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Assume that

$$ W(x_1,...,x_n;k)=\left [ \begin{array}{rrrrrrrr} 1 & x_1 &... & x_1^{n-2} & x_1^k \\ 1 & x_2 &... & x_2^{n-2} & x_k \\ & & \ddots \\ 1 & x_n & ... & x_n^{n-2} & x_n^k \\ \end{array} \right ], $$ where $k\geq n-1, k \in \mathbb N$,


$$ V(x_1,...,x_n)=\left [ \begin{array}{rrrrrrrr} 1 & x_1 &... & x_1^{n-1} \\ 1 & x_2 &... & x_2^{n-1} \\ & & \ddots \\ 1 & x_n & ... & x_n^{n-1} \\ \end{array} \right ] =\prod_{1\leq i < j \leq n} (x_j-x_i) $$ ($V(x_1,...,x_n)$ is the Vandermonde's determinant).

How to prove that in the ring $\mathbb R[x_1,...,x_n]$

$$ V(x_1,...,x_n) | W(x_1,...,x_n;k), $$ that is, there is a polynomial $Q(x_1,...,x_n) \in \mathbb R[x_1.,,,.x_n]$ such that $W(x_1,...,x_n;k)=Q(x_1,...,x_k) \cdot V(x_1,..,x_n)$ ?

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Possible Duplicate. – P.. Apr 4 '13 at 9:29
Let's give a proof not using the UFD property of polynomial rings, for a change. What you want to prove is that $a_{\rho}\mid a_{\mu+\rho}$ for $\mu=\left(k-n+1,0,0,0,...,0\right)$ (with $n-1$ zeroes), in the notations of John R. Stembridge, A Concise Proof of the Littlewood-Richardson Rule, Electronic Journal of Combinatorics vol. 9 (2002) paper N5. But this follows from the first Corollary (the Bi-Alternant formula) in that paper, since $s_{\mu}$ is a polynomial (by definition). True, the paper is a fruit of some 50 years of combinatorics, but the UFD proof always felt like a cheat to me. – darij grinberg Apr 16 '13 at 6:30
up vote 3 down vote accepted

The polynomial $W(x_1,\ldots,x_n;k)$ vanishes whenever one substitutes $x_j:=x_i$ for some pair of indices $i<j$. This means it is divisible by $x_j-x_i$ for every such pair. (To see this, consider $\Bbb R[x_1,\ldots,x_n]$ as $R[x_j]$ where $R=\Bbb R[x_1,\ldots,\widehat{x_j},\ldots,x_n]$; then $x_j-x_i$ is a monic polynomial, by which we can perform Euclidean division, and the remainder of any polynomial $P$ after this division is just the evaluation $P[x_j:=x_i]$ of $P$ at $x_i\in R$, which in our case equals $0$.)

Also all these linear polynomials $x_j-x_i$ are clearly pairwise relatively prime (in fact individually irreducible) in the Unique Factorization Domain $\Bbb R[x_1,\ldots,x_n]$, so the least common multiple is their product, which is $V(x_1,\ldots,x_n)$. Since $W(x_1,\ldots,x_n;k)$ is divisible by each of the factors, it is divisible by this least common multiple.

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