Let $V$ be a vector space over $k$ of dimension $m$. (I'm only interested in the case $k=\mathbb{Q}$.) Let $R:=\Lambda^*V$ be the exterior algebra. It carries the structure of a supercommutative ring: $R=R^+\oplus R^{-}$, where both the odd and the even part have $k$-dimension $2^{m-1}$. We define a multiplication on $R^{\otimes n}$ by $(\alpha_1\otimes\cdots\otimes\alpha_n)\cdot (\beta_1\otimes\cdots\otimes\beta_n) =\pm (\alpha_1\beta_1\otimes\cdots\alpha_n\beta_n)$, where the sign comes from reordering factors of odd degree. The symmetric group $\mathfrak{S}_n$ acts on $R^{\otimes n}$ by permuting the factors such that a transposition of two factors of odd degrees yields a negative sign.

The supersymmetric power $S^nR$ is defined as the quotient of $R^{\otimes n}$ by this action. Over $k=\mathbb{Q}$ this is the same as taking the subspace of invariants. The canonical inclusion of $V$ in $S^nR$ gives $S^nR$ the structure of a $R$-algebra.

Now I conjecture that $S^nR$ is free as a module over $R$ for $n\geq 1$.

Let me give evidence to this. As a $k$-vector space, $ S^nR$ can be decomposed as: $$ S^nR = \bigoplus_{p+q=n} \text{Sym}^p(R^+)\otimes \Lambda^q(R^{-}). $$ The generating function of the dimensions of $S^nR$ is therefore $ \sum_{n\geq 0} \dim_k(S^nR)\, t^n= \left(\frac{1+t}{1-t}\right)^{2^{m-1}}, $ which can be seen to have all coefficients divisible by $2^m=\dim_k R$ except for the first one.

For instance, if $V$ is one-dimensional, then $R$ is isomorphic to $k[x]/(x^2)$ and $S^nR = R$ for all $n\geq1$. If $V$ is two-dimensional with basis $\{\theta_1,\theta_2\}$, then $S^2R$ is generated over $R$ by $1\otimes 1$ and $\theta_1\otimes \theta_2$.

Any idea of proof or references will be gratefully welcomed.