This sounds like something that must have been answered long ago, but for some reason I can find nothing on it in the internet. (There has been lots of recent activity in diagonal covariants, related to the $n!$ conjecture, but invariants seem to have become a stepchild in this process.)

Let $k$ be a field of characteristic $0$. Let $n\in\mathbb N$. The group $S\_n\times S\_n$ acts on the polynomial ring $k\left[X\_1,X\_2,...,X\_n,Y\_1,Y\_2,...,Y\_n\right]$ by

$\left(\sigma,\tau\right)\left(P\right) = P\left(X\_{\sigma\left(1\right)},X\_{\sigma\left(2\right)},...,X\_{\sigma\left(n\right)},Y\_{\tau\left(1\right)},Y\_{\tau\left(2\right)},...,Y\_{\tau\left(n\right)}\right)$.

Thus, the symmetric group $S\_n$ also acts on $k\left[X\_1,X\_2,...,X\_n,Y\_1,Y\_2,...,Y\_n\right]$ due to the diagonal embedding $S\_n\to S\_n\times S\_n$.

Since the action of $S\_n$ is not generated by pseudoreflections, it follows from the [converse of the Chevalley-Shephard-Todd theorem](http://en.wikipedia.org/wiki/Chevalley%25E2%2580%2593Shephard%25E2%2580%2593Todd_theorem) that $k\left[X\_1,X\_2,...,X\_n,Y\_1,Y\_2,...,Y\_n\right]$ is not a free $k\left[X\_1,X\_2,...,X\_n,Y\_1,Y\_2,...,Y\_n\right]^{S\_n}$-module. But it is easy to see that $k\left[X\_1,X\_2,...,X\_n,Y\_1,Y\_2,...,Y\_n\right]$ is a free $k\left[X\_1,X\_2,...,X\_n,Y\_1,Y\_2,...,Y\_n\right]^{S\_n\times S\_n}$-module of rank $n!^2$.

**Question:** Is $k\left[X\_1,X\_2,...,X\_n,Y\_1,Y\_2,...,Y\_n\right]^{S\_n}$ a free $k\left[X\_1,X\_2,...,X\_n,Y\_1,Y\_2,...,Y\_n\right] ^ {S\_n\times S\_n}$-module?