective generator and an injective cogenerator.



                        However, one has to be careful with this approach, since there are still many differences between these categories for different groups $G$. For example, suppose that $R$ is a noetherian $G$-graded ring. If $G$ is finite, then the category of $G$-graded $R$-modules has a noetherian generator, but if $G$ is infinite then it does not.






                        share|cite|improve this answer
























                          0












                          0








                          0






                          Maybe not really what you are after, but anyway: This holds for all those statements that depend only on the structure of abelian category on the category of graded modules. Namely, if $G$ is any commutative group, then the category of $G$-graded $R$-modules, where $R$ is a $G$-graded ring, is abelian and fulfills AB5 and AB4, hence has a projective generator and an injective cogenerator.



                          However, one has to be careful with this approach, since there are still many differences between these categories for different groups $G$. For example, suppose that $R$ is a noetherian $G$-graded ring. If $G$ is finite, then the category of $G$-graded $R$-modules has a noetherian generator, but if $G$ is infinite then it does not.






                          share|cite|improve this answer












                          Maybe not really what you are after, but anyway: This holds for all those statements that depend only on the structure of abelian category on the category of graded modules. Namely, if $G$ is any commutative group, then the category of $G$-graded $R$-modules, where $R$ is a $G$-graded ring, is abelian and fulfills AB5 and AB4, hence has a projective generator and an injective cogenerator.



                          However, one has to be careful with this approach, since there are still many differences between these categories for different groups $G$. For example, suppose that $R$ is a noetherian $G$-graded ring. If $G$ is finite, then the category of $G$-graded $R$-modules has a noetherian generator, but if $G$ is infinite then it does not.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 2 hours ago









                          Fred Rohrer

                          4,37111734




                          4,37111734






























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