# Symmetric polynomials theorem

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.

• How do we find that the $s_j$ satisfy no polynomial relations from the fact that the Galois group is small? Apr 20 '16 at 13:21

## 1 Answer

I hope the following works. Let $A=k[x_1,\ldots,x_n]^{S_n}$, and let $B=k[s_1,\ldots,s_n]$. The polynomial algebra $k[x_1,\ldots,x_n]$ is an integral extension of $B$, and hence, a fortiori, $A$ is integral over $B$. I think your argument proves that $A$ and $B$ have the same fraction field. However, since $B$ is (isomorphic to) a polynomial algebra, it must be integrally closed, whence $A=B$.

• The theorem that you are trying to reprove is true if $k$ is any commutative ring. Jul 4 '10 at 21:36
• I was not trying to reprove any theorems, and in the statement of the question $k$ is a field. Jul 5 '10 at 1:22
• Thank you! It is exactly the kind of argument I was looking for. Jul 5 '10 at 8:27