So we have the situation that $f\in K[X_1,...,X_n]$ is anti-symmetric, which means that $\sigma (f)=\pm f$ where it is a plus if $\sigma$ is an even permutation on the $X_i$ and a minus if it is not an even permutation. Now I have to prove that there exists a $g$ which is symmetric such that $f=g\prod_{1\leq i < j\leq n}(X_i-X_j)$. This is what I thought I should do to prove the statement: If we have permutation $\sigma=(12)$ ,which sends $X_1\mapsto X_2$ and $X_2\mapsto X_1$, then $\sigma(f)=-f$. This implies that $X_1-X_2$ divides $f$ (this is the statement which I am not so sure of). Now we can do this for every permutation $\sigma=(ij)$ where $i\neq j$. Then we have that $\prod_{1\leq i < j\leq n}(X_i-X_j)$ divides $f$. It follows immediately that $g$ must be symmetric.

Is this proof valid? Thanks for looking at it.


1 Answer 1


As I understand, the only question is whether the following statement is true: if $\sigma = (12)$ and $\sigma(f) = -f$ then $X_1 - X_2 | f$.

If $K$ is of chracteristic $2$, then anti-symmetric polynomials are symmetric polynomials. The original statement is false. For example, $X_1 + X_2 + X_3 = - (X_1 + X_2 + X_3)$ is not divided by $X_1 + X_2 = X_1 - X_2$.

Assume that $K$ is not of characteristic $2$. Given $f \in K[X_1,\ldots,X_n]$ anti-symmetric, let $R = K[X_2,\ldots,X_n]$, then $f \in R[X_1]$. In $R[X_1]$, evaluate both $f$ and $\sigma(f) = -f$ at $X_2$ and we get $f(X_2) = -f(X_2)$. So $f(X_2) = 0$, meaning that $X_2$ is a root of $f \in R[X_1]$. So $X_1 - X_2 | f$.

  • $\begingroup$ Thanks, this was the argument I was looking for! $\endgroup$
    – Badshah
    Commented May 23, 2014 at 7:36
  • $\begingroup$ For future visitors: note the end of this uses the Factor Theorem over integral domains for monic polynomials math.stackexchange.com/questions/1559384/… $\endgroup$
    – D.R.
    Commented Sep 8, 2023 at 22:01

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