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. 



Alternating elements in free graded-commutative algebras
If you assume $\ell>n$, then you may as well work over the field $F=\mathbb{Q}$, since the representation theory is exactly the same (the group algebra is semisimple, all Young symmetrisers in the group algebra over $\mathbb{Q}$ have denominators dividing $n!$, etc.). 


Under this (much) simplifying assumption, we have $F[x\_1,\ldots,x\_n]\cong FS\_n\otimes F[x\_1,\ldots,x\_n]^{S\_n}$, and the exterior algebra $\Lambda(a\_1,\ldots,a\_n)$ can be written as $\Lambda(V\oplus F)$, where $V$ is the standard $(n-1)$-dimensional representation of $S\_n$, and $F$ is the trivial one, and that is the same as $\Lambda(V)\otimes\Lambda(F)$, where $\Lambda(F)\cong \Lambda(a\_1,\ldots,a\_n)^{S\_n}$. Now, we see that your ring is, as a representation of $S\_n$, 
 $$
FS\_n\otimes\Lambda(V)\otimes F[x\_1,\ldots,x\_n]^{S\_n}\otimes \Lambda(a\_1,\ldots,a\_n)^{S\_n} . 
 $$
To extract the isotypic component of the sign representation, you should take that component in $FS\_n\otimes\Lambda(V)$, and tensor with the invariants. This already gives freeness as a module over the symmetric polynomials, of course. But also, note that $FS\_n$ is the sum of all irreducible modules with multiplicities equal to dimensions, and $\Lambda(V)$ is the sum of all irreps corresponding to hooks with all multiplicities equal to one. Also, the sign representation multiplicity in $U\otimes V$ is the trivial representation multiplicity in $U\otimes V\otimes\mathrm{sign}$ is the dimension of $\mathop{\mathrm{Hom}}(U,V\otimes\mathrm{sign})$ because of self-duality of representations of symmetric groups. Finally, tensoring with sign replaces each Young diagram by its transpose, which does not change either $FS\_n$ or $\Lambda(V)$ as a representation. Thus, the corresponding multiplicity is equal to the sum of all dimensions of irreps corresponding to hooks; those dimensions are $\binom{n-1}{k}$ for various $k$, so the multiplicity is $2^{n-1}$. Multiplying that by $2=\dim\Lambda(a\_1,\ldots,a\_n)^{S\_n}$, we finally see that the module is free of rank $2^n$. 


I am not sure what to expect in the case $\ell\le n$: most of the statements I made above would fail for rather trivial reasons.


