The eventual Hilbert function of a zero-dimensional projective quotient equals its total length

If $k$ is infinite, set $K=k$ and $S_K=S$; if $k$ is finite, set $K=k(t)$ with fraction field structure as in [L5], so that $K$ is infinite, and set $S_K=S{\otimes }_{k}K$; in the finite case [L3] gives ${\dim }_K(S_K{)}_m={\dim }_k{S}_m$ for every $m$, ${\mathrm{len}}_K(X_K)={\mathrm{len}}_k(X)$ for $X_K=\mathrm{Proj}(S_K)$, and $X_K$ zero-dimensional in the chartwise sense.

Assume now that $k$ is infinite. If $X\mathrm{\ne }\mathrm{\varnothing }$, let $p_1,\dots ,p_r$ be its finitely many points, written as homogeneous primes of $S$ by [L1], and for each $j$ let $V_j=\{a=(a_0,\dots ,a_n)\in k^{n+1}:a_0{x}_0+\cdots +a_n{x}_n\in p_j\}$; each $V_j$ is a proper $k$-subspace, because it is the kernel of the linear map $k^{n+1}\to (S\mathrm{/}{p}_j{)}_1$, which is nonzero as $x_i\mathrm{\notin }{p}_j$ for some $i$; if $X=\mathrm{\varnothing }$ let $L=x_0$, and otherwise [L5] provides $a\mathrm{\notin }{\bigcup }_j{V}_j$ and we set $L=a_0{x}_0+\cdots +a_n{x}_n\in S_1$. In both cases $L\mathrm{\notin }{\mathfrak{p}}_x$ for every point $x\in X$, so $X\subseteq D_{+}(L)$.

Consequently it suffices to prove the displayed equality for the pair $(S_K,K)$: if ${\dim }_K(S_K{)}_m={\mathrm{len}}_K(X_K)$ holds for all $m\ge m_0$, then in the finite case ${\dim }_k{S}_m={\dim }_K(S_K{)}_m={\mathrm{len}}_K(X_K)={\mathrm{len}}_k(X)$ for all $m\ge m_0$, and in the infinite case the equality is the claim itself; from here on we therefore assume that $k$ is infinite.

Since $D_{+}(L)\subseteq X$ on the other hand, we have $D_{+}(L)=X$ as open subschemes; by [L2] the open subscheme $D_{+}(L)$ is the affine scheme $\mathrm{Spec}(A)$ with $A=(S_L{)}_0=\mathrm{\Gamma }(X,{\mathcal{O}}_X)$, so $X=\mathrm{Spec}A$.

By [L4] the space $X$ is the finite disjoint union ${\coprod }_{x\in X}\mathrm{Spec}({\mathcal{O}}_{X,x})$. If $X=\mathrm{\varnothing }$, [L6] gives $A=\mathrm{\Gamma }(X,{\mathcal{O}}_X)=0$, so ${\dim }_{k}A=0={\mathrm{len}}_k(X)$ by the empty-sum convention in [L4]. If $X\mathrm{\ne }\mathrm{\varnothing }$, [L6] applies with the positive number of factors and gives $A={\prod }_{x\in X}{\mathcal{O}}_{X,x}$, a finite-dimensional $k$-algebra with ${\dim }_{k}A={\sum }_{x\in X}{\dim }_k{\mathcal{O}}_{X,x}$; applying [L4] to the affine scheme $\mathrm{Spec}A$ and to the local rings ${\mathcal{O}}_{X,x}$ gives ${\mathrm{len}}_k(X)={\dim }_{k}A={\sum }_x{\mathrm{\ell }}_{{\mathcal{O}}_{X,x}}({\mathcal{O}}_{X,x})[\kappa (x):k]$. Thus $A$ is finite-dimensional and ${\dim }_{k}A={\mathrm{len}}_k(X)$ in either case.

For every index $i$ there is $N_i\ge 1$ with $x_i^{N_i}\in LS$: the open subschemes $D_{+}(x_i)$ and $D_{+}(L{x}_i)=D_{+}(L)\cap D_{+}(x_i)$ of $X$ coincide by step 2.2, so inside the chart $D_{+}(x_i)=\mathrm{Spec}(A_i)$ the localization $A_i\to (A_i{)}_{v_i}$ at the degree-zero dehomogenisation $v_i=L\mathrm{/}{x}_i$ of $L$ on that chart is an isomorphism, $v_i$ is a unit of $A_i$ with inverse $w$, and writing $w=a\mathrm{/}{x}_i^m$ with $a\in S_m$ gives $(L\mathrm{/}{x}_i)(a\mathrm{/}{x}_i^m)=1$ in $(S_{x_i}{)}_0$, hence $La-x_i^{m+1}$ is killed by a power of $x_i$ and $x_i^{N_i}=L\cdot (x_i^{M}a)\in LS$ for $N_i=m+1+M$.

For $m\ge 0$ define the $k$-linear map ${\phi }_m:S_m\to A$, $s\mapsto s\mathrm{/}{L}^m$, using $A=(S_L{)}_0\subseteq S_L$; every element of $A$ is a fraction $s\mathrm{/}{L}^m$ with $s\in S_m$, so $A={\bigcup }_{m\ge 0}{\phi }_m(S_m)$, and since $A$ is finite-dimensional by step 3.1 the ascending chain of images ${\phi }_m(S_m)$ stabilizes: there is $m_0$ with ${\phi }_m$ surjective for every $m\ge m_0$.

Let $K_0={\sum }_{i=0}^n{N}_i$; every monomial in $x_0,\dots ,x_n$ of degree $K_0$ is divisible by some $x_i^{N_i}$ by the pigeonhole principle, and $S_{+}^{K_0}$ is generated by those monomials, so $S_{+}^{K_0}\subseteq LS$.

Let $T=\ker (S\to S_L)=\{s\in S:L^{M}s=0\text{ for some }M\}$, a graded ideal of $S$ by [L7], and put $T_m=T\cap S_m$; then $\ker {\phi }_m=T_m$ for every $m$, because $s\mathrm{/}{L}^m=0$ in $(S_L{)}_0$ exactly when $s$ is killed by a power of $L$.

Consequently $S_m=L{S}_{m-1}$ for every $m\ge K_0$: $S_{+}^{K_0}\subseteq LS$ gives $(S_{+}^{K_0}{)}_m\subseteq (LS{)}_m=L{S}_{m-1}$, and $(S_{+}^{K_0}{)}_m=S_m$ because every degree-$m$ monomial with $m\ge K_0$ is divisible by some degree-$K_0$ monomial.

For every $m\ge K_0$ one has $T_m=L{T}_{m-1}$: if $t\in T_m$, then $t\in S_m=L{S}_{m-1}$ by step 5.2, say $t=La$ with $a\in S_{m-1}$, and $L^{M}t=0$ implies $L^{M+1}a=0$, so $a\in T_{m-1}$; conversely $L{T}_{m-1}\subseteq T_m$. Hence ${\dim }_k{T}_m\le {\dim }_k{T}_{m-1}$ for $m\ge K_0$, and the sequence of dimensions is eventually constant, say equal to $d$ for all $m\ge m_1$.

One has $d=0$: for $m\ge m_1$ the map $L:T_m\to T_{m+1}$ is surjective by step 6.1, so $L^M:T_{m_1}\to T_{m_1+M}$ is surjective for every $M$; but $T_{m_1}\subseteq S_{m_1}$ is finite-dimensional and every element of $T$ is killed by a power of $L$, so some $L^M$ kills all of $T_{m_1}$ and $T_{m_1+M}=0$, forcing $d=0$ and $T_m=0$ for all $m\ge m_1$.

For $m\ge \max (m_0,m_1)$ the map ${\phi }_m:S_m\to A$ is surjective by step 4.1 and has kernel $T_m=0$ by steps 5.1 and 7.1, hence is an isomorphism of $k$-vector spaces and ${\dim }_k{S}_m={\dim }_{k}A={\mathrm{len}}_k(X)$ by step 3.1.

If $k$ is infinite, step 8.1 proves the claim; if $k$ is finite, step 8.1 applied over the infinite field $K=k(t)$ to $S_K$ and $X_K$ gives ${\dim }_K(S_K{)}_m={\mathrm{len}}_K(X_K)$ for all large $m$, and step 2.1 converts this into ${\dim }_k{S}_m={\mathrm{len}}_k(X)$ for all large $m$; the empty case is included, since then $L=x_0$ still gives $A=(S_{x_0}{)}_0=0$, and all steps above remain valid.

The proof is complete: the equality ${\dim }_k{S}_m={\mathrm{len}}_k(X)$ holds for all sufficiently large $m$, no saturation of $I$ was used, the case $X=\mathrm{\varnothing }$ is covered by ${\mathrm{len}}_k(\mathrm{\varnothing })=0$, and the Axiom of Choice is inherited from the finite-chart, base-change and finiteness suppliers of [L3], [L4] and [L5].