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Apr 2, 2020 at 17:15 answer added prochet timeline score: 2
Nov 17, 2010 at 5:02 vote accept Timothy Wagner
Nov 17, 2010 at 5:02 vote accept Timothy Wagner
Nov 17, 2010 at 5:02
Nov 17, 2010 at 5:02 vote accept Timothy Wagner
Nov 17, 2010 at 5:02
Nov 17, 2010 at 5:01 comment added Timothy Wagner @Victor: There is no particularly good reason for this. I was planning an expository talk and needed the equality of dimensions above, but I don't think I could develop the entire dimension theory of the extended Rees ring, so I was looking for alternate approaches.
Nov 17, 2010 at 0:39 answer added Mariano Suárez-Álvarez timeline score: 2
Nov 17, 2010 at 0:29 answer added Daniel Erman timeline score: 4
Nov 13, 2010 at 18:47 comment added Victor Protsak The use of the Rees ring is very natural: after all, $gr_I(R)$ is the special fiber of a flat deformation (given by the Rees ring) with the general fiber R. Why do you want to avoid it?
Nov 13, 2010 at 17:55 comment added Timothy Wagner @Theo. Thanks for that comment. I have modified the post and clarified that I mean the Krull dimension of the ring (the supremum of the length of any chain of prime ideals of the ring).
Nov 13, 2010 at 17:54 history edited Timothy Wagner CC BY-SA 2.5
added 8 characters in body; edited title; added 47 characters in body
Nov 13, 2010 at 17:45 comment added Theo Johnson-Freyd What kind of "dimension" do you mean? If you had said "algebra" instead of "ring", I would assume that "dimension" was the dimension of the algebra as a vector space over the ground field, in which case this has nothing to do with rings. But since you say "ring", I'm going to assume you mean some geometric notion (which the experts may say is the obvious and canonical notion --- I am not an expert).
Nov 13, 2010 at 14:32 history edited Timothy Wagner CC BY-SA 2.5
edited body
Nov 13, 2010 at 4:53 history edited Timothy Wagner CC BY-SA 2.5
edited title
Nov 13, 2010 at 4:04 history asked Timothy Wagner CC BY-SA 2.5