Regular-base acyclicity for finite-length perfect complexes

Let R be a regular local ring of dimension c and let F be a bounded finite free complex, supported only at the closed point. Define the local-cohomology operation explicitly by tensoring with the stable Čech complex $C(x)={\otimes }_{j=1}^c[R\to R[x_j^{-1}]]$, for a regular parameter system. It is the filtered colimit of the dual Koszul complexes on $x_1^k,\dots ,x_c^k$. These powers remain a regular sequence, so their Koszul cohomology is only in degree c; exactness of filtered module colimits gives the same for C(x). For a finite-length module all parameter localizations vanish, so $C(x)\otimes M=M$ in degree zero. Filtering C(x)⊗F by the cohomology of F therefore gives $C(x)\otimes F=F$. Filtering instead by the finite free terms of F shows cohomology only in [a+c,b+c] if its free terms are in [a,b]. The original F has cohomology only in [a,b]. Whenever $c>b-a$, these intervals are disjoint and $F=0$. This supplies the actual local-cohomology argument without an unnamed intersection-theorem input.

More generally, let T be a regular finite-dimensional Noetherian scheme. If a perfect complex on T has support codimension≥c and its local minimal free models have length $<c$, it is zero. If not, localize at a minimal prime of its nonzero cohomological support. This is a regular local ring of dimension at least c, and all its homology has finite length. Cancel unit entries in a free model to retain the prescribed length and apply the previous paragraph. This proves the off-diagonal vanishing criterion needed below without an intersection theorem over a singular ring.

If the support is a single closed point x of a regular Noetherian scheme and the local ring at x has dimension d, a free model in degrees [−d,0] has cohomology only in degree zero: intersect [−d,0] with [0,d] in the same local-cohomology calculation. If its H⁰ is k, the complex is therefore k. This is the diagonal criterion; it needs known regularity, which the geometric Hilbert supplier supplies.