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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.

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  • $\begingroup$ 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). $\endgroup$
    – Theo Johnson-Freyd
    Commented Nov 13, 2010 at 17:45
  • $\begingroup$ @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). $\endgroup$
    – Timothy Wagner
    Commented Nov 13, 2010 at 17:55
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    $\begingroup$ 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? $\endgroup$
    – Victor Protsak
    Commented Nov 13, 2010 at 18:47
  • $\begingroup$ @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. $\endgroup$
    – Timothy Wagner
    Commented Nov 17, 2010 at 5:01

3 Answers 3

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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.

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  • $\begingroup$ Thanks Daniel, for the references. I shall look them up. $\endgroup$
    – Timothy Wagner
    Commented Nov 17, 2010 at 5:01
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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]

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  • $\begingroup$ Is there a reference which deals with the analogous question but for Hochschild cohomological dimension? $\endgroup$
    – ABIM
    Commented Jul 11, 2014 at 3:00
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    $\begingroup$ @CSA, it is not true, so no :-). For example, suppose $2\neq0$ in the field $k$; then the algebra $k[x]/(x^2-1)$ has zero dimension as a bimodule over itself, yet there is an increasing filtration on it such that the associated graded algebra is $k[x]/(x^2)$, which has infinite H dimension. In any case, there are several different things you can mean by «analogous»... $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jul 11, 2014 at 3:44
  • $\begingroup$ hm... then would you know of a technique for calculating the hochschild c. dimension of $k<x,y>/(xy-yx-x)$? Since the result in your paper (from C&E's Homological algebra) does not apply and neither does this (^false hypothesised) result... $\endgroup$
    – ABIM
    Commented Jul 11, 2014 at 15:53
  • $\begingroup$ @CSA, that algebra is the enveloping algebra of the unique non-abelian Lie algebra of dimension 2. It's global dimension is equal to is bimodule projective dimension is equal to 2. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jul 11, 2014 at 18:53
  • $\begingroup$ Cool, thanks Mariano!.. but just so I can understand the nooks and crannies of that argument would you have a reference (that particular algebra is extremely important for a counterexample Im working on at you just confirmed my intuition that it indeed does work) $\endgroup$
    – ABIM
    Commented Jul 11, 2014 at 20:39
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The answer is wrong, one needs the ring to be noetherian, it's false in general. For instance, take a local ring $A$ with a finite type maximal ideal, then one has that $\dim(gr(A))=\dim(gr(\hat{A}))$ and for such a ring the completion is noetherian, so then you get that $\dim(gr(A))=dim(\hat{A})$. But at the same time, by considering valuation rings, you may have $\dim(A)\neq\dim(\hat{A})$

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    $\begingroup$ Or the question is wrong in that the ring is implicitly assumed to be noetherian...! Anyway, it's a useful complementary answer. $\endgroup$
    – YCor
    Commented Apr 2, 2020 at 17:25

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