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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 ?

user1072285
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3 Answers3

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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.

Marc van Leeuwen
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  • Do I understand it right that the trascendence degree is something like the Dimension in linear algebra, and so the Trascendence Degree of $K\subset K^{sym}$ is n because the n elementary symmetric polynomials form a transcendence Basis. If we now have n polynomials which generate the full ring (like the powersums) they have to be algebraic independent (because otherwise we could eliminate some elements in this set to get a algebraic independent set which still generate the full ring and so the transcendence degree would be <n) ? – user1072285 Sep 20 '22 at 14:16
  • Yes that is rather well formulated. I should say I was a bit sloppy in using the transcendence degree for rings; it is really a notion for fields (or rather field extensions) and should here be applied to the fields of fractions of the rings, as extensions of $\Bbb Q$. If you want a proof that does not use the transcendence degree, see the nice argument by Qiaochu Yuan, which is really what I wanted to say, but I was too lazy to get the details right and went for the transcendence degree. Which is a nice notion to know about anyway. – Marc van Leeuwen Sep 21 '22 at 15:14
  • Actually if I'm not wrong even algebraic independence is just defined for field extensions and so my Question doesn't make sense. It would be better to ask if every symmetric polynomial can be written uniquely as Polynomial in the $P_1,...,P_n$ and if we know that the $P_1,...,P_n \in K(X_1,...,X_n)^{sym}$ (so the field of fractions) are algebraically independent over K it follows the uniqueness. Thank you for your help ! – user1072285 Sep 21 '22 at 16:59
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  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.

Anne Bauval
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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.

Qiaochu Yuan
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  • Could you explain why algebraic independence is equivalent to showing that this family is linearly independent ? – user1072285 Sep 21 '22 at 17:06
  • @user1072285: algebraic independence means showing that there isn't a nontrivial polynomial that vanishes when you plug in the $p_i$. A polynomial is a sum of monomials. So...? – Qiaochu Yuan Sep 21 '22 at 17:37
  • You are right, this easy clear now. I have still some problems to understand the proof, maybe you can help me: If I understand it right you want to show that the family $ (\prod_{i=1}^n p_i(X_1,...,X_n)^{m_i}){m_i\in \mathbb{N}) is linearly independent. So we need to show that every finite subfamily is. But you actually just show that $ (\prod{i=1}^n p_i(X_1,...,X_n)^{m_i})_{\sum m_i = N} $ for some N which I don't understand. The N seems to be fixed, but it can't be because it is the sum of the $m_i$ which are running. – user1072285 Sep 21 '22 at 19:48
  • Homogeneous polynomials of different degree are always linearly independent, so to show that a bunch of homogeneous polynomials are linearly independent it suffices to work degree by degree. That's what this argument does. – Qiaochu Yuan Sep 21 '22 at 19:53
  • ahh thank you very much now I think I understand. – user1072285 Sep 21 '22 at 19:54
  • You're welcome! I think it's worth knowing that this proof can be done in a straightforward way using just basic linear algebra and properties of the degree. – Qiaochu Yuan Sep 21 '22 at 19:56