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Let $s_i$ be the symmetric elementary functions. For example, $s_1=x_1+\cdots+x_n$.

Suppose a polynomial $p(z_1,\ldots,z_n)\in R[z_1,\ldots,z_n]$ satisfies $p(s_1,\ldots,s_n)=0$ in $R[x_1,\ldots,x_n]$. How does one prove that $p(z_1,\ldots,z_n)=0$ in $R[z_1,\ldots,z_n]$. I could not find a proof of this that did not gloss over parts of some induction.

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As to the claim in your title, this looks promising: en.wikipedia.org/wiki/Elementary_symmetric_polynomial –  Will Jagy Jan 4 '12 at 5:20
    
So you would like to see a proof without using induction on $n$? –  Paul Jan 4 '12 at 5:20
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in Galois Theory by Joseph Rotman, appendix D, he refers to Hadlock, Field Theory and its Classical Problems. He also says that Waring published an algorithm for finding the representation in 1770. –  Will Jagy Jan 4 '12 at 5:29

1 Answer 1

up vote 2 down vote accepted

I'll suppose $R$ is an integral domain, and therefore contained in an algebraically closed field $K$ (one can do without this assumption, but it simplifies the argument). Suppose $p(s_1,\ldots,s_n)=0$ but $p$ is not the zero polynomial; then certainly there exist constants $a_1,\ldots,a_n\in K$ such that $p(a_1,\ldots,a_n)\neq0$. Now consider $$ Q=X^n+a_1X^{n-1}+a_2X^{n-2}+\cdots+a_{n-1}X+a_n\in K[X]. $$ Since $K$ is algebraically closed, this polynomial splits $Q=\prod_{i=1}^n(X-r_i)$ for roots $r_1,\ldots,r_n\in K$. But that means that if $f:K[x_1,\ldots,x_n]\to K$ is the ring morphism of substituting $-r_j$ for $x_j$, for $j=1,\ldots,n$, then $a_i=f(s_i)$ for $i=1,\ldots,n$. Now $$ 0=f(0)=f(p(s_1,\ldots,s_n))=p(f(s_1),\ldots,f(s_n))=p(a_1,\ldots,a_n), $$ a contradiction. The before-last equality is simply the fact that a ring morphism $f$ commutes with the ring operations used in forming a polynomial expression $p$.

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