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    Bertini's theorem

    T-BERTINI

    General members are smooth: hyperplane sections preserve smoothness generically, and characteristic-zero linear systems do so away from their base locus.Suggest a slogan

    (Bertini's theorem for hyperplane sections)

    Let \(X \subseteq \PP^n\) be a smooth projective variety over an algebraically closed field \(k\) of any characteristic. The hyperplanes \(H \subseteq \PP^n\) with \(H \not\supseteq X\) and \(H \cap X\) smooth form a dense open subset of \(\dualof{(\PP^n)}\). If \(X\) is irreducible of dimension at least \(2\), then \(H \cap X\) is moreover connected, hence irreducible, for \(H\) in that open set.

    (Bertini's theorem for linear systems)

    Let \(X\) be a smooth variety over an algebraically closed field of characteristic zero and let \(\mathfrak{d}\) be a linear system on \(X\). Then a general member of \(\mathfrak{d}\) is smooth away from the base locus of \(\mathfrak{d}\).

    The second statement fails in characteristic \(p\). On \(\AA^2\) the pencil spanned by \(x^p\) and \(y^p\) has base locus the origin, and every member \(a x^p + b y^p = (a^{1/p} x + b^{1/p} y)^p\) is a nonreduced line, singular at each of its points.

    What is Bertini's theorem?

    Dependencies

    • D-DIVLINSYSLinear systems, base loci, and base-point freenessuses

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