{c}

For a Noetherian local ring $(A,\mathfrak m)$, the Krull dimension of $A$, the Krull dimension of the <associated graded ring> $\operatorname{gr}_{\mathfrak m}(A)$, and the order of the pole at $t=1$ of its <Hilbert series> are equal.

Dimension theorem for Noetherian local rings

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

Paper 101 / 4 / ii / a / Solution

The weighted <Hilbert series> of $K[x,y,t_1,\ldots,t_n]$ is
$$
\frac1{(1-z)^2\prod_{i=1}^n(1-z^i)}.
$$
The homogeneous polynomial $x^k-y^k$ has degree $k$ and is a <non-zero-divisor>, so quotienting by it multiplies the series by $1-z^k$. Therefore the requested <Poincare series of a graded module> is
$$
P_R(z)=\frac{1-z^k}{(1-z)^2\prod_{i=1}^n(1-z^i)}.
$$

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 101 / 5 / b / Solution

{wiki=Hilbert_series_and_Hilbert_polynomial}

For a graded module $M=\bigoplus_{n\ge0}M_n$ whose pieces have finite length, its Hilbert series is
$$
H_M(z)=\sum_{n\ge0}\operatorname{length}(M_n)z^n.
$$


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