Hilbert-Serre theorem

Let $S=k[x_1,\dots ,x_r]$ be graded with positive degrees $d_i=\mathrm{deg}{x}_i$, and let $M$ be a finitely generated graded $S$-module with finite-dimensional graded pieces. Thenfor a Laurent polynomial $P(t)\in \mathbb{Z}[t,t^{-1}]$. For a standard grading, ${\dim }_k{M}_n$ consequently agrees with a polynomial for all sufficiently large $n$.

The proof is by induction on $r$. For $x=x_r$ of degree $d$, the exact sequence of graded modulesgivesBoth modules on the right are finitely generated over $k[x_1,\dots ,x_{r-1}]$, so induction supplies the asserted denominator.

Let $S_0$ be a commutative Noetherian ring, let a graded $S$ be generated over it by finitely many positive-degree homogeneous elements, and let $M$ be a finitely generated graded module. For an integer-valued function $\lambda$ additive on exact sequences of finitely generated $S_0$-modules, its graded-piece series has the displayed form. The exact sequence for multiplication by the last variable gives $(1-t^d)P_M=P_{M/xM}-t^d{P}_{(0{:}_{M}x)}$. Those two modules are finite over the ring with one fewer variable, and induction proves rationality. Finite graded pieces and the finite-generation assumptions cannot simply be omitted.

For a homogeneous element $x$ of degree $e>0$ acting on a finite graded module, set $N=(0{:}_{M}x)$ and $C=M/xM$. The multiplication exact sequence $0\to N(-e)\to M(-e)\to M\to C\to 0$ and additivity of component lengths give $(1-t^e)P_M=P_C-t^e{P}_N$. This is the inductive step in the Hilbert-Serre theorem; the annihilator term must be kept unless multiplication by $x$ is injective.

Let a commutative $k$-algebra $R$ be generated by $x_1,\dots ,x_n$, and let $R_j$ be the span of words of total degree at most $j$ in those generators. The associated graded ring of this filtration is a standard graded algebra, andThe Hilbert-Serre theorem therefore makes ${\dim }_k{R}_j$ eventually a polynomial in $j$. Its degree is $\dim R$; this follows by applying the Rees-ring deformation, whose special fibers are $R$ and the associated graded ring, so they have the same Krull dimension.

Forwith the four displayed generators, $R_j$ has basis $Y_1^a{Y}_2^b$ with $|a|+|b|\le j$. The number of integer lattice points in this diamond isIts quadratic growth agrees with $\dim R=2$.