# Very basic question on projective bundles: Why is the fiber a vector space?

Consider an integral scheme $(X,\mathcal O_X)$ of dimension $n$ and a "good" locally free sheaf of $\mathcal O_X$-modules $\mathcal E$ of rank $n+1$. Then, thanks to the "global proj construction" we can construct the projective bundle $\mathbb P(\mathcal E)$ which is a projective $X$-scheme: $$\pi:\mathbb P(\mathcal E)=\text{Proj}(\text{Sym}(\mathcal E))\longrightarrow X\,.$$

Among all the properties of $\mathbb P(\mathcal E)$ on the Dogalchev book (or here ) I read:

For $x\in X$, the fiber $\pi^{-1}(x)$ is the projectivization of the vector space $\mathcal E(x):=\mathcal E\otimes_{\mathcal O_{X,x}}k(x)$ on the field $k(x)$.

I don't understand the tensor product $\mathcal E\otimes_{\mathcal O_{X,x}}k(x)$! It should be the so called "extension by scalars of a module", but $\mathcal E$ is a sheaf, not a module.

Maybe the right tensor product should be $\mathcal E_x\otimes_{\mathcal O_{X,x}}k(x)$.

Moreover, why is this projective vector space (written in the correct form) isomorphic to $\mathbb P^n$?

Yes, it should say $\mathcal E_x\otimes_{\mathcal O_{X,x}}k(x)$. Since $\mathcal{E}$ is locally free of rank $n+1$, $\mathcal{E}_x$ is a free $\mathcal{O}_{X,x}$-module of rank $n+1$, so $\mathcal E_x\otimes_{\mathcal O_{X,x}}k(x)$ is a $k(x)$-vector space of dimension $n+1$. Choosing a basis, its projectivization can be identified with $\mathbb{P}^n_{k(x)}$.