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This seems to be part of folklore, but I can't find a proof anyqhere, only references to "a very nice proof of a special case by Arkady Berenstein in a seminar I attended in 1989".

Okay, here it goes. We fix $n \in \mathbb N$ and $\Lambda = \mathbb C[x_1, \ldots, x_n]$. The symmetric group $S_n$ acts on $\Lambda$ in the obvious way, and the algebra of invariant polynomials $\Lambda^{S_n}$ is generated by $p_i = x_1^i + x_2^i + \cdots x_n^i$ for $1 \leq i \leq n$ as an algebra over $\mathbb C$. Now let $I$ be the ideal of $\Lambda$ generated by the $p_i$'s; by the Chevalley Shephard Todd theorem $\Lambda / I$ is an $S_n$-module isomorphic to the regular representation.

Now fix $v \in \mathbb C^n$ and define an automorphism $\phi_v$ of $\Lambda$ by setting $\phi_v(x_i) = x_i + v_i$. If $f$ is any symmetric polynomial then $\phi_v(f)$ may no longer be symmetric. However, if $g \in S_n$ is such that $g(v) = v$ then $g \phi_v(f) = \phi_v(f)$, and the same is true for its image modulo $I$. Hence there is an induced map $\phi_v: \Lambda^{S_n} \to (\Lambda/I)^H$, where $H$ is the stabilizer of $v$ in $S_n$. In all cases I have checked, this map is surjective, but I have not found a proof for the general case. Does anyone knnow where I can find this result, or give a simple proof?

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