ag.algebraic geometry - Is tensoring with a module representable iff it is locally free of finite rank?

Motivation:

It's nice when you can think of the elements of an $A$-module $M$ as sections some $A$-scheme
$Yto Spec(A)$. That is, maps $Spec(A)to Y$ such that $Spec(A)to Y to Spec(A)$ is the identity.

What's wrong with the "espace étalé":

One way to do this is to consider the associated sheaf $tilde{M}$, and form its "espace étalé" $acute{E}t(tilde{M})$. Observe that this topological space is naturally an $X$-scheme (essentially by its construction, as for $acute{E}t$ of any sheaf of sets), and that $Gamma(U,acute{E}t(tilde{M})) = tilde{M}(U)$ for opens $Usubseteq Spec(A)$.

I'm not happy with this construction in that it has "the wrong fibres": for $Itriangleleft A$, the sections of $acute{E}t(tilde{M})$ over (base changed to) a closed subscheme $Z(I)$, e.g. a point, do not correspond to $widetilde{M/IM}$. This is just an instance of the fact that $acute{E}t$ doesn't respect base change: given $f:Spec(B)to Spec(A)$, in general $acute{E}t(f^* tilde{M})neq f^* acute{E}t(tilde{M})$.

Conclusion:

I want a construction that does respect base change. That is, for any module $M$ on $X$, I want an $X$-scheme $Y$ such that for any $X'to X$, $Gamma(X',Y_{X'}) = tilde{M}_{X'}(X')$. This amounts to finding a scheme which represents the functor $Bmapsto Botimes_A M$ from $A$-algebras to sets.

The question: (updated, thanks to some comments from a fortiori and buzzard)

EGA I (1971) 9.4.10 mentions in passing, without proof, that this functor is representable by a scheme if and only if $M$ is locally free of finite rank.

• If this is correct, does anyone know where to find the proof?




• If not, does anyone know a correct (and useful) equivalent condition on $M$?

So far, I gather that:

• It is not always representable if $M$ is not finitely generated; see this earlier question.




• If $M$ has a pre-dual, say $N^vee = M$, $mathbb{V}(N)=Spec(Sym(N))$ does not generally work (see a fortiori's comment below)

(This may not have a useful answer, or perhaps it has several...)