# Showing that an alternating polynomial is the product of some symmetric polynomial and the Vandermonde polynomial

For simplicity, consider polynomials of two variables. Let $f(x,y)$ be an arbitrary alternating polynomial. I want to show that $f(x, y)$ is the product of some symmetric polynomial and the Vandermonde polynomial. I can show that $f(x, y)$ is divided by the Vandermonde polynomial, $(y - x)$. Let $q(x,y)$ be the quotient. Since $f$ is alternating, I have $q(x, y)(y-x) = -q(y, x)(x-y) = q(y, x)(y-x)$. If I cancelled out $(y-x)$ I would get what I want to, but I think $(y-x)$ doesn't have the inverse in the polynomial ring.

How can I show that an alternating polynomial is the product of some symmetric polynomial and the Vandermonde polynomial?

-
You don't need inverses. Polynomial rings (over integral domains) are integral domains, so you can cancel common factors. (Equivalently, they embed into fields.) –  Qiaochu Yuan May 19 '12 at 15:31
You haven't told us what your underlying ring is. But you can cancel in the integers even though most integers have no inverse in that context ... –  Mark Bennet May 19 '12 at 15:32
I realized I can cancel them out because there are no zero divisors, just after I pushed the `Post your question' button :-( –  Pteromys May 19 '12 at 15:44
If there were an issue, you could always multiply by Vandermonde to give yourself a symmetric polynomial, instead of trying to cancel. –  Mark Bennet May 19 '12 at 22:21