Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A lot of times I see theorems stated for local rings, but usually they are also true for "graded local rings", i.e., graded rings with a unique homogeneous maximal ideal (like the polynomial ring). For example, the Hilbert syzygy theorem, the Auslander-Buchsbaum formula, statements related to local cohomology, etc.

But it's not entirely clear to me how tight this analogy is. I certainly don't expect all statements about local rings to extend to graded local rings, so I'd like to know about some "pitfalls" in case I ever decide to make an "oh yes, this obviously extends" fallacy. What are some examples of statements which are true for local rings whose graded analogues are not necessarily true? Or another related question: what kind of intuition should I have when I want to conclude that statements have graded versions?

There is a notion of "generalized local ring" due to Goto and Watanabe which includes graded local rings and local rings: a positively graded ring that is finitely generated as an algebra over its zeroth degree part, and its zeroth degree part is a local ring, so one possibility is just to see if this weaker definition is enough to prove the statement. Of course the trouble comes when the proofs cite other sources, and become unmanageable to trace back to first principles.

share|improve this question
    
How general a grading are you interested in? Things can be quite bad if you look at rings graded by more general things than the integers I believe. –  Greg Stevenson Oct 19 '09 at 11:11
    
I'm mostly interested in Z-graded rings, and maybe occasionally Z<sup>d</sup>-graded rings. But really I'm interested in homogeneous coordinate rings of projective varieties over some base field. –  Steven Sam Oct 19 '09 at 13:55
    
Another, definition that encompasses both is the following (which I learned in a class taught by Peter May): simply take a "local ring" to be a graded ring with a unique homogeneous maximal ideal. (Basically, in Peter May's approach, you ignore inhomogeneous elements, ideals, etc.) Local rings in the usual sense can be seen as an instance of this by considering them as graded rings concentrated in degree zero. Note that to make this work, one has to allow the maximal ideal to include degree-zero terms (although obviously it won't if the degree-zero ring is a field). –  Charles Staats Jan 25 '12 at 17:25
add comment

2 Answers

up vote 5 down vote accepted

One small thing I know of which changes is that if one has a Z-graded-commutative noetherian ring (where Z is the integers) Matlis' classification of indecomposable injective modules goes through but with one small hiccup.

Every indecomposable injective is isomorphic to E(R/p)[n] for some unique homogeneous prime ideal p but the integer shift n is not necessarily unique although under the hypotheses I think you are interested in one probably gets uniqueness. I can't think of an example where this really causes much of a problem though.

Having thought about this some more I think that non-negative integer graded-local noetherian rings, in particular those generated in degree 1 such that the maximal homogeneous ideal is also maximal if one forgets the grading, are incredibly well behaved and the analogy with local rings is very good. In fact, there is even a version of Nakayama's lemma for such rings (maybe one needs a little more) which is stronger than the usual one in the sense that one can drop the finiteness condition on the module. There are also no problems with graded versions of prime avoidance etc... in general.

I'd recommend section 1.5 of Cohen-Macaulay Rings by Bruns and Herzog where they prove that a bunch of standard facts still go through and one can see what does and doesn't change in the proofs.

As I mentioned in the comment I think one has to be most careful when considering rings graded by things like monoids which aren't as nice as the non-negative integers. In particular, if the grading is not positive (i.e. some elements of the monoid are invertible) and/or if the monoid is not cancellative at the identity (i.e. a+b = a does not imply b is the identity). I think in the non-cancellative case one can construct a counterexample to Nakayama's lemma but I am not 100% sure on this.

share|improve this answer
    
I think the graded Nakayama's lemma is much easier to prove. If the ring is positively graded, we can just use induction: say mM = M where m is the maximal ideal. Then M_0 = 0 since nothing maps to it, and hence M_1 = 0, etc. So maybe I will just reask if I run into a specific statement I am unsure about. Thanks! –  Steven Sam Oct 20 '09 at 23:26
add comment

I will try to provide some geometric intuition, why there should be an analogy between local rings and graded rings with unique homogeneous maximal ideal. Maybe this also helps to guess whether a statement true for local rings should still hold in the graded case. A graded k-Algebra can be thought of as an affine space with $k^*$-action. Homogeneous prime ideals correspond to invariant closed sub-varieties. So your sort of algebras corresponds to spaces with exactly one fixpoint. For example in the case of the polynomial ring its $k^n$ with the obvious action of the multiplicative group and $0$ is the fixpoint. From this example we see, that the action can be used to "contract" the space to the fixed point. Hence the local nature of the space.

share|improve this answer
4  
In a very similar flavor: For a local ring, the closed point is in the closure of any other point. For an $\mathbb N$-graded ring: the central point is in the closure of the $k^\times$-orbit of any other point. –  Allen Knutson Sep 27 '10 at 1:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.