Let $\mathbb{P}(E)$ be a projective bundle over some smooth projective variety $X$, defined over $\mathbb{C}$ for definiteness. Then this bundle is also a smooth projective variety.
Smoothness is clear from the trivialization, and it is also clear using the Segre embedding that every patch can be embedded in some projective space. Does is automatically follow that the entire bundle can be embedded in some projective space? One can definitely glue them to get a variety, by using the triple intersection rules, but is it necessarily projective?
EDIT: in Tyurin's Vector Bundles, one reads that a vector bundle over a complete variety is neither affine nor complete, but the projectivization is an actual projective variety. This is what I am wondering about.
Best Answer
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 $\mathbb{P}(\mathcal{E})$.