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

For an $\mathfrak{m}$-primary ideal $I$ in a Noetherian local ring, the functionagrees for all sufficiently large $n$ with a polynomial in $n$. This is the Hilbert-Samuel polynomial, also called here the characteristic polynomial of $I$. Using $R/I^{n+1}$ instead merely shifts its variable.

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Suppose $I$ has $s$ generators. Its associated graded ringis generated in degree one by their initial forms, so there is a graded surjectionBecause $I$ is $\mathfrak{m}$-primary, $R/I$ has finite length of a module. The degree-$n$ piece on the left has lengthso $\mathrm{length}(I^n/I^{n+1})$ grows with degree at most $s-1$. Summing these lengths shows that $\mathrm{length}(R/I^n)$ has polynomial degree at most $s$.

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The weighted Hilbert series of $K[x,y,t_1,\dots ,t_n]$ isThe 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

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Let $P=p_1\cdots p_n$ and $A=\mathbb{Z}/P\mathbb{Z}=R_0$. By the Chinese remainder theorem, $A\cong {\prod }_i{\mathbb{F}}_{p_i}$, so ${\mathrm{length}}_A(A)=n$.

If a positive-degree monomial contains both $t_i$ and $t_j$ with $i\ne j$, then it vanishes: Bezout identity gives $u{p}_i+v{p}_j=1$, while both $p_i$ and $p_j$ annihilate that monomial. Thus the degree-$d$ component for $d\ge 1$ isand every summand has length one. Hence every $R_d$ has length $n$, including $d=0$, and

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The denominator has degree one, independently of the number $n$ of variables. This reflects the fact that all mixed monomials vanish and each component of the ring supports only one polynomial direction; equivalently, the Krull dimension of this graded ring is one.

Solved by gpt-5.6-sol high.