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clarification "by a scheme"

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edited Dec 6 2009 at 11:16

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 representing 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...)

title fix

edited Dec 6 2009 at 9:37

Andrew Critch
5,856●12●44

Is tensoring with a module representable iff the module it is locally free of finite rank?

reformulated

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edited Dec 6 2009 at 9:30

Andrew Critch
5,856●12●44

When is Is tensoring with a finitely generated module over a Noetherian ring representable by a schemeiff the module is locally free?

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 representing the functor $B\mapsto B\otimes_A M$ from $A$-algebras to sets. What I know:

(SOME EDITS below here...)

It is not always possible if $M$ is not finitely generated; see this earlier

The question.

If $M$ has a pre-dual, say $N^\vee = M$, $\mathbb{V}(N)=Spec(Sym(N))$ does not generally work : (updated, thanks to some comments from a fortiori for pointing this outand buzzard)

EGA I (1971) 9.4.10 mentions in passing, without proof, that this functor is representable if and only possible when if $M$ is locally free of finite rank; I'm not sure if .

If this is correct, but if it is, I would like does anyone know where to see find the proof.

Why these restrictions on $M,A$?

If not, I mostly want this geometric interpretation for finitely generated modules over does anyone know a Noetherian ring correct (and at the moment, the partial answer to Dinakar's question doesn't suit my purposes). useful) equivalent condition on $M$?

So far,

QuestionI gather that:

It is not always representable if $M$ is a not finitely generated$A$-module, ; see this earlier question.

If $A$ NoetherianM$ has a pre-dual, under what necessary and sufficient conditions on say $N^\vee = M$is the functor $\otimes_A M: A$-alg $\to$ Set representable by an , $A$-scheme?\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. Any help is appreciated!several...)

corrections

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edited Dec 6 2009 at 9:15

Andrew Critch
5,856●12●44

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":

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 representing the functor $B\mapsto B\otimes_A M$ from $A$-algebras to sets. What I know:

(SOME EDITS below here...)

It is not always possible if $M$ is not finitely generated; see this earlier question.
If $M$ has a pre-dual: , say $N^\vee = M$, and then $\mathbb{V}(N)=Spec(Sym(N))$ does the trick.
It is not always possible if $M$ is not finitely generated; see generally work (thanks to a fortiori for pointing this earlier question.out)
EGA I (1971) 9.4.10 incorrectly says mentions in passing, without proof, that this is only possible when $M$ is locally free of finite rank; I'm not sure if this is correct, but if it is, I would like to see the proof.

Why these restrictions on $M,A$?

I want the answer to this question to provide a rigorous geometric intuition about the module $M$. For example, I want to view $Spec(A/I)\subset Spec(A)$ as the support of the $A$-module $A/I$, which doesn't often have a predual. In practice, I mostly want this geometric interpretation for finitely generated modules over a Noetherian ring (and at the moment, the partial answer to Dinakar's question doesn't suit my purposes). So,

Question:

If $M$ is a finitely generated $A$-module, $A$ Noetherian, under what necessary and sufficient conditions on $M$ is the functor $\otimes_A M: A$-alg $\to$ Set representable by an $A$-scheme?

This may not have a useful answer, or perhaps it has several. Any help is appreciated!

elaboration

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edited Dec 6 2009 at 7:18

Andrew Critch
5,856●12●44

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.

Some issues

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 $\acute{E}t(f^* \tilde{M})\neq f^* \acute{E}t(\tilde{M})$.

So

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 representing the functor $B\mapsto B\otimes_A M$ from $A$-algebras to sets. What I know:

It is possible if $M$ has a pre-dual: say $N^\vee = M$, and then $\mathbb{V}(N)=Spec(Sym(N))$ does the trick.

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

EGA I (1971) 9.4.10 incorrectly says this is only possible when $M$ is locally free.

Why these restrictions on $M,A$?

I want the answer to this question to provide a rigorous geometric intuition about the module $M$. For example, I want to view $Spec(A/I)\subset Spec(A)$ as the support of the $A$-module $A/I$, which doesn't often have a predual. In practice, I mostly want this geometric interpretation for finitely generated modules over a Noetherian ring (and at the moment, the partial answer to Dinakar's question doesn't suit my purposes). So,

Question:

If $M$ is a finitely generated $A$-module, $A$ Noetherian, under what necessary and sufficient conditions on $M$ is the functor $\otimes_A M: A$-alg $\to$ Set representable by an $A$-scheme?

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asked Dec 6 2009 at 7:10

Andrew Critch
5,856●12●44

When is tensoring with a finitely generated module over a Noetherian ring representable by a scheme?

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.

Some issues:

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 $\acute{E}t(f^* \tilde{M})\neq f^* \acute{E}t(\tilde{M})$.

So 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 representing the functor $B\mapsto B\otimes_A M$ from $A$-algebras to sets. What I know:

It is possible if $M$ has a pre-dual: say $N^\vee = M$, and then $\mathbb{V}(N)=Spec(Sym(N))$ does the trick.
It is not always possible if $M$ is not finitely generated; see this earlier question.
EGA I (1971) 9.4.10 incorrectly says this is only possible when $M$ is locally free.

Why these restrictions on $M,A$?

I want the answer to this question to provide a rigorous geometric intuition about the module $M$. In practice, I mostly want this geometric interpretation for finitely generated modules over a Noetherian ring (and at the moment, the partial answer to Dinakar's question doesn't suit my purposes). So,

Question:

If $M$ is a finitely generated $A$-module, $A$ Noetherian, under what necessary and sufficient conditions on $M$ is the functor $\otimes_A M: A$-alg $\to$ Set representable by an $A$-scheme?

ag.algebraic-geometry ac.commutative-algebra representable-functors