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I am looking for a proof of the statement:

Every antisymmetric (alternating) polynomial is divisible by Vandermonde product, yielding a symmetric polynomial in the result.

The statement is really helpful in constructing further Schur polynomials theory.

I came across it reading David M. Bressoud "Exploiting Symmetries: Alternating-Sign Matrices and the Weyl Character Formulas" and can be seen here in the context http://link.springer.com/chapter/10.1007%2F978-0-387-78510-3_3#page-1

According to the author, the orginal statement has been proven by Cauchy in "Mémoire sur les fonctions qui ne peuvent obtenir que deux valeurs égales et de signes contraires par suite des transpositions opérées entre les variables qu'elles renferment", which can be viewed here: http://math-doc.ujf-grenoble.fr/cgi-bin/oeitem?id=OE_CAUCHY_2_1_91_0

Unfortunately it is only available in French, which makes it impossible for me to digest.

I really appreciate your help!

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  • $\begingroup$ I'll be using the notations of §2.6 of Vic Reiner's "Hopf algebras in combinatorics" ( math.umn.edu/~reiner/Classes/HopfComb.pdf ). Notice first that every alternating polynomial is a linear combination of $a_{\lambda+\rho}$'s, where $\lambda$ ranges over all partitions. Therefore, you only need to prove that the $a_{\lambda+\rho}$'s are divisible by the Vandermonde product (which is $a_{\rho}$). This follows either from Reiner's Corollary 2.37, or from his Proposition 2.35 ... $\endgroup$ Aug 10, 2013 at 13:20
  • $\begingroup$ ... (notice that Proposition 2.35 is only stated for $\mathbf{k}$ being a field or $\mathbb Z$, so we couldn't have applied it right away to any alternating polynomial over any commutative ring; but now that we have reduced the situation to the $a_{\lambda+\rho}$'s, of course we only need to apply it to $\mathbf k = \mathbb Z$). Sorry for brevity. $\endgroup$ Aug 10, 2013 at 13:21
  • $\begingroup$ PS. If Reiner's notes are too long for you, you can also read Stembridge's paper on the Littlewood-Richardson rule ( combinatorics.org/ojs/index.php/eljc/article/view/v9i1n5 ) which is nice and self-contained. It uses the same notations, and Reiner's Corollary 2.37 is called "The Bi-Alternant Formula" in this paper. $\endgroup$ Aug 10, 2013 at 13:23
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    $\begingroup$ An alternating polynomial becomes identically 0 when any two variables are set equal (ignoring characteristic 2). Isn't that sufficient already to prove that the Vandermonde product divides it? $\endgroup$ Aug 10, 2013 at 13:48
  • $\begingroup$ Only over a UFD. (Otherwise one can still reduce to the case of an UFD by writing it as a linear combination of the $a_{\lambda+\rho}$'s.) But that's the argument using Proposition 2.35 I've suggested.) $\endgroup$ Aug 10, 2013 at 13:55

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