Corollary

— Let $\xi_1, \ldots, \xi_n, \eta_1, \ldots, \eta_n$ be elements of $\mathbf{A}$ and suppose that $\mathbf{A}$ is an integral domain.

a) If $s_k(\xi_1, \ldots, \xi_n) = s_k(\eta_1, \ldots, \eta_n)$ for $1 \leq k \leq n$, then there exists a permutation $\sigma \in \mathfrak{S}_n$ such that $\eta_i = \xi_{\sigma(i)}$ for $1 \leq i \leq n$.

b) Suppose that $n! . 1 \neq 0$ in $\mathbf{A}$ and

for $1 \leq k \leq n$. Then there exists a permutation $\sigma \in \mathfrak{S}_n$ such that $\eta_i = \xi_{\sigma(i)}$ for $1 \leq i \leq n$.

Under the hypotheses a) we have $\prod_{i=1}^n (X - \xi_i) = \prod_{i=1}^n (X - \eta_i)$. If we substitute $\eta_n$ for $X$, we find that $\prod_{i=1}^n (\eta_n - \xi_i) = 0$ and since $\mathbf{A}$ is an integral domain, there is an integer $\sigma(n)$ such that $l \ s \ \sigma(n) \leq n$ and $\eta_n = \xi_{\sigma(n)}$. Now the assertion a) follows easily by induction because $\mathbf{A}[X]$ is an integral domain.

Under the hypotheses of b) there exist by Prop. 3 polynomials $\Pi_1, \ldots, \Pi_n$ in $n$ indeterminates, with coefficients in the field of fractions of $\mathbf{A}$ such that $s_k = \Pi_k(p_1, \ldots, p_n)$ for $1 \leq k \leq n$. Now the relation (23) implies

for $1 \leq k \leq n$ and so b) follows from a).