(Krull) dimension of any associated graded ring of a ring R equals the dimension of R 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.
 A: 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.
A: 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]
A: 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 $\dim(A)\neq\dim(\hat{A})$
