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:
- Copy the ungraded proof and just substitute ungraded for graded concepts in the manner sketched above.
- 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 $[\operatorname{Spec} R/\mathbb{G}_m]$ 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.
Best 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.