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Let $X$ be a smooth projective geometrically connected variety over a field $k$.

Let the cyclic group $G=\{e,a\}$ with two elements act on $X \times X$ via $a\cdot (x_1,x_2) = (x_2,x_1)$.

When is the quotient $X\times X/ G$ nonsingular?

I wrote down the case of $X=\mathbf{A}^1$. The answer is that $X\times X/G$ is given by nonsingular scheme $\mathrm{Spec} (k[xy,x+y]) \cong \mathbf{A}^2$, where $\mathbf{A}^1_x\times \mathbf{A}^1_y = X\times X = \mathrm{Spec} (k[x]\otimes k[y])$. (The subscript $x$ and $y$ indicate the coordinate used.)

I'm very interested in the case $\dim X = 1$.

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    Ha! You're right. But can't this calculation be used to show that the 2-symmetric product of $\mathbf{P}^1$ is nonsingular?2012-07-19

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Over an algebraically closed field:

The quotient $X^{(2)}=X\times X/C_2$ is called the symmetric square of $X$.

If $X$ is an irreducible (quasi)projective nonsingular curve, then $X^{(2)}$ is always smooth and, in fact, this generalizes to $X^{(n)}=X\times\cdots\times X/S_n$, the obvious quotient by the symmetric group on $n$ letters.

If $X$ is an irreducible (quasi)projective nonsingular surface, then $X^{(n)}$ is singular for all $n\geq2$.

Lothar Göttsche wrote notes for a short course he gave in ICTP a few years ago on the subject of Hilbert schemes of points, with title Hilbert schemes: local properties and Hilbert scheme of points. There you should find proofs of all the above statements. (The Hilbert scheme in all this is a «less singular» replacement for what I wrote $X^{(n)}$ which, in the case of smooth surfaces, turns out to be itself a desingularization of the symmetric power; in general, it is also singular, though)

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    I found the notes on his website : http://users.ictp.it/~gottsche/2012-07-19