# 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

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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.)

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Okay, so here is a short but unsatisfying answer over $\mathbb{Q}$: write out the invariant monomials as polynomials in the $x_i$. These will be linear combinations of invariant polynomials in $\mathbb{Q}(x_0, ... x_{n-1})$ with coefficients in $\{ 1, \zeta_n, ... \zeta_n^{n-1} \}$ and taking all of the corresponding invariant polynomials gets you generators of the invariant polynomials over $\mathbb{Q}$, but there are relations between them... – Qiaochu Yuan Aug 23 '12 at 22:42
Qiaochu Yuan Thanks for your answer. I really meant $\mathbb{Q}$ not $\mathbb{C}$ , though it is interesting to note how it really matters and I am wondering how it does not matter in the case $S_{n}$. The motivation behind this question is Noether's formulation of inverse Galois problem( see 'Introduction' in Serre's Topics in Galois thory). It says that when a permutation group $G$ acts on $\mathbb{Q}(x_1,..x_n)$ and if its fixed field is rational over $\mathbb{Q}$ then via Hilbert's irreducibility theorem we can realize $G$ over $\mathbb{Q}$. – Dinesh Aug 24 '12 at 9:41
..contd. In fact it gives an explicit polynomial for $G$ (when we have enough information). So if I know the analog of newton's theorem for $C_p$ then I can realize $C_p$ and might also be able to produce an explicit polynomial for it. The reason I chose $n$ to be prime is that, Lenstra showed that the fixed field of $C_8$ is not rational over $\mathbb{Q}$ (Reference: Serre's Topics in Galois theory). – Dinesh Aug 24 '12 at 9:43
@Dinesh: it doesn't matter for $S_n$ because all of the irreducible representations of $S_n$ are defined over $\mathbb{Q}$. What do you mean by rational over $\mathbb{Q}$? – Qiaochu Yuan Aug 24 '12 at 15:30
By a field rational over $\mathbb{Q}$ I mean a field of the form $\mathbb{Q}(t_1,...,t_n)$ where $t_1,...,t_n$ are algebraically independent. – Dinesh Aug 25 '12 at 3:32