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Sep 4, 2015 at 8:18 comment added Pham Hung Quy Thank you Fred, I have a quick look their proof. They used the fact $I$ is irreducible iff the index of reducible of $I$ is one. So their proof need Noetheian condition. My proof is also true for $\mathbb{Z}$-graded (I fell it works for $\mathbb{N}^n$-graded also).
Sep 3, 2015 at 12:54 answer added Thomas Kahle timeline score: 1
Sep 1, 2015 at 12:50 comment added Fred Rohrer The case of noetherian modules over rings graded by the integers is studied in this preprint by Justin Chen and Youngsu Kim.
Sep 1, 2015 at 2:54 answer added Pham Hung Quy timeline score: 3
Feb 5, 2015 at 14:24 history edited user26857 CC BY-SA 3.0
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Apr 7, 2014 at 20:57 history asked user26857 CC BY-SA 3.0