# Dimension of the associated graded module at an ideal

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.

• 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 $\mathrm{Frac}(R)$.