It's nice when you can think of the elements of an $A$-module $M$ as sections some $A$-scheme
$Y\to Spec(A)$. 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 $U\subseteq Spec(A)$.

I'm not happy with this construction in that it has "the wrong fibres": for $I\triangleleft A$, 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 $\acute{E}t(f^* \tilde{M})\neq f^* \acute{E}t(\tilde{M})$.


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 $B\mapsto B\otimes_A M$ 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...)

  • $\begingroup$ It is not sufficient for M to have a pre-dual: V(N) represents Hom_B(N⊗B,B)=Hom(N,B), not Hom(N,A)⊗B. In fact, if Hom(N,A)⊗B=Hom(N,B), this functor is exact in the variable A-module B (consider algebras of the form B'=A⊕B with B²=0), so N is projective. $\endgroup$ – user2035 Dec 6 '09 at 8:21
  • $\begingroup$ @a-fortiori, I'm not sure I understand your argument about projectivity... it seems like the arrows go the wrong way to make use of the $A\oplus B$ algebras you suggest, but maybe I'm missing something... $\endgroup$ – Andrew Critch Dec 6 '09 at 11:22
  • $\begingroup$ By the way, an interesting way to think of more general quasi-coherent sheaves as geometric objects is with $\mathbf{A}^1$-linear Picard stacks (and higer Picard stacks). $\endgroup$ – Jonathan Wise Dec 6 '09 at 18:20
  • $\begingroup$ If the canonical map Hom(N,A)⊗B → Hom(N,B) is an isomorphism for all A-algebras B, this is true in particular for A-algebras of the type A⊕T for some A-module T, so it is true for Hom(N,A)⊗T → Hom(N,T). Exactness of Hom(N,T) for variable T is equivalent to N projective. $\endgroup$ – user2035 Dec 6 '09 at 19:24
  • $\begingroup$ I think Corollary 2 of the article "Nitsure, Nitin: Representability of Hom implies flatness. Proc. Indian Acad. Sci. Math. Sci. 114 (2004), no. 1, 7–14" gives a proof in the case $A$ is noetherian. (He considers the functor on all schemes as in EGA but I think the proof should also work for the functor restricted to affine schemes.) $\endgroup$ – ulrich Jun 15 '11 at 12:27

Say X is a scheme and E is an $\mathbf{A}^1$-linear scheme over X (a group scheme over X with an action of the sheaf of rings $\mathbf{A}^1$). You are asking when it is possible for E to be quasi-coherent. If E is quasi-coherent and of finite type then it admits a surjection from a finite dimensional vector space F over X. For any point $e \in E$,

$\dim_{p(e)} E + \dim_e F = \dim_{p(e)} F$.

This implies that if the dimension of a fiber of E over X jumps, the dimension of the corresponding fiber of F over E must drop, which is impossible.

  • $\begingroup$ This is a nice partial answer; I wonder how one could go about proving (rather than assuming) that the representing scheme $E$ would have to be of finite type... $\endgroup$ – Andrew Critch Dec 6 '09 at 11:07

Suppose the functor is represented by some scheme $G$, let $e\colon\mathrm{Spec}(A)\to G$ be the zero section. As in your previous question, EGA IV, 8.14.2 shows that $G$ is locally of finite presentation over $A$, and the infinitesimal criterion shows that $G$ is smooth over $A$. Therefore, $\Omega^1\_{G/A}$ is a locally free $\mathcal O_G$-module of finite rank, so $e^\*\Omega^1\_{G/A}$ is a projective $A$-module of finite rank, hence $\mathrm{Hom}(e^\*\Omega^1\_{G/A},A)$ is as well. On the other hand, $\mathrm{Hom}(e^\*\Omega^1\_{G/A},A)$ is the $A$-valued points of the Lie algebra of $G$, so can be computed as $\ker(A[T]/(T^2)\otimes M\to A\otimes M)$, and this is isomorphic to $M$. (Check that this is actually an isomorphism of $A$-modules.)


Here is an example where representability fails. If $R$ is an $A$-algebra representating $\otimes_AM$ on $A$-algebras, and if $B\to C$ is an injective map of $A$-algebras, then $R(B)\to R(C)$ will be injective ($R(B)$ is the $A$-algebra homs from $R$ to $B$). But, for example, if $M=A/I$ then "usually" $B/IB\to C/IC$ is not injective (for example if $A$ is the integers, $I=(2)$, $B=A$, $C=A[1/2]$) so you're already dead in the water.

Edit: emphasis of question changed, so ephasis of answer has been changed too.

  • $\begingroup$ Andrew: are you sure EGA is wrong?? Can you justify your assertion? If EGA is right then it provides a beautiful answer to your question. $\endgroup$ – Kevin Buzzard Dec 6 '09 at 8:42
  • $\begingroup$ This only shows that (often) $\otimes A/I$ is not representable by an affine scheme Spec(R); I just added "by a scheme" to a few more places in the question to be clear about this. (Also, I'm starting to believe EGA...) $\endgroup$ – Andrew Critch Dec 6 '09 at 11:20

Contrary to what I guessed initially, I now think the question has a great answer: the functor is representable if and only if $M$ is locally free, and the proof is EGA I, 9.4.10.

Edit: this is an answer to an earlier version of the question.

  • $\begingroup$ So you did check? Could you give some summary of why that is true? $\endgroup$ – Hailong Dao Dec 6 '09 at 9:03
  • $\begingroup$ The proof should go something like this: if it's representable then tensoring with $M$ had better send injections of $A$-algebras to injections. Now some trickery involving $A$-algebras of the form $A+N$ with $N$ an $A$-module shows that tensoring with $M$ sends injections of $A$-modules to injections. So tensoring with $M$ is exact and off you go. Does that sound OK? $\endgroup$ – Kevin Buzzard Dec 6 '09 at 9:10
  • $\begingroup$ Regarding the EGA reference, I'm not sure it's false anymore, but it's only a comment; I haven't seen (or recognized) any sort of proof. $\endgroup$ – Andrew Critch Dec 6 '09 at 9:18
  • $\begingroup$ Is there a problem with my sketch above? Let me know if any of it needs expanding or doesn't pan out. $\endgroup$ – Kevin Buzzard Dec 6 '09 at 9:26
  • $\begingroup$ How does this proof deal with the possibility of a non-affine scheme representing the functor? $\endgroup$ – Jonathan Wise Dec 6 '09 at 10:26

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.