Let $X$ be a smooth projective $\mathbb C$-variety and let $X^{(n)}$ denote the symmetric product $X^n/S\_n$, parametrizing effective $0$-cycles of degree $n$ on $X$.



> 
> **Question**. Let $S$ be a noetherian $\mathbb C$-scheme, $\mathcal F$ a coherent sheaf on $X\times S$, flat and
>  finite (of degree $n$) over $S$. Does its support define a family of
>  $0$-cycles on $X$, i.e. a morphism $S\to X^{(n)}$?
> 
> 
> 


I came up with this question studying the construction of the Hilbert-Chow morphism $X^{[n]}\to X^{(n)}$ in *FGA explained*, where a fundamental tool is the result that any flat sheaf $\mathcal F$ gives a relative effective Cartier divisor on $X\times S$ with good properties (essentially commutation with base change). 


I am aware that the subscheme $\textrm{Supp }\mathcal F\subset X\times S$ may not be flat over $S$ (it *is* flat if $S$ is reduced). I am asking if $\mathcal F$ induces an $S$-valued point $S\to X^{(n)}$, *possibly not* factoring through a point $S\to X^{[n]}$.


Here the problem is that I do not know what the functor of points of $X^{(n)}$ looks like, in other words I do not know how to describe $$\textrm{Hom}(S,X^{(n)})$$ for an arbitrary scheme $S$.


Thanks!



Do finite flat sheaves define families of $0$-cycles?
I quickly reviewed the following article of David Rydh.


David Rydh.   

Families of Cycles.   

2008.   

<https://people.kth.se/~dary/famofcycles20080518.pdf>


Rydh extends to positive characteristic the definition of Angeniol in characteristic $0$. I will explain the construction in your special case. The key algebra lemma has to do with the $n\times n$ determinant. For a unital, commutative ring $R$, for a free $R$-module $F$ of rank $n$, for the associated free $R$-module $E=\text{Hom}\_R(F,F)$ of rank $n^2$ with its structure of associative, unital $R$-algebra, the usual determinant,
$$
\text{det}\_{F}:E \to R,
$$ 
comes from an element
$$
\text{det}'\_F\in \text{Sym}\_R^n(E^\vee),
$$
where for every $R$-module $M$, $M^\vee$ is shorthand for $\text{Hom}\_R(M,R)$. Assuming that $n!$ is invertible in $R$, this element is equivalent to a homomorphism of $R$-modules,
$$
\widetilde{\text{det}}\_F:(E^{\otimes n})^{\mathfrak{S}\_n} \to R,
$$
where for an $R$-module $M$, $M^{\otimes n}$ denotes the $n$-fold tensor product $M\otimes\_R \dots \otimes\_R M$, and where for an $R[\mathfrak{S}\_n]$-module $N$, $N^{\mathfrak{S}\_n}$ denotes the $R$-submodule of $\mathfrak{S}\_n$-invariants. The determinant homomorphism is normalized so that for an ordered basis $(\mathbf{e}\_1,\dots,\mathbf{e}\_n)$ of $F$ and for the associated basis $(\mathbf{e}\_{i,j})\_{1\leq i,j\leq n}$ of $E$, this $R$-module homomorphism sends the $\mathfrak{S}\_n$-invariant, idempotent element $$\epsilon\_{\text{id}} := \sum\_{\sigma \in \mathfrak{S}\_n} \mathbf{e}\_{\sigma(1),\sigma(1)}\otimes \dots \otimes \mathbf{e}\_{\sigma(n),\sigma(n)}$$ to $1$.


The algebra fact is that for a commutative $R$-subalgebra $A$ of $E$, the restriction of $\widetilde{\text{det}}\_F$ to $(A^{\otimes n})^{\mathfrak{S}\_n}$ is a homomorphism of commutative $R$-algebras. For a choice of ordered basis as above, for the subalgebra $A$ spanned by $\mathbf{e}\_{1,1},\dots,\mathbf{e}\_{n,n}$, this can be checked directly. The $R$-algebra $(A^{\otimes n})^{\mathfrak{S}\_n}$ is isomorphic to a direct product of factors of $R$ indexed by the nondecreasing functions $u:\{1,\dots,n\}\to \{1,\dots,n\}$ (with the standard total order). The corresponding idempotent is $$\epsilon\_u := \frac{1}{\# \text{Aut}(u)}\sum\_{\sigma\in \mathfrak{S}\_n} \mathbf{e}\_{u(\sigma(1)),u(\sigma(1))}\otimes \dots \otimes\mathbf{e}\_{u(\sigma(n)),u(\sigma(n))}.$$ The idempotent $\epsilon\_{\text{id}}$ corresponds to the identity function on $\{1,\dots,n\}$. The determinant $R$-module homomorphism sends the idempotent $\epsilon\_{\text{id}}$ to $1$, and it sends every other idempotent to $0$. 


Returning to the setup in the question, denote by $\text{pr}\_S:X\times S \to S$ the projection morphism. By hypothesis, the $\mathcal{O}\_S$-module $F:=\text{pr}\_{S,\*}\mathcal{F}$ is a locally free $\mathcal{O}\_S$-module of rank $n$. For every open affine subscheme $U\subset X$, there exists a maximal open subscheme $V\subset S$ such that $\text{Supp}(\mathcal{F}) \cap (V\times X)$ is contained in $V\times U$. Since $X$ is projective, as $(U,V)$ varies, the open subsets $V$ form an open cover of $S$. Thus, it suffices to construct the "Hilbert-Chow morphisms" on the open subsets $V$, provided those local morphisms satisfy the glueing condition on overlaps. 


For every element $a\in \Gamma(U,\mathcal{O}\_U)$, left multiplication by $a$ on $\mathcal{F}|\_{V\times U}$ defines an element $\widetilde{a}$ in the endomorphism algebra, $$E := \text{Hom}\_{\mathcal{O}\_V}(F|\_V, F|\_V).$$ This defines a homomorphism of $\mathcal{O}\_V$-algebras of quasi-coherent sheaves,
$$ \phi : \Gamma(U,\mathcal{O}\_U)\otimes\_{\mathbb{C}} \mathcal{O}\_V \to E.$$ Taking tensor products, this defines a homomorphism of $n$-fold tensor product $\mathcal{O}\_V$-algebras, $$ \phi^{\otimes n}: \Gamma(U,\mathcal{O}\_U)\otimes\_{\mathbb{C}} \dots \otimes\_{\mathbb{C}}\Gamma(U,\mathcal{O}\_U) \otimes\_{\mathbb{C}} \mathcal{O}\_V \to E\otimes\_{\mathcal{O}\_V} \dots \otimes\_{\mathcal{O}\_V} E.$$ There is a natural action of the finite symmetric group $\mathfrak{S}\_n$ on the domain and the target, and $\phi^{\otimes n}$ is equivariant for these actions. In particular, there is a restriction homomorphism on the subrings of $\mathfrak{S}\_n$-invariants, $$\phi^{\otimes n}\_{\mathfrak{S}\_n} : (\Gamma(U,\mathcal{O}\_U)^{\otimes n})^{\mathfrak{S}\_n}\otimes\_{\mathbb{C}} \mathcal{O}\_V \to (E^{\otimes n})^{\mathfrak{S}\_n}.$$ Finally, the determinant above defines a homomorphism of $\mathcal{O}\_V$-modules,
 $$\widetilde{\text{det}}\_F: (E^{\otimes n})\_{\mathfrak{S}\_n} \to \mathcal{O}\_V.$$ Composing with $\phi^{\otimes n}\_{\mathfrak{S}\_n}$ defines a homomorphism of $\mathcal{O}\_V$-modules, $$\text{det}(\phi):(\Gamma(U,\mathcal{O}\_U))^{\otimes n})^{\mathfrak{S}\_n}\otimes\_{\mathbb{C}} \mathcal{O}\_V \to \mathcal{O}\_V.$$ By the algebra fact above, this is a homomorphism of commutative $\mathcal{O}\_V$-algebras. This algebra homomorphism defines a morphism $V\to U^{(n)} \subset X^{(n)}$. These morphisms glue to a morphism $S\to X^{(n)}.$ 


**Edit. Proof of the Algebra Fact.** For certain choices of $R$, there are some funny commutative subalgebras $A$ of $E$, cf. my answer to the following question: [Simultaneous triangularizability over a commutative ring](https://mathoverflow.net/questions/167811/simultaneous-triangularizability-over-a-commutative-ring/167813). Thus, it makes sense to give a proof of the algebra fact. Actually it follows from a more general fact about $\widetilde{\text{det}}\_F$ on the associative $F$-algebra $(E^{\otimes n})^{\mathfrak{S}\_n}$. 


Since $n!$ is invertible in $R$, there is a basis of $(E^{\otimes n})^{\mathfrak{S}\_n}$ as follows. For every ordered $n$-tuple $(I,J) = ((i\_1,j\_1),\dots,(i\_n,j\_n))$ of pairs $(i,j)\in \{1,\dots,n\}\times \{1,\dots,n\}$, define $$\epsilon\_{I,J} = \sum\_{\sigma\in \mathfrak{S}\_n} \mathbf{e}\_{i\_{\sigma(1)},j\_{\sigma(1)}}\otimes \dots \otimes \mathbf{e}\_{i\_{\sigma(n)},j\_{\sigma(n)}}.$$ The element only depends on $(I,J)$ up to permutation by an element in $\mathfrak{S}\_n$. By direct computation, $\widetilde{det}\_F$ is nonzero on this element if and only if $(I,J)$ is the graph of a bijection, i.e., $((\tau(1),1),\dots,(\tau(n),n))$ for some $\tau\in \mathfrak{S}\_n$. In this case, denote the element by $\epsilon\_\tau$. By direct computation, $\widetilde{\text{det}}\_F(\epsilon\_\tau)$ equals $\text{sgn}(\tau)$. Also $\epsilon\_\tau \cdot \epsilon\_\rho$ equals $\epsilon\_{\tau\cdot \rho}$ for $\rho,\tau\in \mathfrak{S}\_n$. Finally, if $\epsilon\_\tau\cdot \epsilon\_{I,J}$, resp. $\epsilon\_{I,J}\cdot \epsilon\_\tau$, equals $\epsilon\_\rho$, then $\epsilon\_{I,J}$ equals $\epsilon\_{\tau^{-1}\cdot \rho}$, resp. $\epsilon\_{\rho\cdot \tau^{-1}}$. Altogether, this implies that $\widetilde{\text{det}}\_F(\epsilon\_{I,J}\cdot \epsilon\_{I',J'})$ is zero unless both $\epsilon\_{I,J} = \epsilon\_\tau$ and $\epsilon\_{I',J'}=\epsilon\_\rho$ for some $\rho,\tau\in \mathfrak{S}\_n$. Moreover, in this case, $\widetilde{\text{det}}\_F(\epsilon\_\tau\cdot \epsilon\_\rho)$ equals $\text{sgn}(\tau\cdot \rho)$. Since the sign of a permutation is a group homomorphism, it follows that in all cases, $$\widetilde{\text{det}}\_F(\epsilon\_{I,J}\cdot \epsilon\_{I',J'}) = \widetilde{\text{det}}\_F(\epsilon\_{I,J})\cdot \widetilde{\text{det}}\_F(\epsilon\_{I',J'}).$$ Therefore, already the $R$-module homomorphism $$\widetilde{\text{det}}\_F:(E^{\otimes n})^{\mathfrak{S}\_n} \to R,$$ is a homomorphism of associative, unital $R$-algebras. Thus, the restriction to any commutative $R$-subalgebra of $(E^{\otimes n})^{\mathfrak{S}\_n}$ is a homomorphism of commutative $R$-algebras.


