Let $X$ be a projective variety. Symmetric product of $X$ is the quotient of the product $X^n$ by the action of the symmetric group $\Sigma_n$ permuting the factors.
When does it exist (as an algebraic variety)?
Let $X$ be a projective variety. Symmetric product of $X$ is the quotient of the product $X^n$ by the action of the symmetric group $\Sigma_n$ permuting the factors. When does it exist (as an algebraic variety)? 


To fix ideas, let $K$ be a field and $X/K$ be a seperated $K$scheme of finite type. Let $G$ be a finite group operating on $X$ via $K$morphisms. The operation is said to be admissible provided every orbit of $G$ is contained in an open affine subset of $X$. If the operation is admissible, then there is a pair $(Y, p)$ consisting of a seperated $K$scheme $Y$ of finite type and a finite, surjective morphism $p\in Hom(X, Y)^G$, such that the map $$Hom(Y, Z)\to Hom(X, Z)^G,\ f\mapsto f\circ p$$ is bijective for all schemes $Z/K$. Then $(Y, p)$ is said to be the quotient of $X$ mod $G$. (Cf. SGA I, V.1 for such constructions.) From now on assume that $X/K$ is projective. Then $X$ is a closed subscheme of $P_n$ and every finite subset of $X$ is contained in an open affine subset of $X$. We see: If a finite group $G$ acts on $X/K$ by $K$morphisms, then the operation is automatically admissible. Now take a projective $K$scheme $V/K$. Then $V^n=V\times\cdots\times V$ is projective over $K$ (Segreembedding) and hence the natural operation of the symmetric group $S_n$ on $V^n$ will be admissible. Consequently the quotient exists in that case. Aside: It can happen that $V^n/S_n$ is nonsmooth, even if $V^n$ is smooth over $K$. 

