Dimension of Homogenous Symmetric Polynomials with $(k,r)$-Clustering Property

Question

Denote by $\Lambda_N=\mathbb{C}[x_1, \cdots, x_N]^{S_N}$ the ring of symmetric polynomials in $N$ variables. This is graded ring with degree. Denote by $\Lambda_N^M$ the subspace of homogenous symmetric polynomials of degree $M$. We say a polynomial $f\in \Lambda_N$ satisfies a $(k,r)$-clustering property if $$ f(\underbrace{Z,Z,\cdots, Z}_{k\text{ times}}, x_{k+1},\cdots, x_N)=\prod_{i=k+1}^N(Z-x_i)^r g(Z,x_{k+1}, \cdots, x_N) $$ for some $g\in \mathbb{C}[Z, x_{k+1}, \cdots, x_N]$. Define the $\mathbb{C}$-vector space $$ V_{N,M}^{(k,r)}=\mathrm{span}_{\mathbb{C}}\{f\in \Lambda_N^M\mid f\text{ satisfies a }(k,r)\text{-clustering property}\} $$ The vector space $\bigoplus_{M}V^{(k,r)}_{N,M}$ is in fact an ideal of $\Lambda_N$. My question is about the dimension of $V_{N,M}^{(k,r)}$ in special circumstances:

Let $k,r,n$ be positive integers such that $k+1$ and $r-1$ are relatively prime. Let $N=nk$ and $M=n(n-1)nk/2$. For these integers is the following true? $$ \dim V^{(k,r)}_{N,M}=1 $$

Under such circumstances I know that $\dim V^{(k,r)}_{N,M}\neq 0$. This is because letting $\alpha=-\frac{r-1}{k+1}$ and partition $$ \Lambda=\Big(r^{k}(2r)^k(3r)^k \cdots ((n-1)r)^k\Big) $$ the Jack polynomial $P^{\beta}_{\Lambda}(x_1, \cdots, x_N)$ specialized to $\beta=\alpha$ is well-defined (see here) and in fact inside $V_{N,M}^{(k,r)}$ (see here). My question is whether or not that is the only symmetric polynomial with the above properties.

Answer 1

I don't know if I should erase this question or not, but this is not true as I just came up with a counterexample. I'll just post it as an answer because maybe this is helpful to somebody later.

Let $n=2$, $k=r=4$. Consider the two polynomials (here $S$ is symmetrization) $$ P_1(x_1,\cdots, x_4, y_1, \cdots, y_4)=S\left[\prod_{i,j=1}^4(x_i-y_j)\right] $$ and $$ P_2(x_1, \cdots, x_4, y_1, \cdots, y_4)=S \left[\prod_{i=1}^4(x_i-y_i)^2\prod_{i=1}^4(x_i-y_{i-1})^2\right] $$ where $y_{0}$ in the above is understood as $y_4$. Under $x_1=x_2=x_3=x_4=Z$ they both become $$ P_1(Z, \mathbf{y})=P_2(Z, \mathbf{y})=1152(Z-y_1)^4(Z-y_2)^4(Z-y_3)^4(Z-y_4)^4 $$ Hence $P_1, P_2\in V_{8,16}^{(4,4)}$. However $P_1$ and $P_2$ are not proportional to one another as one can check with a simple mathematica code. Therefore $\dim V_{8,16}^{(4,4)}>1$.