# 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
$Y\to Spec(A)$. 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 $\acute{E}t(f^* \tilde{M})\neq f^* \acute{E}t(\tilde{M})$.

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

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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. – user2035 Dec 6 '09 at 8:21
@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... – Andrew Critch Dec 6 '09 at 11:22
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). – Jonathan Wise Dec 6 '09 at 18:20
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. – user2035 Dec 6 '09 at 19:24
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.) – 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.

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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... – 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.)

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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.

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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. – Kevin Buzzard Dec 6 '09 at 8:42
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...) – 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.

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So you did check? Could you give some summary of why that is true? – Hailong Dao Dec 6 '09 at 9:03
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? – Kevin Buzzard Dec 6 '09 at 9:10
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. – Andrew Critch Dec 6 '09 at 9:18
Is there a problem with my sketch above? Let me know if any of it needs expanding or doesn't pan out. – Kevin Buzzard Dec 6 '09 at 9:26
How does this proof deal with the possibility of a non-affine scheme representing the functor? – Jonathan Wise Dec 6 '09 at 10:26