Polarizations generate the ring of invariants? The symmetric group $S_n$ acts on $\mathbb R^n$ by permuting the coordinates and the ring of polynomial invariants is generated by the elementary symmetric polynomials. If we restrict the action to the alternating group $A_n$ then the invariant ring is generated by elementary symmetric polynomials along with the discriminant $\prod_{i <j}(x_i-x_j)$. If we take $m$ copies of $\mathbb R^n$ i.e., $\oplus_m \mathbb R^n$, then $S_n$ acts naturally and a theorem of Hermann Weyl says that the ring of invariants is generated by the polarizations of the elementary symmetric polynomials. Now if we restrict the action to $A_n$ then what is a minimal generating set for the ring of invariants ? Is it generated by the polarizations of the above invariants ? 
 A: To complete Peter's answer, I would like to give a concrete counterexample: by computing (e.g. using GAP) the Molien series for $S_3$ (resp. for $A_3$) acting naturally on $\mathbb{R}^3\oplus \mathbb{R}^3$, one sees that already in degree 2 the dimension of the space of $A_3$-invariant polynomials is bigger (by 1) than the one of $S_3$.
This extra degree 2 invariant cannot come from polarization of $A_3$-invariants in the natural action, as the difference with $S_3$ only starts occurring in degree 3. 
(I also checked for $n=5$ - there the difference starts in degree 6, again too low for polarization to kick in).
A: Maybe the following paper helps:


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*Mark Losik, Peter W. Michor, Vladimir L. Popov: On Polarizations in Invariant Theory , J. Algebra 301 (2006), 406-424.
(pdf)
A: I think the paper 
Gobel, M. (1995). Computing Bases for Permutation-Invariant Polynomials. Journal of Symbolic Computation 19, 285{291 MR 96f:13006 
should be helpful for finding a small set of generators.  It doesn't deal with the polarization aspect of your question though - just the fact that you have a permutation representation.
