Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 / 5

The positive-degree part $A_{+}={⨁}_{n>0}{A}_n$ is an ideal and $A/A_{+}\simeq A_0$, so $A_0$ is Noetherian. Since $A$ is Noetherian, $A_{+}$ has finitely many homogeneous generators $x_1,\dots ,x_s$. Induction on degree shows that every positive-degree homogeneous element is a polynomial in the $x_i$ over $A_0$. Thus

For an additive length function $\lambda$ finite on the graded pieces, define the Poincare series of a graded moduleThe Hilbert-Serre theorem states thatfor a Laurent polynomial $f(t)\in \mathbb{Z}[t,t^{-1}]$.

Induct on $s$. For the last generator $x_s$ of degree $k_s$, multiplication gives an exact sequence whose kernel is the $x_s$-torsion and whose cokernel is $M/x_{s}M$. Additivity of $\lambda$ yieldsBoth modules on the right are finite graded modules over the algebra generated by $x_1,\dots ,x_{s-1}$. The induction hypothesis gives the asserted denominator. The case $s=0$ is a finite Laurent polynomial because $M$ is finitely generated over $A_0$.

When every $k_i=1$, cancel common factors to writewhere $d$ is the pole order at $t=1$. Sincethe coefficient of $t^n$ in $P(M,t)$ is, for all sufficiently large $n$, a fixed linear combination of shifted binomial polynomials. It is therefore a polynomial in $n$ of degree exactly $d-1$ unless $P=0$, in which case its degree is $-1$.

The Krull dimension $\dim R$ is the supremum of lengths $n$ of strict chainsof prime ideals. The transcendence degree ${\mathrm{trdeg}}_{k}L$ is the cardinality of a transcendence basis of $L/k$.

By Noether normalization lemma, there are algebraically independent $y_1,\dots ,y_d\in A$ such that $A$ is finite, hence integral, over $k[y_1,\dots ,y_d]$. Their fraction field has transcendence degree $d$, and $L$ is algebraic over it, so ${\mathrm{trdeg}}_{k}L=d$. Going up and incomparability show that an integral extension preserves Krull dimension, while a polynomial ring in $d$ variables over a field has dimension $d$. Hence

A chain of length $s$ in $R/\mathfrak{p}$ lifts to a chainin $R$. Since $R$ is a domain and $\mathfrak{p}\ne 0$, prepending $(0)⊊\mathfrak{p}$ gives a chain of length $s+1$. Therefore $s+1\le r$ and

Going up lifts every prime chain in $A$ to one in $B$, so $\dim B\ge \dim A$. Conversely, contracting a strict chain of primes of $B$ gives a chain in $A$, and the incomparability theorem for integral extensions ensures that no strict inclusion contracts to equality. Thus $\dim A\ge \dim B$, and