Let $I$ be an ideal of a Noetherian local ring $(R, \mathfrak m)$. Define the associated graded ring $G_I(R):=\bigoplus_{n=0}^\infty I^n/I^{n+1}$. Then $G_I(R)$ is a Noetherian ring of the same dimension as that of $R$ (Theorem 15.7, Matsumura, Commutative Ring Theory).

For a given finitely generated $R$-module $M$, define $$G_I(M):=\bigoplus_{n=0}^\infty \dfrac{I^nM}{I^{n+1}M}.$$ Then, $G_I(M)$ is a finitely generated graded $G_I(R)$-module (Proposition 10.22, Atiyah-Macdonald, Commutative Algebra).

My question is: Is it true that $\dim G_I(M)=\dim M$ ? If needed, I am willing to assume that $I$ is $\mathfrak m$-primary.

Recall that the dimension of a finitely generated module $M$ over a Noetherian ring $R$ is defined as $\dim M:=\dim R/ \operatorname{ann}_R (M)$. So in other words, I'm asking if $$\dim G_I(R)/\operatorname{ann}_{G_I(R)} (G_I(M))=\dim R/\operatorname{ann}_R(M)$$ is true or not.

Any reference would also be very appreciated.

Thanks in advance.

  • $\begingroup$ I guess it's true. It seems to boil down (passing to a quotient) to the case when $M$ has dimension $d$ and every proper quotient of $M$ has dimension $<d$. In this case (modding out by the annihilator, which is then prime) we can suppose that $R$ is a domain, and $M$ is isomorphic to a nonzero finitely generated $R$-submodule of $\mathrm{Frac}(R)$. $\endgroup$
    – YCor
    Mar 29, 2021 at 8:08
  • $\begingroup$ Perhaps Theorem 4.5.8 in Bruns-Herzog, `Cohen-Macaulay Rings', might be of help. $\endgroup$ Mar 30, 2021 at 6:13

1 Answer 1


In Algèbre locale, multiplicités, J.P. Serre shows that the dimension is equal to the degree of the Hilbert--Samuel polynomial. In turn, this polynomial is the same for the module and its associated graded.

Serre has assumed that the ground rind is semi-local. I'd like to know whether this holds, say, over a polynomial ring...


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