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

disjointunion of affine 1-space over P and (affine 1-space x (P-S)) over P-S. Does this satisfy your axioms? It's not flat. If it does satisfy them then there's your example, and if it doesn't then there's your clue as to how to deduce flatness from the other assumptions. $\endgroup$ – eric Dec 5 '15 at 14:22