In the sequel, let $S$ be a scheme, and $X$ a locally of finite type algebraic space over $S$.

In his thesis ([R1-R4]), David Rydh introduces, among several others, the notion of relative cycles on $X\to S$, and, for integers $r\ge 0$, he defines the functor:

$$\text{Chow}_r(X/S) : (\text{Sch}/S)^{\rm opp}\to\text{Set}$$


$$\text{Chow}_r(X/S)(T) := \text{Cycl}^{\rm prop}_r(X\times_ST/T)$$

where the right side is the set of equidimensional relative cycles of dimension $r$ on $X\times_ST\to T$, with proper support (see [R4]).

$\text{Chow}_r(X/S)$ is an fppf sheaf, ultimately because the fppf sheaf of divided powers of $X\to S$ is representable (see [R1]).

In [R4], representability of $\text{Chow}_r(X/S)$ is shown only in those cases when $\text{Chow}_r(X/S)$ can be proved to be isomorphic as a functor to some other functor known to be representable (e.g., Angéniol's Chow space (see [An]), when $S$ is of pure characteristic zero, and $X\to S$ is smooth and separated).

It appears that the best representability result one may extrapolate from [R4] is, therefore, as follows:

Theorem 1. Let $S$ be a scheme of pure characteristic zero, $X\to S$ a smooth and separated algebraic space, $r\ge 0$ an integer. Then the fppf sheaf $\text{Chow}_r(X/S)$ is represented by a separated algebraic space over $S$, locally of finite type.

The algebraic space representing $\text{Chow}_r(X/S)$ will be called Chow space, in the sequel.

It is easy to show, using representability of the Hilbert functor when $X\to S$ is projective, that the following holds:

Theorem 2. Let $S$ be a scheme of pure characteristic zero, $X\to S$ a smooth and projective algebraic space, $r\ge 0$ an integer. Then:

  1. $\text{Chow}_r(X/S)$ is proper.
  2. the Hilbert-Chow morphism: $$\text{Hilb}_r(X/S)\to\text{Chow}_r(X/S)$$ constructed in [R4], is proper.


  1. Is it true that, under the assumptions in Theorem 2, $\text{Chow}_r(X/S)$ in $(1)$ is projective?
  2. Under the assumptions in Theorem 2, what are the currently known properties of the Hilbert-Chow morphism $\text{Hilb}_r(X/S)\to\text{Chow}_r(X/S)$ in $(2)$? Is it surjective/surjective on geometric points/birational?
  3. Upon inspecting Angéniol's proof of representability of his Chow functor, it appears to me the smoothness assumption can be removed, upon appropriately exploiting the deformation theory in vol. 1 of Illusie's thesis. Has this been done anywhere in the literature?
  4. What can one say, to this day, about the case when $S$ is of pure characteristic $p>0$?
  5. Around question $(2)$. Is representability of $\text{Chow}_r(X/S)$ known at least in the case when $S = \text{Spec}(k)$, for $k$ a field of characteristic $p>0$, and $X\to S$ a smooth projective $S$-scheme?


[An] B. Angéniol. Familles de cycles algébriques. Springer.

[R1] D. Rydh. Families of zero-cycles and divided powers: I. Representability.

[R2] D. Rydh. Families of zero-cycles and divided powers: II. The universal family.

[R3] D. Rydh. Hilbert and Chow schemes of points, symmetric products and divided powers.

[R4] D. Rydh. Families of cycles.

  • 6
    $\begingroup$ In characteristic $0$ there is also Barlet's construction. Also, all of the "Chow varieties" constructed by Chow and van der Waerden (which a priori depend on an embedding in projective space) are equal and quasi-projective in characteristic $0$. So I guess you are really asking whether the ample invertible sheaves on the Chow variety (which are studied by Mumford in GIT, by Fogarty, by Knudsen-Mumford, ...) are restrictions of an invertible sheaf defined on Angeniol's Chow scheme. $\endgroup$ Aug 2, 2017 at 10:11
  • 6
    $\begingroup$ For Question 2, you can check surjectivity on geometric points after base change from $S$ to Spec of algebraically closed fields. So long as the cycle contains no irreducible component of $X$, it is straightforward to construct a closed subscheme whose associated cycle is the specified cycle. However, if you take the cycle $2[X]$, for example, that cannot be the underlying cycle of a closed subscheme of $X$. Re birationality, I assume that you are asking whether the Hilbert-Chow morphism is an immersion on the open in Hilb parameterizing integral closed subschemes . . . $\endgroup$ Aug 2, 2017 at 12:23
  • 6
    $\begingroup$ . . . I believe that this is not true. I believe that Mumford's pathological examples of smooth space curves of degree $14$ and genus $24$ give counterexamples. $\endgroup$ Aug 2, 2017 at 12:25
  • 5
    $\begingroup$ Nice question and nice comments. Does the Chow functor satisfy the existence and uniqueness parts of the valuative criterion when $X/S$ is proper/projective? $\endgroup$
    – user95222
    Aug 3, 2017 at 8:32
  • 6
    $\begingroup$ @AG2073951378. In characteristic $0$, as in the first two questions, the reduced scheme of the Chow scheme is the classical Chow variety. The (components of the) classical Chow varieties are quasi-projective, resp. projective, if $X/S$ is quasi-projective, resp. projective. The valuative criterion is stated in terms of morphisms from valuation rings, and these factor through the reduced scheme. $\endgroup$ Aug 3, 2017 at 9:13

1 Answer 1


Mathoverflow answer

In my thesis [R4], I gave an ad hoc definition of a Chow functor ($\mathrm{Chow}_r$ above) that was meaningful also in characteristic p and close to Barlet's and Angéniol's definitions. It is "ad hoc" because the definition involves specifying zero-cycles over every suitable projection to an r-dimensional smooth space and imposing compatibility conditions, whereas Barlet (reduced, char 0) and Angéniol (char 0) have a single object representing the family of cycles.

Theorem 1: I could only show that $\mathrm{Chow}_r$ coincides with Barlet/Angéniol's functor when X/S is smooth and separated for (a) reduced parameter schemes, (b) families of "multiplicity-free" cycles (the multiplicity of every component is 1), and (c) families of Weil divisors, see [R4, Thm 16.2]. In particular, Thm 1 above is, as far as I know, not known in general. It is possible that my ad-hoc definition needs to be modified slightly to agree in general (see introduction and discussion after 5.4).

Theorem 2: $\mathrm{Chow}_r(X/S)$ is not proper: it is (typically an infinite) disjoint union of projective schemes $\mathrm{Chow}_{r,d}(X/S)$ where you fix the degree d of the cycles. The morphism $\mathrm{Hilb}_r \to \mathrm{Chow}_r$ is also not proper for similar reasons, but $\mathrm{Hilb}_P\to \mathrm{Chow}_r$ is projective where $P$ is a Hilbert polynomial of degree $r$ and leading coefficient $dr!$.

Q1: In characteristic zero, it is known that $(\mathrm{Chow}_r)_\mathrm{red}$ coinicides with the reduction of Angéniol's functor (see above) and for projective schemes the latter is known to coincide with the classical Chow variety [R4, Cor 16.3], which is projective after fixing the degree $d$. There is also an explicit map $\mathrm{CH}\colon \mathrm{Chow}_{r,d}(X/S)\to \mathrm{Div}^d(G/S)$ where $G$ is a suitable Grassmannian [R4, Def 17.4]. Since $\mathrm{CH}_\mathrm{red}$ is a closed immersion, $\mathrm{CH}$ is at least finite and hence $\mathrm{Chow}_{r,d}$ is projective (if representable, otherwise Angéniol's variant).

Q2: As Jason mentioned, you need a tiny bit to ensure that the Hilb-Chow morphism is surjective: a geometric $k$-point of Chow is simply a cycle and such can be lifted to a subscheme if it has at least codimension $1$ (in the local ring of every generic point of the support of the cycle, choose a closed subscheme with length equal to the multiplicity and take the closure).

The restriction of $\mathrm{Hilb}_r(X/S)\to \mathrm{Chow}_r(X/S)$ to the open subscheme parametrizing families of normal equidimensional subschemes is an open immersion [R4, Cor 12.9]. At least in characteristic zero: I think there is a characteristic zero assumption missing in Cor 12.9 (12.8 and 12.9 implicitly rely on Thm 9.8) but with the "correct" definition of $\mathrm{Chow}$ this should work out (Thm 9.8 should then hold in any characteristic).

The restriction of $\mathrm{Hilb}_r(X/S)\to \mathrm{Chow}_r(X/S)$ to the open subscheme parametrizing reduced equidimensional subschemes is a monomorphism [R4, Cor 9.10]. The point is that the flat subscheme $Z$ can be recovered from its associated family of cycles. Although [R4, Cor 9.10] assumes characteristic zero (via Cor 9.9 which for any family of multiplicity-free cycles gives a unique potentially non-flat $Z$), one can in any characteristic prove [R4, Cor 9.10] directly from [R4, Prop 9.2] ($Z$ is in fact unique).

I do not know if $\mathrm{Hilb}_P(X/S)\to \mathrm{Chow}_r(X/S)$ is an immersion after restricting to reduced equidimensional subschemes or even integral subschemes. This is claimed, without proof, in [R4, Rmk 9.11] (it also does not follow from the second sentence of the remark as claimed even if we knew that Chow_r was representable). Indeed, it could very well be false as Jason indicates (I don't know if Mumford's example gives a counter-example though). One would obtain a counter-example if one has a non-flat family $Z\to T$ with $Z$ and $T$ reduced such that the fibers are irreducible and generically reduced and their reductions all have the same Hilbert polynomial.

Q3: Angéniol requires $X\to S$ to be smooth (which I think can be generalized to smoothly embeddable). He heavily relies on using cotangent sheaves but one could perhaps replace these with dualizing complexes but I think a lot of work would be required (IIRC, the most involved part is how the cohomology class representing the cycle behaves under change of projections and this could be difficult to generalize).

Q4: Unfortunately, even now I can say nothing about the representability of $\mathrm{Chow}_r$ in positive characteristic. The key would be to understand the deformation theory of families of cycles and at the time I wrote my thesis I knew very little about deformation theory. I thought, and still think, that it can be done with a reasonable amount of work for (1) multiplicity-free cycles and (2) relative Weil-divisors such that every irreducible component meets the smooth locus (e.g., families of semi-log canonical pairs).

Q5: No, not as far as I know. I think this case is as difficult as the general case.

  • $\begingroup$ Great. Thank you! $\endgroup$
    – user87684
    Oct 2, 2017 at 10:48

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