# Are “vector spaces” over a smooth scheme with constant fiber dimension locally free?

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

• I have not thought remotely seriously about what I'm about to say, but it might be instructive to work out the following example: let S be affine 1-space, let P be a point in S, let V be the disjoint union 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. – eric Dec 5 '15 at 14:22
• @eric: Your example does not admit a zero section. – Jason Starr Dec 5 '15 at 14:23
• There is a tiny issue in positive characteristic $\ell$, but from your formulation I imagine that you are aware of it. You should assume that the stabilizer in $\mathbb{A}^1$ (for the scalar product) of the complement of the zero set in $V$ is just the zero section of $\mathbb{A}^1$. That rules out things like "scaling" by $t*v = t^\ell v$. – Jason Starr Dec 5 '15 at 14:28
• Since $V$ is a group scheme over $S$ under addition, $\Omega_{V/S}$ equals $p^*0^*\Omega_{V/S}$ compatibly with base change of $S$. Thus, if the fiber of $0^*\Omega_{V/S}$ over a closed point of $S$ has rank larger than the rank over the generic point, you can conclude that the fiber of $p$ over that point also jumps dimension. Therefore $0^*\Omega_{V/S}$ is a coherent sheaf on a smooth scheme $S$ with constant rank, hence it is locally free (Hartshorne, Exercise II.5.8). Thus $\Omega_{V/S}$ is locally free of rank equal to the relative dimension. Now use III.10 to prove $p$ is flat. – Jason Starr Dec 5 '15 at 14:59