This is really a followup to Why are Jucys-Murphy elements' eigenvalues whole numbers? , specifically to Igor Makhlin's beautiful answer. I'm trying to make it even more beautiful by getting rid of the eigenvalues and turn it into a sequence of algebraic manipulations. Here is what I think it would boil down to:
Question. Let $k$ be a nonnegative integer and $x, y, s$ be three elements of a noncommutative ring satisfying \begin{align} xy &= yx; \\ s^2 &= 1; \\ sy &= xs + 1; \\ \prod_{i=-k+1}^{k-1} \left(x-i\right) & = 0. \end{align} Does it then follow that $\prod\limits_{i=-k}^{k} \left(y-i\right) = 0$ ? Barring that, does it follow that some power (say, the square) of $\prod\limits_{i=-k}^{k} \left(y-i\right)$ is $0$?
For example, for $k = 2$, the last assumption is $\left(x+1\right)x\left(x-1\right) = 0$, and we are trying to prove that either the product $\left(y+2\right)\left(y+1\right)y\left(y-1\right)\left(y-2\right)$ or at least some of its powers is $0$.
If this can be done, then the use of eigenvalues in https://mathoverflow.net/a/83493/ can be avoided (although we might still need to use the $A^T A = 0 \implies A = 0$ theorem from real linear algebra to get rid of the power). Specifically, we apply the question to $x = X_k$, $y = X_{k+1}$ and $s = s_k$.
Of course, in this application, the noncommutative ring is actually a finite-dimensional algebra over $\mathbb{Q}$, and this is required to make the eigenvalue argument work, but I feel it has a good chance of having a more algebraic reason to be and thus avoiding these requirements. At the very least, I know that it holds for $k = 1$, but this is perhaps too easy an example...
Note that the relations \begin{align} xy &= yx; \\ s^2 &= 1; \\ sy &= xs + 1 \end{align} are the defining relations of the degenerate affine Hecke algebra $H\left(2\right)$. Thus, this question is concerned with a certain quotient of this algebra.
