Projective bundle is projective?

Solution 1:

A projective bundle $\mathbb{P}(\mathcal{E})$ over a smooth projective variety $X$ (over any base field $k$) is indeed a smooth projective variety.

Such a scheme $X$ is noetherian. By Exercise II.7.10. in Hartshorne, for a locally free sheaf of rank $n+1$ on $X$, its projectivization $\mathbb{P}(\mathcal{E})$ is always a $\mathbb{P}^{n}$-bundle over $X$, and conversely, since $X$ is also regular, every $\mathbb{P}^{n}$-bundle over $X$ arises in this way.

By definition (see here) the morphism $\pi \colon \mathbb{P}(\mathcal{E})\to X$ is projective in the sense of EGA. But $X$ admits an ample invertible sheaf, so in this case EGA-projective imlpies Hartshorne-projective (see the same reference a bit below). Since composition of Hartshorne-projective morphisms is Hartshorne-projective and the structure morphism $X\to \text{Spec}(k)$ is Hartshorne-projective, so is $\mathbb{P}(\mathcal{E})\to \text{Spec}(k)$ and therefore $\mathbb{P}(\mathcal{E})$ is a closed subset of some projective space over your base field $k$.

Smoothness of $\mathbb{P}(\mathcal{E})$ over $k$ follows as you say from the trivializations: smoothness is a local property and $\mathbb{P}(\mathcal{E})$ is locally a product of an open set of $X$ (smooth) and a projective $n$-space over $k$ (also smooth).

Irreducibility can be shown as follows: if $U$ is a trivializing open in $X$, by irreducibility of $X$, $U$ is also irreducible. Now $\pi^{-1}(U)$ is the product of two (irreducible) quasi-projective varieties over $k$, hence also irreducible. But in fact $\pi^{-1}(U)$ is dense in $\mathbb{P}(\mathcal{E})$, so we get that $\mathbb{P}(\mathcal{E})$ is irreducible.

To see that $\pi^{-1}(U)$ is dense in $\mathbb{P}(\mathcal{E})$ you have two arguments:

• The (topological) map $\pi$ is open and therefore the preimage of a dense subspace is dense.

• For any other trivialising open $V$, the intersection $U\cap V$ is dense (again by irreducilibity of $X$) and so the preimage $\pi^{-1}(U\cap V)=\pi^{-1}(U)\cap \pi^{-1}(V)$ is dense in $\pi^{-1}(V)$. But these sets cover $\mathbb{P}(\mathcal{E})$, hence $\pi^{-1}(U)$ is dense in $\mathbb{P}(\mathcal{E})$.