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    Serre's theorems A and B

    T-MODSERRE

    Twist a coherent sheaf far enough and two things happen: it becomes globally generated and all higher cohomology vanishes.Suggest a slogan

    (Serre A and B)

    Let \(X\) be projective over a noetherian ring \(A\), with \(\OO_X(1)\) very ample, and let \(\mcf\) be coherent. Then there is \(N_0\) such that for all \(n \geq N_0\):

    1. \(\mcf(n)\) is generated by finitely many global sections;

    2. \(H^i(X, \mcf(n)) = 0\) for all \(i > 0\).

    [@Har10a, Theorems II.5.17 and III.5.2]

    (Dimension shifting on projective space)

    On \(\PP_A^r\), part (A) gives a surjection from a finite sum of twists to any coherent sheaf \(\mathcal K_0\). Iterating on the kernels gives exact sequences \[ 0\longrightarrow\mathcal K_{j+1}\longrightarrow\mathcal E_j\longrightarrow\mathcal K_j\longrightarrow0, \] where \(\mathcal E_j\) is a finite sum of twists and \(\mathcal K_j\) is coherent. Take the first \(r\) such sequences. For sufficiently large \(n\), the positive-degree cohomology of every \(\mathcal E_j(n)\) in this finite list vanishes by the cohomology calculation for \(\OO(d)\) [@Har10a, Theorem III.5.1]. The long exact sequences then give, for \(i>0\), \[ H^i(\PP_A^r,\mathcal K_0(n))\cong H^{i+r}(\PP_A^r,\mathcal K_r(n))=0. \] The last vanishing follows from the standard cover by \(r+1\) affine opens: its intersections are affine, so its Cech complex computes quasi-coherent cohomology and has no terms in degrees greater than \(r\) [@Har10a, Theorem III.4.5]. For \(r=0\), affineness gives the vanishing directly. Thus a finite initial part of the resolution suffices for this dimension-shifting proof, even when the resolution does not terminate.

    (A coherent sheaf with no finite resolution by twists)

    Let \(k\) be a field, \(A=k[\varepsilon]/(\varepsilon^2)\), and \(X=\PP_A^0\cong\Spec A\). Every twist of \(\OO_X\) is trivial. The finite \(A\)-module \(k=A/(\varepsilon)\) has the free resolution \[ \cdots\xrightarrow{\varepsilon}A\xrightarrow{\varepsilon}A\xrightarrow{\varepsilon}A\longrightarrow k\longrightarrow0. \] It is exact because the kernel and image of multiplication by \(\varepsilon\) are both \((\varepsilon)\). Tensoring with \(k\) gives zero differentials, so \(\operatorname{Tor}_i^A(k,k)\cong k\) for every \(i\ge0\). A finite free resolution would make these groups zero in sufficiently high degrees. Therefore the associated coherent sheaf has no finite resolution by finite sums of twists.

    (An explicit twist bound)

    Let \(k\) be a field, let \(m\ge0\) be an integer, and take \(\mcf=\OO_{\PP_k^1}(-m)\). For every \(n\ge m\), the sheaf \(\mcf(n)=\OO(n-m)\) is generated by its degree-\((n-m)\) monomials, and its positive-degree cohomology vanishes. Indeed, \(H^1(\PP_k^1,\OO(q))=0\) for \(q\ge-1\), and all higher groups vanish [@Har10a, Theorem III.5.1]. Thus \(N_0=m\) is a numerical bound for both (A) and (B). There is no common vanishing bound for all these sheaves on \(\PP_k^1\): for any fixed \(n\), choosing \(m\ge\max(0,n+2)\) gives \(\dim_k H^1(\PP_k^1,\mcf(n))=m-n-1>0\).

    (Finiteness over the base)

    By finiteness of coherent cohomology, each \(H^i(X,\mcf)\) is a finite \(A\)-module. When \(A=k\) is a field, the dimensions \(h^i(X,\mcf)=\dim_k H^i(X,\mcf)\) are finite and the Euler characteristic \(\chi(X,\mcf)=\sum_i(-1)^i h^i(X,\mcf)\) is defined. For all sufficiently large \(n\), part (B) gives \(\chi(X,\mcf(n))=h^0(X,\mcf(n))\).

    State Serre's theorems A and B.
    What do they say about twisting a coherent sheaf far enough?
    What are they used for?

    Dependencies

    • D-MODAMPLEAmple and very ample invertible sheavesuses
    • D-MODGGGlobally generated sheavesuses

    Backlinks

    • T-MODRESCoherent sheaves on projective space have finite resolutions by twistsuses

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