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Let $k=\mathbb{F}_q$ be a finite field with $q$ elements and let $X$ be a quasi-projective $k$-scheme. I saw somewhere claims the following results (without explanation):

  1. Let $N$ be a positive integer and let $i: Z\hookrightarrow X$ be the closed immersion of the (finite) disjoint union of ${\rm Spec}(\kappa(x))$ for all the closed points $x\in X$ of degree $[\kappa(x):k]<N$. For $n< N$, we have a bijection $${\rm Sym}^ni: ({\rm Sym}^nZ)(k)\xrightarrow{\cong}({\rm Sym}^nX)(k).$$
  2. Suppose $X={\rm Spec}(L)$ for a finite extension $L/k$ of degree $d$. Then $({\rm Sym}^nX)(k)$ is empty if $d\nmid n$, and is a singleton if $d\mid n$.

I don't have any idea with them. Can anyone help me to prove these results? Thanks also for any idea/hint/reference!

For the construction of symmetric products of varieties, one can see Section 3.1 in Hilbert and Chow Schemes of Points, Symmetric Products and Divided Powers.

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There are probably many ways to answer your question depending on what your preferred point of view on schemes and symmetric products are. Let me offer the following approach.

Forget that $k$ is a finite field, let it just be a perfect field, call $\newcommand{\alg}{\operatorname{alg}}k^{\alg}$ a fixed chosen algebraic closure and $\Gamma := \newcommand{\Gal}{\operatorname{Gal}}\Gal(k^{\alg}/k)$ its absolute Galois group. We see a $k$-scheme $X$ of finite type through its set $X(k^{\alg})$ of geometric points endowed with a (continuous) action of $\Gamma$; for $k \subseteq k' \subseteq k^{\alg}$, the set $X(k')$ is the set of $\Gamma_{k'}$-fixed points of $X(k^{\alg})$ where $\Gamma_{k'} := \Gal(k^{\alg}/k')$. We note that $\newcommand{\Sym}{\operatorname{Sym}}\Sym^n(X)$ has the set $\Sym^n(X(k^{\alg}))$ (of $n$-element multisubsets of $X(k^{\alg})$) as geometric points, with the obvious Galois action. In particular, $(\Sym^n(X))(k)$ is the set of $\Gamma$-stable $n$-element multisubsets of $X(k^{\alg})$. Also note that $\newcommand{\Spec}{\operatorname{Spec}}\Spec(L)$ is seen as the homogeneous $\Gamma$-set $\Gamma/\Gamma_L$ of left cosets of $\Gamma_L$ under $\Gamma$.

Now remember that $k$ is a finite field. Then $\Gamma$ is procyclic with progenerator given by the Frobenius $\sigma \colon x \mapsto x^q$ (i.e., a $\Gamma$-set is just a set with a finite-order permutation $\sigma$). When $L$ is the extension of degree $d$, the $\Gamma$-set $\Spec L$ is given by a $d$-cycle; a closed point in $X$ with residue field $L$ is such a $d$-cycle in the Galois action on the geometric points of $X$.

With this in mind, claim (1) essentially says that if $\sigma$ is cyclic permutation on an $n$-element multiset (a $\sigma$-stable $n$-element multisubset of $X(k^{\alg})$), we can see it as a multiset sum of $d$-cycles for $d\leq n$, which we then see as a single $n$-element $\sigma$-stable multisubset of the disjoint union of “archetype” $d$-cycles (one for each $d\leq n$).

As for (2), it says that a $d$-cycle has a single $\sigma$-stable $n$-element multiset when $d|n$ (namely, take every element of the cycle $n/d$ times), and none if $d$ does not divide $n$.

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  • $\begingroup$ Thank you very much for your answer. I have some questions: 1) you wrote that $\newcommand{\Sym}{\operatorname{Sym}}\Sym^n(X)$ has the set $\Sym^n(X(k^{alg}))$ (of $n$-element multisubsets of $X(k^{alg})$) as geometric points. How can we see this easily? 2) I'm not familiar with multisets, what do you mean by "$d$-cycles"? Is it in the sense of permutation? $\endgroup$
    – Lao-tzu
    Commented Mar 28, 2022 at 20:08
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    $\begingroup$ (1) This comes from the fact that the symmetric powers are defined as geometric quotients (definition 3.1.1 in the paper by Rydh cited in your question) and the fact that geometric points of geometric quotients are orbits of geometric points as part of the definition of geometric quotients (Mumford Fogarthy Kirwan, Geometric Invariant Theory (3rd ed 1994), definition 0.6(ii) and the following remark and footnote). And of course orbits of $\mathfrak{S}_n$ acting on $Z^n$ (where $Z$ is any set) are exactly $n$-element multisubsets of $Z$. $\endgroup$
    – Gro-Tsen
    Commented Mar 29, 2022 at 8:41
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    $\begingroup$ (See also ¶3.6 in the paper by Rydh cited in your question, which makes the point that the $n$-th symmetric power and the Chow variety of $0$-cycles of degree $n$ have the same points over any field.) ❧ As for your question (2), yes, by a “$d$-cycle” in my answer I meant one in the sense of permutations: that is, a set with $d$ elements (seen as a multiset with $d$ distinct elements) which are acted upon cyclically by $\sigma$; not “cycles” in the sense of Chow varieties (sorry about the confusion!). $\endgroup$
    – Gro-Tsen
    Commented Mar 29, 2022 at 8:51
  • $\begingroup$ Thank you very much for your great answer and follow-up explanation as well as precise references! I would never imagine the answer is so involved (and I didn't go into the paper by Rydh cited in my question, I found it by a random Google search and used it only to confirm those who read my question what are the symmetric products I'm using; it seems quite interesting and I will read it later). $\endgroup$
    – Lao-tzu
    Commented Mar 29, 2022 at 9:44
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    $\begingroup$ Quotients in algebraic geometry are very complicated (and I can't claim to understand much), but here you shouldn't think of the situation as too involved: think of the affine line $\mathbb{A}^1$ for intuition: a point over $k$ of its $n$-th symmetric power is a monic polynomial of degree $n$ over $k$, which is the same as giving a multiset of $n$ roots (geometrically!) together with Galois action. This is the picture I have in mind (even if $\mathbb{A}^1$ is, of course, very specific). $\endgroup$
    – Gro-Tsen
    Commented Mar 29, 2022 at 9:55

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