Search the corpus

    Serre's cohomological criterion for affineness

    T-5IOUR

    For Noetherian schemes, affine means no higher quasicoherent cohomology; it already suffices to kill $H^1$ of coherent ideals.Suggest a slogan

    Let \(X\) be a Noetherian scheme. The following are equivalent:

    1. \(X\) is affine.

    2. \(H^p(X, \mcf) = 0\) for all quasicoherent \(\mcf\) and all \(p > 0\).

    3. \(H^1(X, \mci) = 0\) for every coherent sheaf of ideals \(\mci\).

    [@Har10a, Theorem III.3.7]

    (Scope of the hypotheses)

    Weakening Noetherian. The criterion holds for quasicompact quasi-separated schemes, with \(\mci\) ranging over quasicoherent ideals rather than coherent ones. Noetherian is used only to have coherent ideals available and to extract finite subcovers, and both can be arranged directly.

    Without quasicompactness. The conclusion fails. An infinite disjoint union of affine schemes has vanishing higher cohomology for every quasicoherent sheaf, being a disjoint union of affines, but it is not affine: its ring of global sections is an infinite product, and the comparison map is not an isomorphism. The finite subcover in the proof is where quasicompactness enters, and without it the \(f_i\) need not be finite in number.

    State Serre's criterion for affineness.
    Can the Noetherian hypothesis be weakened?
    What fails without quasicompactness?

    Dependencies

    • T-SK599$\Spec$ is an anti-equivalence, and the adjunction that recognises affinesuses
    • D-PTIW0Čech cohomology and the comparison with sheaf cohomologyuses

    Backlinks

    • T-COHAFFHigher cohomology vanishes on an affine schemerelated-to

    What links to this