Algebraic Independence of symmetric power sum polynomials

Question

Let $P_k(X_1,...,X_n) = X_1^k+...+X_n^k$. My Question is how to proof that the Polynomials $(P_1,...,P_n)$ are algebraically independent. My first try was to imitate the proof of the algebraic independence in of elementary symmetric functions given by Artin, but this doesn't work (because $P_n(X_1,...,X_{n-1},0 ) \not= 0$ for elementary symmetric polynomials this holds).

I saw that in the book "Symmetric Functions" from Macdonald, he uses the Newton identities and argue that we can represent each $P_k$ as polynomial in the elementary symmetric functions and also as Polynomial in the homogenous complete symmetric polynomials and this are both algebraic independent families, which generate the Ring of symmetric functions so $P_k$ does. But to me, this doesn't make any sense (at least not the part for the algebraic independence). Am I right ? And how could I proof this ?

Answer 1

The important point here is not that one can express the power sums as polynomials in the elementary symmetric polynomials, as that can be done for any symmetric polynomials. The important point is that the Newton identities allow each elementary symmetric polynomials to be expressed in terms of the first $n$ power sums. If there were any algebraic relation between the first $n$ power sums, the transcendence degree of the ring they generate (over a base field that can taken to be $\Bbb Q$) would be strictly less than$~n$, but each of the elementary functions lying in that ring and them generating the full ring of symmetric polynomials, this would imply that the first $n$ elementary symmetric polynomials are algebraically dependent, while we know they are not.

Answer 2

1. Yes.

2. To prove this (over $\mathbb Q$ of course, not in characteristic $p>0$, where $P_p=P_1^p$), you can apply the determinant test, which says that a sufficient condition for polynomials $g_1,\dots,g_n\in K[X_1,\dots,X_n]$ to be algebraically independent over a field $K$ of characteristic $0$ is that their Jacobian determinant $\det\left(\partial g_i/\partial X_j\right)$ is nonzero. In our application, this Jacobian determinant is a Vandermonde one times a factorial.

Answer 3

Algebraic independence is equivalent to showing that the monomials $\prod_{i=1}^n p_i^{m_i}$ are linearly independent. This monomial has degree $N = \sum_{i=1}^n m_i$, which is the same as the degree of the corresponding monomial $\prod_{i=1}^n e_i^{m_i}$ in the elementary symmetric polynomials. If we know that the $e_i$ are algebraically independent and generate the ring of symmetric polynomials, then we know that the monomials $\prod_{i=1}^n e_i^{m_i}$ satisfying $\sum m_i = N$ form a basis of the subspace of symmetric polynomials of degree $N$. If we know that the $p_i$ also generate the ring of symmetric polynomials (e.g. via the Newton identities), then we know that the monomials $\prod_{i=1}^n p_i^{m_i}$ satisfying $\sum m_i = N$ span the subspace of symmetric polynomials of degree $N$. But since there are exactly as many of them as the size of a basis, they must be a basis, and hence must be linearly independent.