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

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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 $d$.

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  • $\begingroup$ There is a problem: in general, m is not the only maximal ideal in the associated ring . I think we should do localization at m on the associated ring, then show that it is the maximal ideal of the maximal chain of prime ideals... $\endgroup$
    – user571299
    Aug 27, 2020 at 1:41

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