Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am not sure if this is appropriate for MO. If not, I shall be happy to take it to SE.

For a local ring $(R,m)$, given any proper ideal $I$, the (Krull) dimension (from here on dimension means Krull dimension) of the associated graded ring of $R$ with respect to $I$, $gr_I(R)=\oplus_{n\geq 0}\frac{I^n}{I^{n+1}}$ is equal to the dimension of $R$ itself. The only proof I know for this involves writing the associated graded ring as a quotient of the extended Rees ring $R[It,t^{-1}]$ and using dimension formulas for the latter. I was wondering if anyone was aware of a proof that does not route via the extended Rees ring. Any references would be appreciated. I googled, but could not stumble upon anything useful.

share|improve this question
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). –  Theo Johnson-Freyd Nov 13 '10 at 17:45
@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). –  Timothy Wagner Nov 13 '10 at 17:55
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? –  Victor Protsak Nov 13 '10 at 18:47
@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. –  Timothy Wagner Nov 17 '10 at 5:01
add comment

2 Answers

up vote 3 down vote accepted

Though I heartily agree with Victor Protsak's comment, I will add some references. These might be useful for you, at least if you haven't seen them before. The references add a restriction, however, by assuming that $I$ is an ideal of finite co-length.

Then Corollary 12.5 of Eisenbud's Commutative Algebra uses the theory of Hilbert-Samuel polynomials to prove that $\text{dim}(R)=\text{dim}\text{ gr}_I(R)$.

Alternately, you might also be interested in Corollary 10.12 of the same book. This second corollary assumes that $I=\mathfrak m$, but the proof makes use of "Going down for flat extensions", which has a somewhat different flavor than the Rees ring approach.

share|improve this answer
Thanks Daniel, for the references. I shall look them up. –  Timothy Wagner Nov 17 '10 at 5:01
add comment

This is dealt with in the generality of non-commutative rings in [McConnell, J. C.; Robson, J. C. Noncommutative Noetherian rings. With the cooperation of L. W. Small. Revised edition. Graduate Studies in Mathematics, 30. American Mathematical Society, Providence, RI, 2001. xx+636 pp. MR1811901]

share|improve this answer
Thanks for the reference, Mariano. –  Timothy Wagner Nov 17 '10 at 5:02
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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