I asked this question some months ago on math.stackexchange.com:


It would be great (for me) to get it resolved.

Let $R$ be a non-zero $\bf{Z}$-graded ring (with no commutativity assumptions whatsoever). Recall that $R$ is "graded-local" if the following equivalent conditions hold:

  1. $R$ has a unique homogeneous left ideal that is maximal among the proper homogeneous left ideals;

  2. $R$ has a unique homogeneous right ideal that is maximal among the proper homogeneous right ideals;

  3. The sum of two homogeneous non-units is again a non-unit.

Note, in particular, that (3) tells us that $R_0$, the ring of degree 0 elements, is a local ring.

But it seems to me that the converse is true. That is, it seems to me that a $\bf{Z}$-graded ring $R$ is "graded-local" if and only if $R_0$ is a local ring. Since this seems a bit suspicious, I have come here to find out if this is really the case.

Here is my argument:

Let $J^g(R)$ denote the intersection of all the homogeneous left ideals that are maximal among the proper homogeneous left ideals. (This is the "graded Jacobson radical". Note that there should be at least one such "maximal homogeneous left ideal" because $R\neq 0$). As an intersection of proper homogeneous left ideals, $J^g(R$) is a proper homogeneous left ideal. I claim that it is maximal among the proper homogeneous left ideals, and hence is the unique such "maximal homogeneous left ideal".

Consider a homogeneous left ideal $I$ with $J^g(R) \subsetneq I$. Since $I$ is homogeneous there exists a homogeneous element $a\in I$ with $a \not\in J^g(R)$. Since $a$ is not in $J^g(R)$ there exists a maximal homogeneous left ideal $\frak{m}$ with $a \not \in \frak{m}$. Well, $R a + \frak{m}$ is a homogeneous left ideal, so by maximality of $\frak{m}$, $1=ra + m$ for some $r\in R$ and $m \in \frak{m}$. Taking the degree 0 components we have $1=r_0a + m_0$ where $m_0$ is a degree zero element in $\frak{m}$ and $r_0a$ is also homogeneous of degree 0. Thus, since $R_0$ is local, either $m_0$ or $r_0a$ is a unit. But $m_0$ cannot be a unit (since it would contradict the properness of $\frak{m}$). Hence $r_0a$ is a unit. In particular, $r_0a$ is left invertible, and thus $a$ is also left invertible. Thus, $I=R$. We conclude that $J^g(R)$ is a "maximal homogeneous left ideal" and hence is the unique such.

Can you see any problems with this? Is it really true that a $\bf{Z}$-graded ring $R$ is graded-local iff $R_0$ is local?

Note: It is clear that an $\bf{N}$-graded ring $R$ is "graded local" iff $R_0$ is local. (If $m_0$ is the unique maximal left ideal of $R_0$ then $m := m_0 \oplus \bigoplus_{d>0} R_d$ is the unique maximal left ideal of $R$.) But this "easy" argument for $\bf{N}$-graded rings doesn't work for $\bf{Z}$-graded rings.

If my claim is false, then a follow up question is: to show that a (not necessarily commutative) $\bf{Z}$-graded ring is graded-local, does it suffice to show that the sum of two homogeneous non-units of the same degree is again a non-unit. This is true for commutative $\bf{Z}$-graded rings, but I'm getting stuck up on the non-commutative ones.

Thank you for reading.

Update The paper by Huishi Li mentioned below convinces me that the result is true (and that my proof is fine and can be simplified as mentioned in the comments). The paper by Huishi Li hasn't been published (beyond being put on the arXiv) so I still wonder if there is a clearly stated reference for this result in the published literature. I've checked the following three books:

  • Nastasescu & van Oystaeyen -- Graded and filtered rings and modules (LNM758, 1979)

  • Nastasescu & van Oystaeyen -- Graded ring theory (1982)

  • Nastasescu & van Oystaeyen -- Methods of graded rings (LNM1836, 2004)

and the result isn't stated in any of them. The second one mentions that $R$ gr-local implies that $R_0$ is local but doesn't say anything about the converse.

I'll leave the question open for a bit longer to see if anyone else chimes in, and then accept Gjergji's answer. Thanks!

  • $\begingroup$ Have you consulted the book "Methods of graded rings" (Springer Lecture Notes in Mathematics) by Constantin Nastasescu and Fred Van Oystaeyen? Chapter 2.9 deals with the "graded Jacobson radical" and in Corollary 2.9.3 they show that $J^g(R) \cap R_0 = J(R_0)$ (in fact they show this for gradations by arbitrary groups). I don't know if this is of any help to you though, but I thought I'd better mention it. $\endgroup$ Jun 27, 2012 at 10:03
  • $\begingroup$ Is it clear that if the sum of any <i>two</i> homogeneous non-units is a non-unit, then the sum of any three homogeneous non-units is a non-unit? $\endgroup$
    – YCor
    Jun 27, 2012 at 22:08
  • $\begingroup$ In my opinion your proof works fine. The writing can be a bit simplified by taking some max. hom. left ideal $\mathfrak{m}$ (in place of the graded Jacobson ideal). Then, if $I \not\subseteq \mathfrak{m}$ is any hom. ideal, your proof shows $I=R$. Hence $\mathfrak{m}$ is the only max. hom. left ideal. $\endgroup$
    – Ralph
    Jun 27, 2012 at 23:20
  • $\begingroup$ @Johan and Ralph: Thanks for the pointers. I'm now pretty convinced that my proof is fine. $\endgroup$ Jun 28, 2012 at 8:28
  • $\begingroup$ @Yves: Good point. It isn't directly clear why the sum of two homogeneous non-units being a non-unit would imply that the sum of three homogeneous non-units is a non-unit. But if we believe the result, it looks like this turns out to be the case. $\endgroup$ Jun 28, 2012 at 8:31

2 Answers 2


I don't have my hands on the book of Nastasescu and Van Oystaeyen on graded rings, which I believe contains this result, but here is a reference to a paper by Huishi Li, "On Monoid Graded Local Rings", where the author proves, in section 2, that in a $\Gamma$-graded ring, $$A=\bigoplus_{\gamma\in\Gamma}A_{\gamma},$$where $\Gamma$ is a cancellation monoid with neutral element $e$, one has that $A_e$ is a local ring if and only if the graded two-sided ideal generated by homogeneous non-units is proper.

  • $\begingroup$ Nice answer! Just a curiosity... Huishi Li, whose paper you refer to, is the same guy that wrote the book "Zariskian Filtrations" (K-Monographs in Mathematics) together with Van Oystaeyen. $\endgroup$ Jun 27, 2012 at 20:51
  • $\begingroup$ @JohanÖinert The answer to your question is yes. $\endgroup$
    – user26857
    Dec 25, 2012 at 20:20

An elementary argument seems to work as well, even for monoids as mentioned previously.$\dagger$

Suppose $M$ and $N$ are distinct maximal homogeneous right ideals. Then $M+N=R$, and there exists $m+n=1$ with $m\in M$ and $n\in N$. Because of the grading, the grade zero parts must be such that $m_0+n_0=1$, and because $M$ and $N$ are both proper and homogeneous, neither $m_0$ nor $n_0$ can be units of $R_0$. This implies $R_0$ is not local.

By contrapositive then, we have shown if $R_0$ is local, then $R$ is graded local.

$\dagger$ I convinced myself that there are maximal proper homogenous ideals, and that the sum of homogeneous ideals is again homogeneous. I hope those lemmas were not heat induced delirium.


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