I posted this question on Mathematics Stack Exchange ([link](http://math.stackexchange.com/questions/104548/generators-of-a-certain-ideal)), 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](http://math.stackexchange.com/a/102905/660), 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.)

**EDIT.** 

(a) In view of the comments made by Martin Brandenburg (whom I thank for his interest), it might be worth writing down the first non-trivial identity mentioned above. If $a,b$ and $c$ are indeterminates, then we have 
$$
\frac{1}{a-b}\ \frac{1}{a-c}+\frac{1}{b-a}\ \frac{1}{b-c}+\frac{1}{c-a}\ \frac{1}{c-b}=0.
$$
(b) The case $n=2$ is trivial, and I'm unable to handle the case $n=3$. 

(c) Martin thinks that the homological algebra tag is inappropriate. He is probably right, but here is why I thought it was. The "model" I have in mind is the Koszul complex, viewed as a free resolution of $K$ viewed as a $K[X_1,\dots,X_n]$-module on which $X_i$ acts by $0$. (By the way, I'll be happy to remove this tag if it is indeed inappropriate.)