ag.algebraic geometry - 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.