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In Nagata's book on field theory it is mentioned that the Vandermonde polynomial is not a zero-divisor in the polynomial ring $R[X_1,...,X_n]$ over a ring $R$ in which 2 is not a zero-divisor. I tried using the fact that the Vandermonde matrix has trivial kernel iff the Vandermonde polynomial is not a zero divisor (which follows from the observation in this post: necessary and sufficient condition for trivial kernel of a matrix over a commutative ring) but I wasn't able to get anywhere. How do I prove this fact?

millskum
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  • What is your definition of "the Vandermonde polynomial"? – Leobeth Apr 08 '25 at 09:24
  • @Leobeth I mean $\prod_{1\leq i<j\leq n}(X_i-X_j)$ – millskum Apr 08 '25 at 12:54
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    But that polynomial has coefficients $\pm 1$, so it should be a non-zerodivisor over any ring, no? A (nonzero) polynomial is a zero-divisor if and only if it is killed by a nonzero element from the base ring. – Leobeth Apr 08 '25 at 13:27

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