Is the dimension of a Noetherian local ring equal to its associated graded ring?

For a noetherian local ring $$A$$ with maximal ideal $$\mathfrak{m}$$, let $$I$$ be a primary ideal in $$A$$, the associated graded ring is $$\bigoplus_{n=0}^{\infty} I^n/I^{n+1}$$

Yes. First we deal with the case $$I=\mathfrak m$$.
If we define $$f(n)=l(A/\mathfrak{m}^n)$$ where $$l$$ denotes the length of a module, then for large $$n$$, $$f(n)$$ equals a polynomial of degree $$d=\dim A$$ in $$n$$. Replacing $$A$$ by its associated graded ring with $$I=\mathfrak m$$ does not change $$f(n)$$, and so does not change $$d$$.
Now consider general $$I$$. Then $$\mathfrak{m}\supseteq I\supseteq \mathfrak{m}^r$$ for some positive integer $$r$$. Therefore $$g(n)=l(A/I^n)$$ satisfies $$f(n)\le g(n)\le f(rn)$$. Again, $$g(n)$$ is a polynomial for large $$n$$, and these inequalities imply that it has the same degree $$d$$ as $$f(n)$$.
If we consider the graded ring $$R=\bigoplus(I^n/I^{n+1})$$, this is also a Noetherian local ring with maximal ideal $$\mathfrak{m}' =\mathfrak{m}/I\oplus (I/I^2)\oplus (I^2/I^3)\oplus\cdots$$ and having $$I'=0\oplus (I/I^2)\oplus (I^2/I^3)\oplus\cdots$$ as a primary ideal. Then $$l(R/I'^n) =g(n)$$ also and $$R$$ also has dimension $$d$$.