I posted this question on Mathematics Stack Exchange (link), but got no answer so far.

Let $K$ be a commutative ring, and let $X_1,\dots,X_n$ be indeterminates. Here $n$ is an integer $\ge2$. For $1\le i\neq j\le n$ put $$ x_{ij}:=\frac{1}{X_i-X_j}\quad, $$ and let $Y_{ij}$ be an indeterminate.

Let $I$ be the kernel of the $K$-algebra morphism $$ \varepsilon:K[(Y_{ij})_{1\le i\neq j\le n}]\to K[(x_{ij})_{1\le i < j\le n}],\quad Y_{ij}\mapsto x_{ij}. $$ Obviously $I$ contains $$ y_{ij}:=Y_{ij}+Y_{ji} $$ for $1\le i < j\le n$.

But $I$ contains also less trivial elements.

Indeed, for each $n$-tuple $m=(m_1,\dots,m_n)$ of positive integers put $$ y_m:=\sum_i\ (-1)^{m_i}\sum_{u\in S(i)}\ \prod_{j\neq i}\ \binom{m_j-1+u_j}{m_j-1}\ Y_{ij}^{m_j+u_j}, $$ where $S(i)$ is the set of those maps $$ u:\{1,\dots,n\}\setminus\{i\}\to\mathbb N,\quad j\mapsto u_j $$ which satisfy $$ \sum_{j\neq i}\ u_j=m_i-1. $$ We claim that $y_m$ is in $I$.

Indeed, in view of this Mathematics Stack Exchange answer, if $P(T)$ is defined by $$ P(T):=(T-X_1)^{m_1}\cdots(T-X_n)^{m_n}, $$ where $T$ is an indeterminate, then we have $$ 1=\sum_i\ \sum_{k=0}^{m_i-1}\ \frac{a_{ik}\ P(T)}{(T-X_i)^{m_i-k}}\qquad(*) $$ with $$ a_{ik}=(-1)^k\sum_{u\in S(i,k)}\ \prod_{j\neq i}\ \binom{m_j-1+u_j}{m_j-1}\ x_{ij}^{n_k+u_k}, $$ where $S(i,k)$ is the set of those maps $$ u:\{1,\dots,n\}\setminus\{i\}\to\mathbb N,\quad j\mapsto u_j $$ which satisfy $$ \sum_{j\neq i}\ u_j=k. $$ Then the claim follows from the fact that $\varepsilon(y_m)$ is the coefficient of $T^{\deg(P)-1}$ in the right-hand side of $(*)$.

Question. Is the ideal $I$ generated by the $y_{ij}$ and the $y_m$?

(I put the homological algebra tag because the ultimate goal is to know whether there is a functorial free resolution of $K[(x_{ij})_{1\le i < j\le n}]$, viewed as a $K[(Y_{ij})_{1\le i\neq j\le n}]$-module, and, if it exists, what can be said about it.)