Polynomials invariant under the action of $S_m \times S_n$

Let $\mathbb{K}$ be a commutative ring. Let $n$ and $m$ be nonnegative integers. The group $G = S_n \times S_m$ acts on the polynomial ring $\mathbb K[X_{i,j}]_{1\leq i\leq n,1\leq j\leq m}$.

The set $M$ of monomials $\prod_{i,j}X_{i,j}^{a_{i,j}}$ (where $(a_{i,j})$ is an arbitrary matrix in ${\cal M}_{n,m}(\mathbb N)$) is invariant by the action of $G=S_n \times S_m$.

Further, for any ${\mathfrak m}\in M$, the orbit $\omega_G({\mathfrak m})$ of ${\mathfrak m}$ under $G$ is finite, so that we may define $i({\mathfrak m})=\sum_{{\mathfrak m}' \in \omega_G({\mathfrak m})} {\mathfrak m}'$ and this polynomial is $G$-invariant.

When ${\mathfrak m}$ is square-free (i.e. all exponents in ${\mathfrak m}$ are $0$ or $1$), we say that $i({\mathfrak m})$ is an $\bf{elementary}$ $G$-invariant polynomial. We denote by ${\cal E}_{n,m}$ the set of all elementary $G$-invariant polynomials (note that this set is finite).

As for ordinary symmetric polynomials, we have :

$\bf Theorem$. If $I(G)$ denotes the ring of all $G$-invariant polynomials in $R=\mathbb K[X_{i,j}]_{1\leq i\leq n,1\leq j\leq m}$, then $I(G)={\mathbb K}[{\cal E}_{n,m}]$.

$\bf Proof.$ One of the inclusions is obvious, and we only need to show that $i({\mathfrak m})\in {\mathbb K}[{\cal E}_{n,m}]$ for an arbitrary (not necessarily square-free ) monomial ${\mathfrak m}$, because an arbitrary $G$-invariant polynomial $P$ can be written as a finite linear combination $\sum \lambda_{{\mathfrak m}}i({\mathfrak m})$ (indeed, any polynomial $P$ is initially a linear combination of monomials $\sum \mu_{{\mathfrak m}}{\mathfrak m}$, and $P$ is $G$-invariant iff $\mu_{{\mathfrak m}}=\mu_{{\mathfrak m}'}$ whenever ${\mathfrak m}' \in \omega_G({\mathfrak m})$, in other words iff $\mu$ is constant on the orbits).

For ${\mathfrak m}=\prod_{i,j}X_{i,j}^{a_{i,j}}$, let $b_{i,j}={\sf max}(a_{i,j}-1,0)$ and define the complexity $c({\mathfrak m})$ of ${\mathfrak m}$ to be $c({\mathfrak m})=\sum_{i,j}b_{i,j}$. Thus $c({\mathfrak m})$ is always nonnegative, and is zero precisely when ${\mathfrak m}$ is square-free.

When $c({\mathfrak m})=c\gt 0$, if ${\mathfrak m}'=\prod_{i,j}X_{i,j}^{b_{i,j}}$, then $i({\mathfrak m})-i({\mathfrak m}')i(\frac{{\mathfrak m}}{{\mathfrak m}'})$ is a $G$-invariant polynomial all of whose homogeneous parts have complexity strictly less that $c$. We are now done by induction on $c({\mathfrak m})$.

Note that the group $G$ is smaller than the full symmetric group on the $nm$ indeterminates ; this results in smaller orbits, so there will be more orbits and more elementary polynomials than in the symmetric case. We deduce $|{\cal E}_{n,m}| > nm$, and the equality we have shown does not extend to an isomorphism property, unlike the fully symmetric case.

An interesting sub-question is : Is $I(G)$ isomorphic to the polynomial ring in $nm$ variables ? My guess is that the answer is NO.