# Families of ideal sheaves: What's the correct definition?

I'm looking at Bridgeland's paper "Flops and Derived categories" and I got confused on what he meant by a family of ideal sheaves.

Let $Y$ be a scheme, and let $S$ be another scheme. A family of sheaves $\mathcal E_Y$ on $Y$ over $S$ is a sheaf on $S\times Y$, flat over $S$. Two such families are equivalent if they differ by tensoring by pullback of a line bundle on $S$.

When we talk about ideal sheaves $\mathcal{I}$ of $Y$, usually an inclusion to the structure sheaf $\mathcal{O}_Y$ is taken as part of the data. Now when we talk about a family of ideal sheaves, what do we really mean by that hidden part of the data? At least a family of sheaves which are flat over $S$, but there should be more. More specifically, when we say "let $\mathcal E_Y$ be a family of ideal sheaves on $Y$ over $S$", is there an inclusion of $\mathcal E_Y$ into $\mathcal O_{S\times Y}$ given part of the data?

(EDIT: according to MartinG's answer, this attempt of definition doesn't seem to be right, some functoriality is missing. Now let me replace my question by the following:

1. What is the functor $M_I(X)$? What's the definition and why is it functorial? (I'm guessing MartinG's suggestion is right, namely rank 1 sheaves with trivial determinant line bundle, but I'm not 100% sure.)

2. Why does $M_I(X)$ exist as a scheme under that definition? (In the paper he seems need this fact in a crucial way.)

3. What's the relation between $M_I(X)$ and $\text{Hilb}(X)$?

End of EDIT)

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I just had a look at the paper: Maybe this is also stated more explicitly elsewhere, but I think at least the proof of Lemma 6.3 reveals that an ideal sheaf means a torsion free sheaf for which there exists a nonzero map to $\mathcal{O}_X$, that for families, this condition is just imposed fibrewise (as Sasha suggested) and moreover that this is a closed condition. Anyway, I am leaving my answer as it is, covering the smooth case. (And I still have no idea whether the natural map could fail to be an iso in the singular case.) – MartinG Sep 19 '12 at 12:34
Thanks for pointing that out. I think rank one is also part of the definition. That's specified by the numerical class. – 36min Sep 19 '12 at 17:06
Rank one and non-zero ==> generic isomorphism, and torsion-free ==> injective (The kernel is supported on a proper closed subset, thus torsion.) – Yuhao Huang Oct 5 '12 at 2:10

## 3 Answers

I suppose one issue is that in those two things different objects are required to be flat (if anything). For $\mathrm{Hilb}(X)$ you want the family of the objects defined by the ideals to be flat, while for $\mathscr M_I(X)$ you may (only) want the family of the ideals to be flat, or even less.

In other words, you have a short exact sequence, $$0\to \mathscr I_Z\to \mathscr O_{S\times X}\to \mathscr O_Z\to 0,$$ and this gives an $S$-valued point in $\mathrm{Hilb}(X)$ if $\mathscr O_Z$ is flat over $S$ and an $S$-valued point in $\mathscr M_I(X)$ if $\mathscr I_Z$ is flat, or even under more general conditions.

If $\mathscr O_Z$ is flat over $S$ then so is $\mathscr I_Z$, but not vice versa.

Here is a simple example: Let $X=S=\mathbb P^1$ and $Z\subset S\times X=\mathbb P^1\times \mathbb P^1$ the union of one member of each of the rulings of $S\times X$. Then $\mathscr I_Z$ is a line bundle and hence flat over $\mathscr O_{S\times X}$ and hence over $S$, but $\mathscr O_Z$ and accordingly $Z$ is not flat. This shows why we like $\mathrm{Hilb}(X)$ much more than $\mathscr M_I(X)$.

Of course if $S$ is a closed point, then this issue does not arise.

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What does the last sentence mean? How can two functors agree if they only agree on closed points, but not $S$-points? I think you pointed out they don't agree on $S$-points... Am I missing something? – temp Sep 18 '12 at 15:48
Also, why does $M_I(X)$ exist as a scheme? It seems larger than $\text{Hilb}(X)$. Will there be any non-separatedness? It seems one can also define a similar scheme for Quot, namely one that classifies subsheaves which are flat over the base. Does that exist too? – temp Sep 18 '12 at 15:55
@temp: No, $M_I(X)$ does not exist: it is not a functor. – MartinG Sep 18 '12 at 18:42
OK, you guys have a point. I just tried to make sense of the issue.... – Sándor Kovács Sep 18 '12 at 19:16
...and that point is not closed! :) – Sándor Kovács Sep 18 '12 at 19:19

[This is now an answer to the edited question(s), with some details added. My answer to the original question is kept at the very end.]

Firstly: The question is a good one, and it is not easy to find references on this. I had spent too much time pondering about the failure of the double dual argument (see below) before I finally heard the arguement given in the last section below, indirectly from Fantechi, via Faber.

Assume $X$ is smooth projective.

Definition: An $S$-valued point in $M_I(X)$ is an $S$-flat coherent sheaf on $S\times X$, with stable fibres of rank one, and with determinant line bundle isomorphic to $\mathcal{O}_{S\times X}$, modulo isomorphism.

(I do not know if this is what Bridgeland meant, but to me this is resonably standard.)

Comment: Stability for rank one means torsion free.

Existence: Let $M(X)$ be the (Simpson) moduli space for stable rank one sheaves. Then $M_I(X)$ is the fibre over $\mathcal{O}_X$ for the determinant map $M(X) \to \mathrm{Pic}(X)$. This map sends a sheaf $I$ (stable rank one fibres) on $S\times X$ to the determinant line bundle $\det(I)$, and it is trivial as a point in $\mathrm{Pic}(X)$ if it is of the form $p^*L$ with $L\in\mathrm{Pic}(S)$. Then $I\otimes p^*L^{-1}$ is equivalent to $I$ in $M(X)(S)$, and it has trivial determinant. This shows that $M_I(X)$ indeed is a fibre of the determinant map.

Of course the determinant of an ideal $I_Y\subset \mathcal{O}_X$ is nontrivial if $Y$ is a non principal divisor, so you cannot map such ideals to $M_I(X)$. In any case, the ideal of a divisor, without the embedding, would only remember the linear equivalence class.

For brevity, let $\mathrm{Hilb}(X)$ be the part of the Hilbert scheme parametrizing subschemes $Y\subset X$ of codimension at least $2$. Then there is a natural map $F: \mathrm{Hilb}(X) \to M(X)$ that sends an ideal $I_Y\subset\mathcal{O}_{S\times X}$ to $I_Y$, forgetting the embedding. Since $Y$ is flat, so is $I_Y$, and its fibres are torsion free (by flatness again) of rank one. By the codimension assumption, the determinant of $I$ is trivial.

Theorem: $F$ is an isomorphism.

Comment: In the literature one sometimes finds the argument that if $I$ is a rank one torsion free sheaf with trivial determinant, then $I$ embeds into its double dual, which coincides with its determinant $\mathcal{O}_X$. This establishes bijectivity on points. (For Hilbert schemes of points on surfaces this is enough to conclude, since you can check independently that both $\mathrm{Hilb}(X)$ and $M_I(X)$ are smooth, and that the induced map on tangent spaces is an isomorphism.) I do not know how to make sense of this argument in families.

Sketch proof of theorem: The essential point is to show that every $I$ in $M_I(X)(S)$ has a canonical embedding into $\mathcal{O}_{S\times X}$ such that the quotient is $S$-flat.

Let $U\subset S\times X$ be the open subset where $I$ is locally free. Its complement has codimension at least $2$ in all fibres. By the trivial determinant assumption, the restriction of $I$ to $U$ is trivial. By codimension $2$, the trivialization extends to a map $I\to \mathcal{O}_{S\times X}$. This map is injective, in fact injective in all fibres: The restriction to each fibre $\{s\}\times X$ is nonzero (as $U$ intersects all fibres) and hence an embedding ($I$ is torsion free in fibres). It follows that the quotient is flat. There are some details to check, but this is the main point, I think.

[End of new answer, here is the original one:]

If we attempt to define $M_I(X)(S)$ as the set of $S$-flat ideals $I_Z$ in $\mathcal{O}_{S\times X}$, then that would not be functorial in $S$, as the inclusion $I_Z \subset \mathcal{O}_{S\times X}$ may not continue to be injective after base change (in the counter example in the other answer, restriction to the problematic fibre gives the zero map). We could impose "universal injectivity", but that is just another way of requiring the quotient $\mathcal{O}_Z$ to be $S$-flat, so then we have (re)defined the Hilbert scheme.

Another common way of defining moduli of ideals is as the moduli space for rank one stable sheaves (i.e. torsion free) with trivial determinant line bundle. The resulting moduli space is isomorphic to the Hilbert scheme of subschemes of codimension at least 2.

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Now I'm more confused... In the Inventiones paper by Bridgeland, he really said "the scheme $M_I(X)$ is the moduli space of ideal sheaves on $X$". (See 7.3 of that paper.) He didn't say what that is and how it is constructed... I thought he meant some well-known thing. – 36min Sep 18 '12 at 22:34
I think he may be taking the thing you mentioned in the second paragraph as definition. And the only case of interest is the moduli space of ideal sheaves that are numerically equivalent to a point in a 3-fold. (He quoted Simpson's paper a lot. That's where the definition of moduli of ideal sheaves is from.) However if the moduli space is isomorphic to the Hilbert scheme then I don't see why it is necessary to consider it. – 36min Sep 19 '12 at 6:29
@36min: And to answer your new question 2: With this definition, you can realize M_I as the fibre over $\mathcal{O}_X$ for the determinant map $M\to \mathrm{Pic}(X)$, where $M$ is the Simpson moduli space for stable/torsion free rank one sheaves. – MartinG Sep 19 '12 at 6:42
There is a problem with the determinant map. You know, the scheme $X$ in Bridgeland's paper is NOT smooth and ideal sheaves he considers are not perfect complexes, so they don't have a locally free resolution and so their determinant is just NOT DEFINED. – Sasha Sep 19 '12 at 9:14
It appear to me the determinant of a coherent sheaf can be defined over a normal scheme. Just push-forward the determinant of the restriction of the coherent sheaf to the smooth locus (complement of a codimension 2 thing). And then $M_I(X)$ makes sense. Did I miss something here? – temp Oct 2 '12 at 1:02

Bridgeland uses another point of view on moduli spaces. His definition of a moduli functor (Def. 3.7) does not uses any assumption of flatness. Instead, he considers DERIVED restriction of the family object to fibers, and the condition which replaces flatness is that these derived restrictions are PURE in the t-structure he considers. If the t-structure is the usual one then this definition is equivalent to the standard definition (flatness = no higher derived pullbacks).

The moduli space $M_I(X)$ of ideal sheaves is defined by the functor which associates to any scheme $S$ the set of (equivalence classes of) all sheaves on $S\times X$ which after (derived) restriction to fibers over $S$ are ABSTRACTLY isomorphic to sheaves of ideals. On a contrary, the Hilbert scheme functor associates to a scheme $S$ the set of sheaves on $S\times X$ with a FIXED EMBEDDING into $O_{S\times X}$ (again up to an appropriate embedding). Forgetting the embedding we obtain a morphism $Hilb(X) \to M_I(X)$.

Of course, if $S$ is a closed point, and $I$ is isomorphic to an ideal sheaf, then there is a unique (up to a constant) embedding of $I$ into $O_X$. This shows that the map above is a bijection on closed points. On the other hand, globally it is not clear why a family of sheaves isomorphic to ideal sheaves has an embedding into $O_{S\times X}$. I think that Bridgeland just does not want to deal with this subtle question, which has nothing to do with his goal, so he just ignores it.

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Well, he said if the derived restrictions are sheaves, then you do get a flat family of sheaves before definition 3.7. How on earth can one forget the inclusion in the definition of an ideal sheaf? Any maximal ideal of $\mathbb{Z}$ are abstractly isomorphic as $\mathbb{Z}$-modules, so it is essential to remember the embedding. It is not just a subtle thing. – 36min Sep 19 '12 at 6:18
These are just different questions. Sometimes you need to remember the embedding, sometimes you don't. – Sasha Sep 19 '12 at 6:26
You always need remember the embedding. What you said is sometimes we can just remember the image. That's what you mean by up to a constant. And I don't think my question is answered. E.g. What is $M_I(X)$? What's the functoriality? – 36min Sep 19 '12 at 6:32
@Sasha: This viewpoint makes sense, except that for the embedding of $I$ into $\mathcal{O}_X$ to be unique, you need to assume codimension at least 2. This is what 36min complained about with maximal ideals in $\mathbb{Z}$. So we have to restrict to codimension at least 2 to get a bijection. (What I wrote in my answer, then, is basically that this $M_I(X)$ can be defined without reference to ideals, and that the morphism from the Hilbert scheme is in fact an isomorphism, although that is not entirely obvious and probably independent from what Bridgeland is doing, as you say.) – MartinG Sep 19 '12 at 8:34
The argument in Lemma 5.3, Prop. 5.4 and Thm. 5.5 with $Y$ replaced by $X$ and $L$ eplaced by $O_X(1)$ proves that $M_I$ is a scheme. – Sasha Sep 22 '12 at 7:26