Suppose $X\to Y$ is a morphism between varieties with all fibers isomorphic to $\mathbf{P}^n$, is $X$ a projective bundle over $Y$, i.e. $X=\mathbf{P}(E)$ for some vector bundle over $Y$?



Is a morphism whose all fibers are $\mathbf{P}^n$ a projective bundle?
No that is not true, and I am confident somebody has written the following example previously on MO. Let $\mathbb{A}^2$ have coordinates $s$ and $t$. Let $Y\subset \mathbb{A}^2$ be the singular plane curve with defining equation $$f(s,t) = t^2-s^2(1-s) = 0.$$ Let $u$ be a coordinate on $\mathbb{A}^1$. The normalization of $Y$ is $$\nu:\mathbb{A}^1 \to Y, \ \nu(u) = (1-u^2,(1-u^2)u).$$ Let $V\subset \mathbb{A}^1$ be the open complement of the closed point $p$ where $u=1$. Let $X$ be $V\times \mathbb{P}^n$. Denote by $$\text{pr}\_V:V \times \mathbb{P}^n\to V,$$ the usual projection. Then one counterexample is $\pi = \nu\circ \text{pr}\_V$. 


**Edit.** One reference for this problem is Exercise II.7.10 of Hartshorne's "Algebraic Geometry". If $\pi$ is a proper, flat morphism $\pi$ as in the question, then Ariyan's answer is perfectly correct and gives étale local triviality. If, further, $Y$ is Noetherian and regular (the hypothesis in Exercise II.7.10), then generic triviality of $\pi$ (i.e., the OP's hypothesis at the generic point of $Y$) implies Zariski local triviality of $\pi$. There should be an easy direct argument for this, but this certainly follows from the known cases of the Grothendieck-Serre conjecture (there is a direct argument after all, see the hint in that exercise). Finally, Exercise II.7.10 implies that $X$ is isomorphic to $\mathbb{P}(E)$.


