# Why is the symmetric power a quasi-projective variety?

Let $X$ be a quasi-projective variety over the complex numbers. Its symmetric power $S^nX$ is defined as follows: Let the symmetric group $S_n$ act on the power $X^n$ by permutation of coordinates and let $S^nX$ be the quotient by this group action. Can you tell about some book, where I can find a proof of the fact that $S^nX$ is also a quasi-projective variety?

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Is it not sufficient to prove the result for the projective space? – Piotr Achinger Jul 4 '11 at 20:51
it is in SGA1.V.1 Basically the criterion for the existence of finite quotient is that each orbit should lie in some affine open. quasi-projective varieties have this property. – Dima Sustretov Jul 4 '11 at 20:52
Thank you for your reply. In SGA it is shown that it is a variety, but where can I find a proof that it is a quasi-projective one? More generally: Can you tell me why a finite quotient of a quasi-projective variety is always quasi-projective? – nick Jul 10 '11 at 13:02
Once you know that $S^nX$ exists, you can see that it is quasiprojective because it carries an ample line bundle induced from $X$. Alternatively, take a look at Knutson's Algebraic Spaces p 180 for a detailed construction (although the result was surely long before). – Donu Arapura Jul 10 '11 at 14:05