Let $F$ be a field of characteristic $\neq 2$ and $E = F(x_1,\ldots,x_n)$, i.e. the fraction field of the polynomial ring $F[x_1,\ldots,x_n]$. We know that $G = \text{Gal} \left( E/F(p_1,\ldots,p_n) \right)$ is $S_n$, where the $p_i$ are the elementary symmetric polynomials. Let $H$ be the subgroup of $G$ corresponding to $A_n$ - explicitly $H$ is the set of all automorphisms $f(x_1,\ldots, x_n) \mapsto f(x_{\pi(1)}, \ldots, x_{\pi(n)})$ such that $\pi \in A_n$.

How to prove that the fixed field of $H$ is $F(p_1,\ldots,p_n,\Delta)$, where $\Delta = \prod_{i<j} (x_i - x_j)$?

I suppose I should prove that $f$ is either symmetric or anti-symmetric. I know how to finish it if I can prove that $f$ is anti-symmetric e.g.

What I found is that all the elements of $S_n \setminus A_n$ act on $f$ in the same way, i.e. $f(x_{\sigma(1)}, \ldots, x_{\sigma(n)}) = f(x_{\pi(1)}, \ldots, x_{\pi(n)})$ for any two $\sigma, \pi \in S_n \setminus A_n$. It remains to show that this constant value must be $\pm f(x_1,\ldots, x_n)$, if I am not mistaken.

I couldn't come up with something else and I'm stuck.

  • $\begingroup$ How does a single transposition act on the product? That should be enough to show an even number of them fix the product. $\endgroup$
    – Steve D
    Aug 20, 2017 at 16:36

1 Answer 1


The fixed field of $S_n$ is $F(p_1,\ldots,p_n)$.

By the Fundamental theorem is Galois Theory the fixed field of $A_n$ is a quadratic extension of $F(p_1,\ldots,p_n)$. So adjoining any element fixed by $A_n$, but not $S_n$ to $F(p_1,\ldots,p_n)$ will do. Clearly $\Delta$ is such an element (as the characteristic is not $2$).


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