I've got the follow question which drives me almost crazy as the answer seems to be simple. Given a morphism $p:V\to S$ of schemes of finite type over some base field. Assume that $p$ has all the properties of a vector space over $S$ that is, there should be morphisms $+:V\times_S V \to V, \cdot: A^1_S \times_S V \to V, 0:S\hookrightarrow V$ and $-:V\to V$ over $S$ satisfying the axioms of a module over the ring object $A^1_S$. If $p:V\to S$ is locally trivial, $p$ is called a vector bundle, and the dimensions of the fibers are (locally) constant. My question is whether or not the converse also holds under the assumption that $S$ is smooth, that is, if $S$ is smooth and if the dimensions of the fibers of $p$ are (locally) constant, then $p$ is a vector bundle.
What is know: If $p$ is affine, then $V=Spec_S( Sym^* E)$ is the relative spectrum of the symmetric algebra of some coherent sheaf $E$ on $S$. The fibers of $E$ must be constant by assumption, and since $S$ is smooth, $E$ and hence $p$ is locally free.
For general $p$ there is a $A^1_S$-linear morphism $f:V\to Spec_S(p_\ast \mathcal{O}_V)$ over $S$ with $p_\ast \mathcal{O}_V\cong Sym^* E$ for $E$ being the subsheaf of fiberwise linear functions. If $p$ is flat, $f$ is an isomorphism on fibers, and $E$ must be locally trivial again by assumption. In particular, $f$ is a bijection on points, and using some version of Zariski's main theorem, one should be able to show that $f$ is an isomorphism. But how can we deduce flatness of $p$ form the assumption on the fiber dimension?
I would be very grateful if anyone can provide some solution.
Best wishes, Sven
