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Let $X$ be a quasiprojective variety(all varieties are over a field $k$).

  • Is an algebraic vector bundle $E$ over $X$ quasiprojective? If $X$ is affine, is $E$ affine?
  • Is a projective bundle $P$ over $X$ quasiprojective? If $X$ is projective, is $P$ projective?
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2 Answers

Yes. The easiest case is when $X$ is affine, say $X=\mathrm{Spec}(A)$. Then $E$ is associated to some locally free $A$-module $M$, and can be realized as $E = \mathrm{Spec} (Sym_A(M^\vee))$.

More generally, if $X$ is quasiprojective, take an ample line bundle $L$ on $X$. Then $P(E) = P(E\otimes L)$, and replacing $E$ by a sufficiently high twist by $L$, the line bundle $O_{P(E)}(1)$ is ample.

Note that you can realize $E$ as an open subset of the projective bundle $P(E\oplus 1)$, so (quasi)projectivity of the latter implies that of the former.

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Can you give an explicit embedding of the bundle $E$ to a projective space? –  Liu Hang Jan 26 '11 at 7:05
Not sure how explicit you want it, but if say $E^*$ is globally generated by $n$ sections, then the surjection of sheaves $O_X^{\oplus n} \to E^*$ defines an embedding of varieties $E \to O_X^{\oplus n} = X\times {\Bbb A}^n$. –  Dave Anderson Jan 26 '11 at 8:03
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If $a:E\to X$ is a vector bundle, then $a$ is an affine morphism and similarly, if $p:P\to X$ is a projective bundle, then it is a projective morphism. As composition of affine (and respectively projective morphisms) is affine (resp. projective) $E$ is affine (resp. projective) if $X$ is. For the quasi-projective question, note that an affine bundle maybe embedded as an open set into a projective bundle.

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