By Lennart Meier


2019-01-07 08:31:52 8 Comments

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.

3 comments

@Konstantinos Kanakoglou 2019-01-08 02:44:33

Since the OP is asking for some "class of statements" which can be readily transferred from the ungraded to the graded case, let me outline a couple of thoughts, which are not really tied to neither the commutative case, nor the $\mathbb{Z}$-gradation itself:

  • On the one hand, if $R$ is a $k$-algebra and $G$ is a group, then a $G$-gradation on $R$ can be equivalently viewed either as a $(kG)^*$-action or as a $kG$-coaction (where $kG$ is the group hopf algebra and $(kG)^*$ its dual hopf algebra). So, the graded algebra $R$ is alternatively viewed as an algebra in the category of $(kG)^*$-modules or an algebra in the category of $kG$-comodules. In this sense, the statements and the properties which are "naturally" transferred between the graded and the ungraded case are those which are in some sense "compatible" or "invariant" with respect to the $kG$-coaction (or the $(kG)^*$-action). I do not know if it sounds like an abuse of terminology to speak about

    properties in the category of $(kG)^*$-modules or properties in the category of $kG$-comodules

  • on the other hand, the Category $R_{gr}$-$mod$ of graded modules over the $G$-graded ring $R$, is equivalent (actually i think it is isomorphic in most cases) to a closed subcategory of modules (in the ungraded sense) over the smash product algebra $R\sharp kG$. (Here the smash product is understood in the sense of the one associated to a Hopf module algebra).
    In this sense, we could say that the statements and properties which one might expect to "naturally transfer" over between graded and ungraded, are

    the ones which are preserved under the smash product between $R$ and the group hopf algebra $kG$.

    (In some sense, this implies that it might be meaningful to consider properties which "transfer" between the $R$-modules and the $R\sharp kG$-modules).

One might reasonably expect that

all those statements that depend only on the structure of abelian category on the category of graded modules

mentioned in Fred Rohrer's answer fall into the classes mentioned above, however i can't tell for sure on the exact overlap between these classes.

@P. Grape 2019-01-07 14:37:16

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 $A\in\mathcal C$, $M$ a finite $A$-module, $\mathbb P(A,M)$, and $A\rightarrow B$ is a regular local homomorphism essentially of finite type, then $B\in\mathcal D$ and $\mathbb P(B,B \otimes_AM)$ holds.

(2) If $A$ is a Noetherian local ring and $M$ a finite $A$-module, $A\rightarrow B$ is a regular local homomorphism essentially of finite type, $B\in\mathcal D$ and $\mathbb P(B,B\otimes_AM)$ holds, then $A\in\mathcal D$ 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 $R_{P^*}\in\mathcal C$ and $\mathbb P(R_{P^*},N_{P^*})$ holds, then $R_P\in\mathcal D$ 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.

@Fred Rohrer 2019-01-07 12:24:33

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.

Related Questions

Sponsored Content

1 Answered Questions

0 Answered Questions

Hironaka decomposition over $\mathbb{Z}$?

1 Answered Questions

[SOLVED] Serre condition $(S_n)$

0 Answered Questions

0 Answered Questions

1 Answered Questions

Sponsored Content