Hello all, I would appreciate comments on the following question:

A main theorem of symmetric functions might be formulated: Let k be a field of char. 0. Then $k[x_1,...,x_n]^{S_n} = k[s_1,...,s_n]$, i.e. symmetric polynomials can be written as polynomials in the elementary symmetric polynomials. Moreover, $s_1,...,s_n$ satisfy no polynomial relations.

I want to see how to prove it using Galois theory. I thus consider the field $M=k(x_1,...,x_n)$ and its subfields $K=k(s_1,...,s_n)$ and $L$, the subfield of symmetric functions. Thus $K \subset L \subset M$. I then consider the polynomial $G(t)=(t-x_1)...(t-x_n)$. It has coefficients in $K$. $M$ is the splitting field of $G$ over $K$. Hence $[M:K] \leq n!$. From this we already see that $s_1,...,s_n$ satisfy no polynomial relations. On the other hand, $M$ has $n!$ different automorphism over $L$, which are permuting the $x_i$. Hence from Galois theory we can conclude that $L=K$.

My question is: How can I deduce the claim $k[x_1,...,x_n]^{S_n} = k[s_1,...,s_n]$ from the corresponding one for rational functions: $L=K$.

Thanks.