Let $\text{char}\,k = 0$ and $n \ge 2$. What is the easiest way to see that $k[x\_1, \dots, x\_n]$ is a free $k[x\_1, \dots, x\_n]^{S\_n}$-module with basis$$x\_2^{m\_2}x\_3^{m\_3} \dots x\_{n-1}^{m\_{n-1}} x\_n^{m\_n},\text{ }m\_2 \in [0, 1],\text{ }m\_3 \in [0, 2], \dots,\text{ }m\_n \in [0, n-1]?$$



Free $k[x_1, \dots, x_n]^{S_n}$-module?
It is enough to show that the given generators span $k[x\_1, \ldots, x\_n]$ as a $k[x\_1, \ldots, x\_n]^{S\_n}$ module. Once we've shown this, it is easy to see that $k(x\_1, \ldots, x\_n)$ is dimension $n!$ as a $k(x\_1, \ldots, x\_n)^{S\_n}$ vector space, so $n!$ vector which span must be linearly independent.


**Notation:** For every $r \in \left\{0,1,\ldots,n\right\}$, let $e^r\_k$ be the $k$-th elementary symmetric polynomial in $x\_1$, ..., $x\_r$. Let $R = k[x\_1, \ldots, x\_n]$ and let $A = R^{S\_n}$.


**Lemma 1** The polynomial $e^r\_k$ is in the ring $A[x\_{r+1}, \ldots, x\_n]$. 


**Proof** Induction on $n-r$. The base case, $r=n$, is that $e\_k(x\_1, \ldots, x\_n) \in A$; this is true. Now, $e\_k^r = e\_k^{r+1} - x\_{r+1} e\_{k-1}^{r
+1}$, so the result follows by induction. $\square$.


**Lemma 2** The monomial $x\_r^r$ is in the $A[x\_{r+1}, \ldots, x\_n]$-linear span of monomials of the form $x\_r^s$ with $0 \leq s < r$.


**Proof** Expand $(x\_r-x\_1) (x\_r-x\_2) \cdots (x\_r-x\_r) = 0$ to get $x\_r^r - e^r\_1 x\_r^{r-1} + e^r\_2 x\_r^{r-2} - \cdots = 0$ or, in other words, $x\_r^r = \sum\_{s=0}^{r-1} (-1)^{r-s+1} x\_r^s e^r\_{r-s}$. Now apply Lemma 1. $\square$


We now prove the following result by induction on $r$: 



> 
> The ring $R$ is spanned as an $A[x\_{r+1}, \ldots, x\_n]$-module by monomials of the form $x\_1^{d\_1} x\_2^{d\_2} \cdots x\_r^{d\_r}$ with $0 \leq d\_j < j$.
> 
> 
> 


The base case $r=0$ is trivial (it says $R = A \cdot 1$); the case $r=n$ is the claim.


We now do the inductive step. Since $R$ is spanned as an $A[x\_{r}, \ldots, x\_n]$-module by monomials of the form $x\_1^{d\_1} x\_2^{d\_2} \cdots x\_{r-1}^{d\_{r-1}}$ with $0 \leq d\_j < j$, we know that $R$ is spanned as an $A[x\_{r+1}, \ldots, x\_n]$-module by monomials of the form $x\_1^{d\_1} x\_2^{d\_2} \cdots x\_{r-1}^{d\_{r-1}} x\_r^t$, with $0 \leq d\_j < j$, and no bound on $t$. But Lemma 2 allows us to replace $x\_r^t$ by lower powers of $x\_r$ if $t \geq r$. **QED**


