TITLE: Smoothness of Symmetric Powers
QUESTION [11 upvotes]: 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?

REPLY [11 votes]: I also wanted to mention a `high-technology' answer to (1). Namely, if $C$ is a smooth algebraic curve, its $n$-th symmetric power coincides with the variety of all degree $n$ effective divisors on $C$ (that is, we consider $n$-element subsets of $C$, and then allow points to merge). In a fancier language, we are looking at the Hilbert scheme of length $n$
subschemes of $C$.
Note that this agrees with the explicit choice of coordinates if $C={\mathbb A}^1$ (and also $C={\mathbb P}^1$): namely, if you want to parametrize $n$-element subsets of ${\mathbb A}^1$ with multiplicities, you do so (essentially by unique factorization) by saying that such a subset is the zero locus of a degree $n$ monic polynomial $p$, so it is uniquely determined 
by coefficients of $p$ (which of course are the elementary symmetric polynomials in terms of
original $n$ points).
OK, so why do finite subschemes of $C$ form a smooth variety? Because deformation theory
says that at a particular finite subscheme $D\subset C$, the obstructions to smoothness belong to $Ext^2_C(O_D,O_D)$. This vanishes because $C$ is a smooth curve, and so there are no $Ext$'s beyond first (OK, no local $Ext$'s, but $D$ is finite, so it's the same thing).
I think this is a useful point of view. For instance, if $S$ is a smooth surface, $Sym^n S$
is singular, but one can check that the Hilbert scheme $Hilb_n S$ of degree $n$ finite subschemes is still smooth; it provides a resolution of singularities of $Sym^n S$. One can thus argue that the Hilbert scheme is a 'better behaved' object compared to the symmetric power...

REPLY [3 votes]: Edit: It seems appropriate to recall here Chevalley-Shephard-Todd's Theorem.
It says  the quotient of $\mathbb A^n_k$ by a finite linear group $G$ with order prime to the characteristic of $k$ is smooth (i.e. the algebra of invariants is polynomial) if and only if $G$ is generated by pseudo-reflections (codimension one fixed point set).
Once one localizes the problem at a point, as VA and Ben Webster did, this settles both 1 and 2 over arbitrary fields of characteristic zero. Of course VA argument is preferable as it is more direct/elementary.

Original answer. Below is my original answer commented by David Speyer below:
Over the category of smooth real manifolds the symmetric power of smooth curves is not smooth in general. The second symmetric power is a smooth surface with boundary and starting from the third symmetric power what we get are varieties with corners at the boundary.
Symmetric powers of smooth surfaces are still smooth as they are locally diffeomorphic to complex curves.
If nothing else these examples show that smoothness might mean different things for algebraic geometers and differential geometers in algebraic geometry over $\mathbb R$ and in differential geometry.