At the moment I'm mostly interested in commutative graded rings (and in particular those graded over $\mathbb{N}$), but any comments or references about more general graded rings would also be appreciated.

  1. What is the name for ring homomorphisms which preserve the gradation? What about ring homomorphisms which respect the gradation, e.g. the map $k[x] \to k[x]$ defined by $x \mapsto x^2$?
  2. It is clear that the category of graded rings and homomorphisms which preserve the gradation contains the usual category of rings as a full (even reflective) subcategory. As far as I can tell, the category is complete and the forgetful functor to $\textbf{CRing}$ is is continuous. The construction given by Mac Lane [CWM 2nd ed., p. 123] seems to give a solution set, so that means there is a left adjoint. What is it?
  3. In some contexts, e.g. the Proj construction, it seems that non-homogeneous elements never enter the picture and only get in the way. It is not hard to construct a definition of something akin to a graded ring without non-homogeneous elements (simply take a $\mathbb{N}$-indexed sequence of abelian groups $S_\bullet$ and define multiplication maps $\mu_{n,m} : S_n \otimes S_m \to S_{n+m}$ satisfying an associative law and require $S_0$ to be a ring and $\mu_{0,0} : S_0 \otimes S_0 \to S_0$ to be the ring multiplication), but perhaps there is an advantage of working with non-homogeneous elements that I'm not seeing. Is there?
  4. The irrelevant ideal seems troublesome. Can we define it away somehow? For example, the improper ideal is not prime because its residue ring is the zero ring, which is defined to not be an integral domain.
  5. If $S$ is a graded ring and $\mathfrak{m}$ is a maximal homogeneous ideal, what properties does $S / \mathfrak{m}$ have? Is there an analogue of the non-graded result that $A / \mathfrak{a}$ is a field if and only if $\mathfrak{a}$ is maximal? [Edit: It turns out this is not the question I meant to ask.]
  6. Why are modules graded over $\mathbb{Z}$ but rings over $\mathbb{N}$? Are there difficulties with $\mathbb{Z}$-graded rings?
  7. Finally, why is $S$ the canonical name for a graded ring?
  • $\begingroup$ When you say "graded commutative" do you really just mean graded and commutative with respect to the trivial symmetric monoidal structure, or do you mean "graded-commutative," which means graded and commutative with respect to the nontrivial symmetric monoidal structure? $\endgroup$ – Qiaochu Yuan Jul 13 '11 at 13:53
  • $\begingroup$ @Qiaochu: I mean a graded ring which has commutative multiplication. What monoidal structure are you referring to? $\endgroup$ – Zhen Lin Jul 13 '11 at 13:55
  • $\begingroup$ The one given by $a \otimes b \mapsto (-1)^{|a| |b|} b \otimes a$ on homogeneous elements. For whatever reason, this is what people generally mean when they say "graded-commutative" (which seems mildly confusing to me). $\endgroup$ – Qiaochu Yuan Jul 13 '11 at 13:58
  • $\begingroup$ @Qiaochu: Ah. That is confusing, yes. I'll have to remember that one. I'm not convinced that the forgetful functor has any adjoints (either left or right) at all, but the conditions for Freyd's adjoint functor theorem seem to be satisfied. The trivial grading is the inclusion functor, but I don't believe it's right adjoint to the forgetful functor, since the category consists of gradation-preserving arrows only, meaning if an element has grade $n$, then its homomorphic image also has grade $n$. There is a bigger category (hinted in question 1), but I'm not considering it at the moment. $\endgroup$ – Zhen Lin Jul 13 '11 at 14:04
  • $\begingroup$ yes, I was confused about which forgetful functor you were using ("the degree-zero part" is also a forgetful functor). $\endgroup$ – Qiaochu Yuan Jul 13 '11 at 14:28

Regarding #3, there is no disadvantage (that I can see) in working with non-homogeneous elements, so I don't think it really matters either way in the sense that there are two categories one could write down and they are equivalent, if not even isomorphic.

Regarding #4, I don't see why you would want to do this.

Regarding #6, I don't see a difficulty with $\mathbb{Z}$-graded rings: for example the ring of Laurent polynomials is $\mathbb{Z}$-graded. Probably for whatever applications you're looking at they are just unnecessary. As far as the Proj construction goes, for example, the irrelevant ideal is not an ideal in a $\mathbb{Z}$-graded ring...!

The reason you could still have $\mathbb{Z}$-graded modules over an $\mathbb{N}$-graded ring is that modules have an obvious notion of degree shift, whereas the multiplication on a ring prevents the degree shift of a graded ring from being a graded ring (with the same multiplication, anyway).

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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