For the maximal ideal $\mathfrak m$ of $(A,\mathfrak m)$, the <associated graded ring> is
$$
G_{\mathfrak m}(A)
=\operatorname{gr}_{\mathfrak m}(A)
=\bigoplus_{n\geq0}\mathfrak m^n/\mathfrak m^{n+1}.
$$
If $x+\mathfrak m^{r+1}$ and $y+\mathfrak m^{s+1}$ are homogeneous classes, their product is
$$
xy+\mathfrak m^{r+s+1}.
$$
It is a graded algebra over the residue field $k=A/\mathfrak m$.

The <Hilbert series> is
$$
H_{G_{\mathfrak m}(A)}(t)
=\sum_{n\geq0}
\dim_k(\mathfrak m^n/\mathfrak m^{n+1})t^n.
$$
The Hilbert-Serre theorem makes this rational. The number $d(G_{\mathfrak m}(A))$ is the order of its pole at $t=1$, as recorded by the <pole dimension of an associated graded ring>.

The <Dimension theorem for Noetherian local rings> states
$$
\boxed{\dim A
=\dim G_{\mathfrak m}(A)
=d(G_{\mathfrak m}(A))}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 101 / 4 / ii / a / Solution

{title2=$d(\operatorname{gr}_{\mathfrak m}(A))$}

For a Noetherian local ring $(A,\mathfrak m)$, the pole dimension $d(\operatorname{gr}_{\mathfrak m}(A))$ is the order of the pole at $t=1$ of
$$
\sum_{n\geq0}\dim_{A/\mathfrak m}(\mathfrak m^n/\mathfrak m^{n+1})t^n.
$$
The Hilbert-Serre theorem makes this series rational, so the order is defined.


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 Incoming links: Pole dimension of an associated graded ring