The symmetric product of a variety $M$ is the quotient of $M^n/S_n$ where $S_n$ is the symmetric group permuting components of nfold product $M^n$. IF $M$ is an affine plane $C^k$ over complex numbers, the coordinate ring of the symmetric product is the invariant polynomials in $R:=C[x^1_1,...,x^1_k, x^2_1,...,x^2_k,... ,x^n_1,...,x^n_k]$ under the action of $S_n$ where $S_n$ permutes the variables $x_i^1,...,x_i^n$ simultaneously for $i=1,...,k$. I want to know the invariant subring $R^{S_n}$ in terms of generators and relations. Could anybody help me?
3 Answers
Those invariant polynomials are called multisymmetric functions. There are several papers on them; you could start with J. Dalbec, Multisymmetric functions, Beiträge Algebra Geom. 40(1) (1999), 2751 http://www.emis.de/journals/BAG/vol.40/no.1/b40h1dal.ps.gz.
The relations might be complicated. The multisymmetric functions of degree up to n generate the ring, but very redundantly. In Lemma 2.2 of
http://annals.princeton.edu/annals/2006/1632/p11.xhtml
Venkatesh and I show that you can get by with using many fewer of these multisymmetric functions, if you are content to generate a subring of R^{S_n} whose fraction field is finiteindex in the fraction field of R^{S_n}.
I may be making a very trivial mistake [Edit: yes, indeed], but isn't it just that:
$(\mathbb{A}^k)^{(n)}=(\mathbb{A}^k)^n/S_n\cong(\mathbb{A}^n)^k/S_n\cong(\mathbb{A}^n/S_n)^k\cong(\mathbb{A}^n)^k$
with affine coordinate ring $\mathbb{C}[\sigma_1,\cdots,\sigma_n]^{\otimes k}$ (where $\sigma_d=\sigma_d(x_1,\cdots,x_n)$ is the degree$d$ symmetric function in the $n$ variables $x_1,\cdots, x_n$) ?

$\begingroup$ Your third isomorphism requires an identification between Sn and (Sn)^k. $\endgroup$– S. Carnahan ♦May 14, 2010 at 16:49

$\begingroup$ I mean, if $G$ acts on $X$ and $Y$, and diagonally on $X\times Y$, then $(X\times Y)/G\cong (X/G)\times (Y/G)$. Is it correct, right? $\endgroup$– QfwfqMay 14, 2010 at 17:07

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1$\begingroup$ The equality I wrote above absolutely doesn't work! For example, if $G$ has positive dimension, the "expected" dimentions of the two sides do not match. $\endgroup$– QfwfqMay 14, 2010 at 17:50