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

For the maximal ideal $\mathfrak{m}$ of $(A,\mathfrak{m})$, the associated graded ring isIf $x+{\mathfrak{m}}^{r+1}$ and $y+{\mathfrak{m}}^{s+1}$ are homogeneous classes, their product isIt is a graded algebra over the residue field $k=A/\mathfrak{m}$.

The Hilbert series isThe 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

Solved by gpt-5.6-sol high.

Put $\overline{A}=A/(x)$ and $\overline{\mathfrak{m}}=\mathfrak{m}/(x)$. Applying the dimension theorem to $A$ and $\overline{A}$ givesSince $x$ is a non-zero-divisor, it belongs to no minimal prime of the Noetherian ring $A$. Any chain of primes in $A/(x)$ lifts to a chainof primes of $A$ containing $x$. A minimal prime $\mathfrak{q}\subseteq {\mathfrak{p}}_0$ cannot contain $x$, so the inclusion is strict. Prepending $\mathfrak{q}$ gives a chain of length $r+1$ in $A$. Thus the dimension drop by a non-zero-divisor givesCombining these equalities proves

Solved by gpt-5.6-sol high.