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May 10, 2020 at 13:45 comment added Noam D. Elkies Good question; I expect it works, but I don't have a proof (except when char$(F) > k$). NB It's not quite the same question as the one about the function fields. On the one hand, inseparable proper extensions can be OK (e.g. in characteristic $2$, use $p_2 = p_1^2$ istead of $p_1$). On the other, the function-field condition only promises that $\{p_{m_i}(A) : 1 \leq i \leq k \}$ characterize $A$ generically, and there might be exceptional loci. For example, $p_1,p_2,p_4$ characterize three-element subsets of $\bf C$, except those with $p_1 = 0$ (for which $p_4 = p_2^2/2$).
May 10, 2020 at 10:35 comment added Sean Eberhard Does this work over arbitrary characteristic, if I replace $1, 3, \dots, 2k-1$ with the first $k$ positive integers $m_1, \dots, m_k$ not divisible by $\text{char}\,F$? In other words, the question is whether $F(p_{m_1}, \dots, p_{m_k})$ is equal to the field of symmetric rational functions over $F$. I have checked this in a few small cases from Newton's identities. (I wonder what general conditions on $m_1, \dots, m_k$ suffice for this.)
May 10, 2020 at 1:05 history answered Noam D. Elkies CC BY-SA 4.0