In view of Mariano Suárez-Alvarez's answer I see how badly phrased my question was, and decided to rewrite it. The drawback is that some comments of Martin Brandenburg are now incomprehensible, but I thought it would suffice to say here that Martin made some legitimate constructive criticisms to the original wording of the question. By the way, thank you also to Vladimir Dotsenko for his comments.

Let $K$ be a commutative ring, and let $X_1,\dots,X_n$ be indeterminates. Here $n$ is an integer $\ge3$. For $1\le i < j\le n$ put 
$$
x_{ij}:=\frac{1}{X_i-X_j}
$$
and let $Y_{ij}$ be an indeterminate. Let $I$ be the kernel of the $K$-algebra morphism 
$$
\varepsilon:K[(Y_{ij})]\to K[(x_{ij})],\quad Y_{ij}\mapsto x_{ij}.
$$
> Is $I$ finitely generated? If it is, can one give an explicit finite set of generators?

Note that the identity
$$
\frac{1}{a-b}\ \frac{1}{a-c}+\frac{1}{b-a}\ \frac{1}{b-c}+\frac{1}{c-a}\ \frac{1}{c-b}=0.
$$
shows that $I$ is nonzero.

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

The question had been posted before on Mathematics Stack Exchange ([link](https://math.stackexchange.com/questions/104548/generators-of-a-certain-ideal)).