Let $G$ be a reflection group acting on $\mathbb{K}^n$ (where $\mathbb{K}$ is a field of characteristic zero). Then $G$ induces an action on the ring of polynomials $\mathbb{K}[x_1, \ldots, x_n]$ by acting linearly on the $x_i$. Chevalley's theorem says that there are $n$ polynomials $f_1, \ldots, f_n$ which are invariant under the action of $G$ such that the space of all $G$ invariant polynomials can be written as $$\mathbb{K}[x_1, \ldots, x_n]^G = \mathbb{K}[f_1, \ldots, f_n].$$ A further version of this theorem (see Humphreys "Reflection Groups and Coxeter Groups", Proposition 3.13) says that the alternating polynomials under the $G$-action can be written as $$Alt_G[x_1, \ldots, x_n] = \mathbb{K}[f_1, \ldots, f_n] \cdot J$$ where $J$ is the Jacobian determinant $J = det ( \frac{\partial f_i}{\partial x_j} )_{1 \leq i,j \leq n}$. That is, a polynomial $f$ is alternating iff it can be written as a product of an invariant polynomial and $J$.
I'm interested in the case $G = S_n$. Now simply plugging $S_n$ in for $G$ one can calculate $J$ to be the Vandermonde determinant $\prod_{1 \leq i < j \leq n}{x_j - x_i}$ and the $f_i$ can be chosen to be either the elementary symmetric polynomials or for example $f_i = x_1^{i+1} + \cdots + x_n^{i+1}$ (which makes the calculation of $J$ easier).
The symmetric group $S_n$ however, also acts on $\mathbb{K}^{n-1}$ (since the action of $S_n$ on the subspace generated by $(1, \ldots, 1)$ is trivial). The space of invariant polynomials under this action can then be expressed as $$\mathbb{K}[f_2, \ldots, f_n]$$ where $f_2, \ldots, f_n$ are the $2^{\text{nd}}$ to $n^{\text{th}}$ elementary symmetric polynomials in $n$ variables. But how can I calculate the alternating polynomials under this action?
I mean, one can't just plug in $f_2, \ldots, f_n$ in the formula for $J$ cause these are $n-1$ polynomials in $n$ variables ($\rightarrow$ non-square matrix).
Suggestions: There are two ways to proceed that I can think of but I'm not sure which one leads to the right answer.
Just take $J$ to be the Vandermonde determinant as above and write the alternating polynomials under the $S_n$ action on $\mathbb{K}^{n-1}$ as $$\mathbb{K}[f_2, \ldots, f_n] \cdot J.$$ I have no idea if this is true or not and why...
Proceed as Humphreys does in his Example 3.12: The action of $S_n$ on $\mathbb{K}^{n-1}$ induces an action on $\mathbb{K}[x_1, \ldots, x_n]$ by permuting $x_1, \ldots, x_n$ subject to the relation $x_n = -(x_1 + \cdots + x_n)$.He then uses the $n-1$ polynomials $$g_i := x_1^{i+1} + \cdots + x_n^{i+1} \hspace{2mm} (1 \leq i \leq n-1)$$ and calculates the corresponding determinant $$J = det ( \frac{\partial g_i}{\partial x_j} )_{1 \leq i,j \leq n-1} = (n+1)! \prod_{1 \leq i,j \leq n-1}(x_j - x_i) \prod_{i=1}^{n-1}(x_i+z)$$ where $z := x_1 + \cdots + x_{n-1}$. So far, so good, but what do I get in the end? Are the alternating polynomials something like $$\mathbb{K}[g_1(x_1, \ldots, x_{n-1}, -z), \ldots, g_{n-1}(x_1, \ldots, x_{n-1}, -z)] \cdot J \hspace{1mm}?$$
What is going on is that you restrict the reflection representation of $S_n$ on $K^n$ to the hyperplane given by $x_1+\cdots+x_n=0$. This hyperplane is stable under all reflections of the representation, and not itself a reflection hyperplane. It is then sufficient to apply the projection of $K[x_1,\ldots,x_n]$ to $K[x_1,\ldots,x_n]/I$ where $I$ is the ideal generated by $x_1+\cdots+x_n$. That generator was the first elementary symmetric function$~f_1$, so it simple disappears in the quotient; all other things are replaced by their image in the quotient (which quotient is clearly isomorphic to $K[x_1,\ldots,x_{n-1}]$).