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Mar 29, 2022 at 10:02 comment added Lao-tzu Thanks! I will think more about them!
Mar 29, 2022 at 9:55 comment added Gro-Tsen 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).
Mar 29, 2022 at 9:45 vote accept Lao-tzu
Mar 29, 2022 at 9:44 comment added Lao-tzu 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).
Mar 29, 2022 at 8:51 comment added Gro-Tsen (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!).
Mar 29, 2022 at 8:41 comment added Gro-Tsen (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 $Z$.
Mar 28, 2022 at 20:08 comment added Lao-tzu 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?
Mar 28, 2022 at 12:22 history answered Gro-Tsen CC BY-SA 4.0