If $C$ is a smooth curve over a field $k$, then from lots of references, e.g. Janos Kollar, Rational Curves on Algebraic Varieties, exercise 1.4.1, that the Hilbert scheme of $n$ points is \begin{equation} \text{Hilb}^n(C)=\text{Sym}^n(C) \end{equation} where $\text{Sym}^n(C):=C \times \cdots \times C/S_n$.

First, I do not know how to show $\text{Sym}^n(C)$ is smooth, I only know how to show $\text{Sym}^n(\mathbb{A}^1)$ is smooth.

Second, how to show $\text{Sym}^n(C)$ is actually the Hilbert scheme of $n$ points?

  • $\begingroup$ Intuitively, smoothness if a local condition. Over $\mathbb C$ for example, $\mathbb A^n/\mathfrak S_n \cong \mathbb A^n$ by the fundamental theorem of symmetric polynomials, so it should work for any smooth curve. $\endgroup$ – user171326 Mar 19 '17 at 20:11

The symmetric product $Sym^n(C)$ is smooth:

For the smoothness part we may assume the field is algebraically closed. We will use the following characterization of smoothness:

A scheme $X$ locally of finite type over an algebraically closed field $k$ is smooth at the closed point $x \in X$ of dimension $n$ iff the completed local ring there satisfies: $\widehat{\mathcal{O}}_{X,x} \cong k[[x_1,...,x_n]]$

We see now that it suffices to convince yourself that completion at a point commutes with forming quotients by finite group actions. It will then follow that for any $x \in Sym^n(C)$ the following holds:

$$\widehat{\mathcal{O}}_{Sym^n(C),x} = {\widehat{\mathcal{O}}_{C^{\times n},x}}^{S_n} \cong k[[x_1,...,x_n]]^{S_n} \cong k[[x_1,...,x_n]]$$

But know recall that the construction of the quotient by a finite group goes through taking invariants of the group action in the category of rings. Invariants are a limit construction and therefore commutes with arbitrary limits. In particular it will commute with the above completion. QED.

The symmetric product is the Hilbert scheme:

There's a canonical map from $Sym^n(C)$ to the Hilbert scheme corresponding to the natural family $\mathcal{X} \subset C \times Sym^n(C)$ over $Sym^n(C)$ (where the fiber is just the disjoint union of the points possibly with multiplicities).

We can now use the fact that over an algebraically closed field maps are determined by where geometric points go. It is now clear that any flat family $\mathcal{X} \subset C \times T \to T$ parametrizing $n$-points in $C$ determines a unique map $T \to Sym^n(C)$ by sending a closed (geometric) point $t \in T$ to the projection of the fiber $\mathcal{X}_t$ back to $C$ (which gives a collection of $n$-points). This map is unique and is a bijection on $k$-points ($k$-algebraically closed) and therefore an isomorphism by Zariski main theorem.


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