Skip to main content
14 events
when toggle format what by license comment
Jun 20, 2017 at 20:25 comment added Johannes Hahn I just want to point out that $\mathbb{Z}$-graded is entirely the wrong assumption here and should be replaced by $\mathbb{N}$-graded. This is because of the counterexample of the group ring $K[\mathbb{Z}]$ for any field $K$. It is like any group ring trivially graded by the group itself, but the only homogeneous non-units are zero so that the only 0 and $K[\mathbb{Z}]$ are graded ideals.
Apr 13, 2017 at 12:19 history edited CommunityBot
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Jul 2, 2012 at 17:33 answer added rschwieb timeline score: 2
Jun 29, 2012 at 6:55 comment added Fred Rohrer Dear @Victoria, the following paper (and especially its Theorem 1) might also be of interest to you: V.P.Camillo, K.R.Fuller, On graded rings with finiteness conditions, Proc. Amer. Math. Soc. 86 (1982), 1-5.
Jun 28, 2012 at 17:21 vote accept Victoria Flat
Jun 28, 2012 at 10:30 comment added Gjergji Zaimi Btw, Li's paper is published sciencedirect.com/science/article/pii/S0022404912001193
Jun 28, 2012 at 8:31 comment added Victoria Flat @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.
Jun 28, 2012 at 8:28 comment added Victoria Flat @Johan and Ralph: Thanks for the pointers. I'm now pretty convinced that my proof is fine.
Jun 28, 2012 at 8:26 history edited Victoria Flat CC BY-SA 3.0
added 900 characters in body
Jun 27, 2012 at 23:20 comment added Ralph 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.
Jun 27, 2012 at 22:08 comment added YCor 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?
Jun 27, 2012 at 10:40 answer added Gjergji Zaimi timeline score: 12
Jun 27, 2012 at 10:03 comment added Johan Öinert 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.
Jun 27, 2012 at 8:28 history asked Victoria Flat CC BY-SA 3.0