Here's something that's been bothering me, and that's come up again for me recently while reading some stuff about Hilbert schemes of points (Nakajima's lectures, specifically):


Let $C$ be an algebraic curve. Define $S^nC$ to be $C\times\ldots\times C/S\_n$, the symmetric power.


Now, over $\mathbb{C}$, I can show that $C$ a complex manifold implies that $S^nC$ is, and that if $X$ is a variety with $S^nX$ smooth, then $X$ is one dimensional, but the argument I have involves looking in analytic open sets and reducing to the case of $\mathbb{C}^n$, and additionally is fairly unhelpful for identifying the total space (ie, that $S^n\mathbb{P}^1\cong\mathbb{P}^n$)


So here's my question: how can we, in a fairly quick and natural way, show that


1. If $C$ is a smooth, 1-dimensional variety over an algebraically closed field $k$,, then $S^nC$ is smooth.
2. If $X$ is a smooth variety over an algebraically closed field $k$, and $S^nX$ is smooth, then $X$ is one dimensional.


Now, I don't want any projectivity hypotheses here, and I'm curious, with the more arithmetically inclined, is this still true over an arbitrary field?



Smoothness of Symmetric Powers
Just convince yourself that if $(C,P\_1,\dots,P\_n)$ and $(D,Q\_1,\dots,D\_n)$ are analytically isomorphic -- $C$ at the point $P\_i$, $D$ at the point $Q\_i$ -- then $Sym^n C$ at the point $(P\_1,\dots,P\_n)$ is analytically isomorphic to $Sym^n D$ at the point $(Q\_1,\dots,Q\_n)$. For this, you need to see that completion commutes with taking $S\_n$-invariants. Let me hedge and say that this is clear if $P\_2,\dots,P\_n$ are smooth points.


(When I say "analytically isomorphic", I don't mean working over $\mathbb C$. Instead, as customary, I mean that the completion of the corresponding rings are isomorphic, for example $\widehat{\mathcal O}\_{C,P\_i}\simeq \widehat{\mathcal O}\_{D,Q\_i}$.)


Thus for 1 you can reduce to $D=\mathbb A^1$, and you are done.


1 is true over any field, without restrictions on characteristic. Indeed, the ring corresponding to $(\mathbb A^1)^n/S\_n$ is the ring of invariants $k[x\_1,\dots, x\_n]^{S\_n}$, and it is a polynomial ring on $n$ variables, the elementary symmetric polynomials in $x\_i$. That is true over any ring $k$ (commutative, with identity).


Part 2 is reduced to the case of $\mathbb A^2$ by the same trick, and then it is an explicit computation. I will illustrate the $(\mathbb A^2)^n/S\_n$ case. We need to find the ring of invariants $k[x\_1,y\_1,x\_2,y\_2,\dots,x\_n,y\_n]^{S\_n}$. Now there are $2n$ elementary polynomials in $x\_i$, resp. in $y\_i$, of course. But there is more. For example $x\_1y\_1 + \dots + x\_ny\_n$ is an invariant.


To prove that $X=(\mathbb A^2)^n/S\_n$ is singular at the point $P=(0,\dots,0)$ is equivalent to showing that $\dim T\_{P,X} = \dim m/m^2 > 2n = \dim X$, where $m$ is the maximal ideal in $\mathcal O\_{P,X}$. It is an exercise that the above $2n+1$ polynomials are linearly equivalent in $m/m^2$, which would complete the proof.


