TITLE: Ring of polynomials as a module over symmetric polynomials
QUESTION [13 upvotes]: Consider the ring of polynomials $\mathbb{k} [x_1, x_2, \ldots , x_n]$ as a module over the ring of symmetric polynomials $\Lambda_{\mathbb{k}}$. Is $\mathbb{k} [x_1, x_2, \ldots , x_n]$ a free $\Lambda_{\mathbb{k}}$-module?
  Can you write down "good" generators explicitly? (I think that it has to be something very classical in representation theory).

Comment:
My initial question was whether this module flat. But since all flat Noetherian modules over polynomial ring are free (correct me if it is wrong), it is the same question.
There is much more general question, which seems unlikely to have good answer. Let $G$ be a finite group and $V$ finite dimensional representation of G. Consider projection $p: V \rightarrow V/G$. When is $p$ flat?

REPLY [2 votes]: Another perspective:
The ring extension $S=k[s_1,\dots,s_n]\subset k[x_1,\dots,x_n]=R$ is finite, and since $R$ is Cohen-Macaulay it is free of rank say $m$. We thus have a Hironaka decomposition $R=\bigoplus_{i=1}^m S\eta_i$ for some homogeneous $\eta_i\in R$. Using this the Hilbert series of $R$ is $$H_R(t)=\sum_{i=1}^mH_S(t)t^{\deg\eta_i}=\sum_{i=1}^mt^{\deg\eta_i}/\prod_{i=1}^n(1-t^i).$$ On the other side, $H_R(t)=1/(1-t)^n$ and therefore $\sum_{i=1}^mt^{\deg\eta_i}=\prod_{i=1}^{n-1}(1+t+\cdots+t^i)$ which for $t=1$ gives $m=n!$.