Many theorems in commutative algebra apply in one ($ mathbb {Z} $-) graduated context. More specifically, we can use any sentence in commutative algebra and replace any occurrence of the word

- Commutative ring by commutative grading ring (unsigned for commutativity)
- Module through graduated module
- Element by homogeneous element
- Ideal by homogeneous ideal (that is, ideally generated by homogeneous elements)

This leads to further substitutions, eg. on $ ast $Local ring is a graded ring with a unique maximum homogeneous ideal, we get an idea of graded depth, etc. After all these substitutions we can ask if the sentence is still true.

A book that takes a few steps in this direction is *Cohen-Macaulay rings* Bruns and Herzog, especially section 1.5. For example, in Exercise 1.5.24 they have the following graded analogue of the Nakayama lemma:

To let $ (R, mathfrak {m}) $ be a $ ast $local ring, $ M $ to be a finitely graded one $ R $Module and $ N $ a graduated submodule. Accept $ M = N + mathfrak {m} M $, Then $ M = N $,

A student of mine has recently shown that the graduated analogue of Lazard's theorem (a module is flat even if it is a filtered colimit of free modules) is also true.

Normally, this kind of theorem is essentially proved by a combination of two techniques:

- Copy the ungraded proof and replace graded terms in the manner outlined above.
- If you get annoyed about the length of the resulting argument, use some abbreviations for some translations between unrated and graded. (For example, a noetheric ring graded after Cohen-Macaulay is also unnamed Cohen-Macaulay.)

Sometimes you can be lucky, and the statement is suitably an algebra geometry that you can argue geometrically with the stack $[Spec R/mathbb{G}_m]$ for a stepped ring $ R $with this a $ mathbb {Z} $-grading corresponds to one $ mathbb {G} _m $-Action.

In any case, my question is this:

Is there a class of statements that automatically recognizes that the graded analog is true if the original statement in the unclassified commutative algebra is true without going through all the proof?

I am not sure if one can hope for a model-theoretical approach here, since I know almost nothing about model theory, but such a statement could save a lot of work in the detection of graded analogues of known theorems.