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.



(Krull) dimension of any associated graded ring of a ring R equals the dimension of R
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.


