Symmetric polynomials theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T07:04:09Z http://mathoverflow.net/feeds/question/30555 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/30555/symmetric-polynomials-theorem Symmetric polynomials theorem Sasha 2010-07-04T20:55:49Z 2010-07-04T21:12:23Z <p>Hello all, I would appreciate comments on the following question:</p> <p>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.</p> <p>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 $L=K$.</p> <p>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: $L=K$.</p> <p>Thanks.</p> http://mathoverflow.net/questions/30555/symmetric-polynomials-theorem/30558#30558 Answer by senti_today for Symmetric polynomials theorem senti_today 2010-07-04T21:12:23Z 2010-07-04T21:12:23Z <p>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 $A=B$.</p>