Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 101 / 5

This is the Artin--Tate lemma. Choose $A$-algebra generators $x_1,\dots ,x_r$ of $C$ and $B$-module generators $e_1,\dots ,e_s$ of $C$. Writewith coefficients $a_{ij},b_{ij\ell }\in B$, and let $B_0$ be the $A$-subalgebra of $B$ generated by these finitely many coefficients.

The module ${\sum }_j{B}_0{e}_j$ contains the $x_i$, is closed under multiplication, and contains $A$ after including an expression for $1$ among the chosen coefficients. It therefore equals $C$. Thus $C$ is a finite $B_0$-module. The ring $B_0$ is Noetherian by the Hilbert basis theorem, and $B\subseteq C$ is a $B_0$-submodule, so $B$ is a finite $B_0$-module. It follows that $B$ is a finitely generated $A$-algebra.

Let $K$ be a field finitely generated as a $\mathbb{Z}$-algebra. If $K$ has positive characteristic $p$, it is a finitely generated algebra over ${\mathbb{F}}_p$; Zariski lemma makes it a finite algebraic extension of the finite field ${\mathbb{F}}_p$, so $K$ is finite.

Suppose instead that $K$ has characteristic zero. Then $K$ is a finitely generated $\mathbb{Q}$-algebra and Zariski lemma makes it a number field. For algebra generators ${\alpha }_1,\dots ,{\alpha }_r$, choose a nonzero integer $N$ such that every ${\alpha }_i$ is integral over $\mathbb{Z}[1/N]$. The whole algebra $K=\mathbb{Z}[{\alpha }_1,\dots ,{\alpha }_r]$ would then be integral over $\mathbb{Z}[1/N]$. But for a prime $q\nmid N$, the element $1/q\in K$ is not integral over the integrally closed domain $\mathbb{Z}[1/N]$, a contradiction. Hence every field finitely generated over the integers is finite, and in particular no infinite field has that property.

The Poincare series of a graded module isIt is the generating function of the Hilbert function $n\mapsto {\dim }_k{A}_n$.

A Hilbert polynomial of $A$ is a polynomial $h_A(X)\in \mathbb{Q}[X]$ such thatfor every sufficiently large integer $n$. Eventual equality makes this polynomial unique.

Take $A=k[x]$ with $\mathrm{deg}x=2$. This is a Noetherian graded algebra with $A_0=k$, butNo polynomial can agree eventually with these alternating values, so $A$ has no Hilbert polynomial.

A sufficient condition is that $A$ be a standard graded algebra: it is generated as a $k$-algebra by finitely many elements of degree one. The Hilbert-Serre theorem then makes $P_A(t)$ a rational function whose denominator, after cancellation, is a power of $1-t$. When the eventual Hilbert polynomial is nonzero,where the right side uses the order of the pole at $t=1$.

The degree-$n$ component isDimensions therefore satisfy the Cauchy product rule, giving