Analog of Newton's theorem for symmetric polynomials Newton's theorem of symmetric polynomials says that every symmetric polynomial can be written as a polynomial in elementary symmetric polynomials. Hence when $S_n$ acts on $\mathbb{Q}(x_1,...,x_n)$ naturally then the fixed field under this action is $\mathbb{Q}(e_1,...,e_n)$ where $e_1,...,e_n$ are elementary symmetric polynomials. Let $C_p$(realised as a permutation group generated by a $p$-cycle), where $p$ is a prime act on $\mathbb{Q}(x_1,...,x_p)$ what can be said about the fixed field under this action? Is there an analog of Newton's theorem for this?
Thanks
 A: Let me pretend that when you said $\mathbb{Q}$ you actually meant $\mathbb{C}$. You did mean $\mathbb{C}$, right? In that case, fix a primitive $n^{th}$ root of unity $\zeta_n$. Let me rename your variables to $x_0, x_1, ... x_{n-1}$, then ignore them to introduce a different set of variables
$$y_i = \zeta_p^0 x_0 + \zeta_p^i x_1 + \zeta_p^{2i} x_2 + ... + \zeta_p^{(n-1)i} x_{n-1}, 0 \le i \le n-1.$$
Then a generator of $C_n$ sends $y_i$ to $\zeta_p^i y_i$. (This is essentially the discrete Fourier transform.) Consequently, a generator of $C_n$ acts on a monomial by
$$\prod y_i^{m_i} \mapsto \prod \zeta_p^{i m_i} \prod y_i^{m_i}.$$
Thus it fixes precisely those monomials which satisfy $\sum i m_i \equiv 0 \bmod n$. Linear combinations of these monomials precisely describe the polynomials in $\mathbb{C}[x_0, ... x_{n-1}]$ invariant under $C_n$, and quotients of these polynomials precisely describe the invariant subfield. (There is no need for the additional hypothesis that $n$ is prime.) 
Example. When $n = 3$, we look at monomials in $y_0, y_1, y_2$. The first few invariant monomials are $y_0, y_1 y_2, y_1^3$ and products of these and their inverses give all invariant monomials (exercise), so these generate the invariant subfield. 
Example. When $n = 4$, we look at monomials in $y_0, y_1, y_2, y_3$. The first few invariant monomials are $y_0, y_1 y_3, y_2^2, y_1^2 y_2$ and products of these and their inverses give all invariant monomials (exercise), so these generate the invariant subfield. 
(The answer over $\mathbb{Q}$ seems more complicated to me. If I finish writing it up, it will be in a separate answer.) 
