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Let $f(x_1,...,x_n) \in \mathbb{Z}[x_1,...,x_n]$ be homogeneous of degree $d>n(n-1)$, i.e. $f$ is the sum of monomials of degree $d$. I am looking for a hint to prove that $f$ is in the ideal generated by the elementary symmetric polynomials $s_1,...,s_n$.

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    Hint 1: The actual bound is much sharper than that: If $d > n(n-1)/2$ then you are already in the ideal. Hint 2: One way to do this problem is to prove that the Hilbert series of $\mathbb{Z}[x_1, \ldots, x_n] / \langle s_1, \ldots, s_n \rangle$ is $\prod_{i=1}^n (1-q^i)/(1-q)^n$.2011-11-23
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    Are Hilbert series enough over $\mathbb Z$? Also, with that Hilbert series, shouldn't it be $n(n+1)/2$?2011-11-23
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    The problem is actually trivial (note that every monomial of degree more than $n\left(n-1\right)$ must have at least one exponent $\geq n$; now use Viete). I am wondering how to prove David's version, though.2011-11-23
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    OK, now I can prove the $n\left(n-1\right)/2$ version using an algorithm which, in each step, replaces a monomial by a list of monomials it (strictly) majorizes (where we identify monomials with their exponent sequences, and "majorize" means majorization as in majorization theory, also called the domination order) such that the original monomial is equivalent to the negative of the sum of these new monomials modulo the ideal generated by the $s_1,s_2,...,s_n$. I hope this is written up somewhere, since I don't have the time to do this...2011-11-23
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    Thanks for making this question an interesting one, David!2011-11-23
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    Note on the algorithm: It only works if our monomial has a "gap", i. e., if it satisfies the following property: If $e_1,e_2,...,e_n$ are the exponents of this monomial *in ascending order*, then some $i$ satisfies $e_i > e_{i+1}+1$. For this condition to be satisfied for every monomial of degree $d$, we need $d$ to be greater than $n\left(n-1\right)/2$.2011-11-23
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    @darij: In your "trivial" solution, when you say use Viete, what exactly do you mean?2011-11-23
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    @darij: and also: consider two monomials $\alpha$ and $\beta$ with $\alpha$ strictly majorizing $\beta$. Then this implies that all variables $x_i$ appear in $\beta$ with non-zero exponents. Correct?2011-11-23
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    Nope, Manos, it does not. Majorization in my sense is this: http://en.wikipedia.org/wiki/Majorization .2011-11-23
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    @Viete: I mean $x_i^n = s_1 x_i^{n-1} - s_2 x_i^{n-2} + s_3 x_i^{n-3} \pm ... \pm s_n$.2011-11-23
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    @darij: got it :-)2011-11-23
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    @darijgrinberg Please consider converting your comments into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-26

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