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 ? 



Polarizations generate the ring of invariants?
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).


