It is classical that every alternating polynomial is (uniquely) the product of a symmetric polynomial with the Vandermonde polynomial, in particular the alternating polynomials are a free rank-one module over the symmetric polynomials.

For some cohomology computations I am doing, I would like to know some analogous statement, not for the polynomial ring, but for the free graded-commutative algebra $$\mathfrak{F}(n)=\mathbb{F}_\ell[X_1,\dots,X_n]\langle A_1,\dots,A_n\rangle,$$ i.e., polynomial variables $X_i$ and exterior variables $A_j$. For peace of mind, $\ell$ is a prime different from $2$, but I would be willing to assume $\ell> n$. Now consider the "diagonal permutation" action of the symmetric group $\Sigma_n$, i.e., an element $\sigma$ acts simultaneously on $\{X_1,\dots,X_n\}$ and $\{A_1,\dots,A_n\}$ via the natural permutation of indices. I am interested in the submodule of $\mathfrak{F}(n)$ where $\Sigma_n$ acts via the sign representation (may I call the elements alternating superpolynomials?).

I would like to know that this submodule is free over the ring of symmetric polynomials in $\{X_1,\dots,X_n\}$, and I would like to have a general method of computing the rank and explicit generators.

So far, the case $n= 2$ is clear and I also worked out the case $n=3$. This is based on first working out the $\Sigma_3$ decomposition of the polynomial ring and the exterior algebra separately and then tensoring the representations (using the assumption $\ell\neq 2,3$). The result is a free rank 8 module. This could of course be done in further cases, but will probably fail to produce a general result because it involves tensoring representations of the symmetric group. So the parts of my question are:

Has this question already been considered somewhere in the literature? (I would guess, it seems natural enough).

Is there a slick general argument dealing with all $n$ at once (and possibly with $\ell\leq n$ as well), like in the classical case of alternating polynomials?

Any ideas or literature references would be much appreciated.