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

- If $C$ is a smooth, 1-dimensional variety over an algebraically closed field $k$,, then $S^nC$ is smooth.
- 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?