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The section ring of an ample invertible sheaf is finitely generated

Statement

Assume AC as inherited from the projective and sheaf-cohomology suppliers. Let X be a projective scheme over C and L an ample invertible sheaf on X (Absolute ampleness by affine section opens). Then the section ring R(X,L)=⨁n≥0Γ(X,L⊗n) is a finitely generated graded C-algebra, hence a Noetherian ring.

Facts & Assumptions

Given: A projective C-scheme X, an ample invertible sheaf L on X, and the section ring R(X,L).

[F1]

Very ample positive power. Applied to the proper finite-type morphism X→Spec⁡C and the ample sheaf L, the very-ampleness theorem gives an integer m≥1, an integer N≥0 and global sections of L⊗m generating L⊗m whose associated morphism is a closed immersion i:X↪PCN with i∗O(1)≅L⊗m. (High powers of an ample line bundle embed a proper scheme, Generating line-bundle sections define a morphism to projective space, Global generation by the evaluation map)

[F2]

Graded sections of a coherent sheaf. For a coherent sheaf F on PCN the graded S-module Γ∗(F)=⨁k≥0Γ(PN,F(k)) has a finitely generated tail: there is k0 with ⨁k≥k0Γ(PN,F(k)) finitely generated over S=C[x0,…,xN]; the extended sheaf i∗G along the closed immersion i is coherent. Each individual space Γ(PN,F(k)) is finite-dimensional over C. (High-degree section module is finite graded, Finite-dimensional coherent cohomology over a field)

[F3]

The coordinate ring image. Let S=C[x0,…,xN] and let A be the homogeneous coordinate ring of the closed immersion i:X↪PN, namely the image of the graded restriction map S→M0=⨁k≥0Γ(X,OX(k)). It is a finitely generated graded C-algebra, being a quotient of S; the restriction maps need not be surjective onto every space of global sections. (Closed subschemes of projective space and saturated ideals)

[F4]

Hilbert basis. A finitely generated algebra over a field is Noetherian. (Hilbert basis theorem: if R is Noetherian then R[x] is Noetherian)

Proof

technique · direct
1.1F1given

If X=∅ then R(X,L)=0 is generated by the empty set and is Noetherian; assume henceforth X≠∅. Fix the integer m≥1 and the closed immersion i:X↪PCN of [F1], so that OX(1) in the following denotes i∗OPN(1)≅L⊗m and L⊗(mk+j)≅i∗O(k)⊗L⊗j for all k≥0, 0≤j<m.

2.1F1F2step 1.1

The graded modules Mj. For 0≤j<m put Mj=⨁k≥0Γ(X,L⊗(mk+j)). Under the identification of step 1.1 and the projection formula for the finite morphism i, Mj=Γ∗(Fj) for the coherent sheaf Fj=i∗(L⊗j) on PCN; hence by [F2] each Mj has a finitely generated tail over S=C[x0,…,xN]. A graded S-module whose tail is finitely generated is finitely generated: the missing finite initial part is a finite-dimensional C-vector space by [F2], and an extension of a finitely generated module by a finite-dimensional one is finitely generated. So each Mj is a finitely generated graded S-module.

3.1F2F3step 2.1algebra

The section modules over the coordinate ring image. Let A⊆M0 be the image in [F3]. It is a finitely generated C-algebra. For p∈S in the kernel of S→A, its restriction is the zero section on X, so multiplication by p is zero on every Γ(X,L⊗(mk+j)); hence the S-action on each Mj factors through A. The finite S-module generators of step 2.1 therefore also generate Mj as an A-module. In particular M0 and each of the finitely many Mj are finite A-modules.

4.1F3F4step 3.1algebra∎

Conclusion. The decomposition R(X,L)=⨁j=0m−1Mj and step 3.1 show that R(X,L) is a finite module over the finitely generated C-algebra A. A finite set of algebra generators of A together with a finite set of A-module generators of R(X,L) generates R(X,L) as a C-algebra. Hence R(X,L) is a finitely generated C-algebra, and it is Noetherian by [F4].

Remarks

  • The proof uses only the ample power. Neither the very-ampleness of L itself nor a Hilbert--Mumford criterion is used; the finite generation comes from the graded-module theorem on projective space applied to the coherent sheaves i∗(L⊗j).
  • No separatedness issue. X is a projective C-scheme, in particular proper and of finite type over Spec⁡C; all cited suppliers are stated in that register.

Depends on

Used by

Dependency tree · two levels

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Sources