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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).

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1) Can you give some specific examples for $y_m$? (because my first guess was that $I$ is generated by the $y_{ij}$ and I'm still not convinced of the contrary - I don't want to get through all these indices). 2) Have you tried the cases $n=2$ and $n=3$? – Martin Brandenburg Feb 3 '12 at 14:05
PS: The homological-algebra tag is not appropriate because you don't resolve a module by free modules, but rather you want to resolve an algebra by free algebras aka find a presentation of it. – Martin Brandenburg Feb 3 '12 at 14:07
Dear @Martin: Thanks for your comments. I edited the question. – Pierre-Yves Gaillard Feb 3 '12 at 15:12
Thank you for your obligingness. Is the relation you have written down for $n=3$ the only one coming from the $y_m$ or is this just an example? After all I could take any $3$-tuple of positive integers? – Martin Brandenburg Feb 3 '12 at 16:24
@Martin: while I too think that the "homological algebra" tag might be slightly misleading, Pierre-Yves is quite right saying that finding a presentation for an augmented algebra is intimately related to finding the first two levels of a resolution of the trivial module by free modules. I personally would think that "syzygies" or something alike would be the most instructive tag. – Vladimir Dotsenko Feb 3 '12 at 17:15
up vote 4 down vote accepted

A polynomial $f\in K[\underline Y]$ is in the kernel of your map iff $f$ is zero in the quotient $$\frac{k[X,Y]}{\bigl((X_i-X_j)Y_{i,j}-1:1\leq i<j\leq n\bigr)}.$$In other words, your kernel is the intersection of the ideal in the denominator with the ring $k[Y]$, $$\ker\varepsilon=k[Y]\cap\bigl((X_i-X_j)Y_{i,j}-1:1\leq i<j\leq n\bigr).$$This intersection is generated by the elements of a Groebner basis which only contain $Y$s, assuming you are using a monomial order which eliminates the $X$s; this is explained in the book by Cox, Little and O'Shea, for example.

Doing small examples shows that

$(\star)$ the intersection is generated by all polynomials of the form $$Y_{i,j} Y_{i,k}+ Y_{j,k}Y_{j,i}+Y_{k,i}Y_{k,j}$$ with $i$, $j$ and $k$ distinct. (I am identifying $Y_{i,j}$ with $-Y_{j,i}$ here when $i\neq j$)

Ordering the variables as in $$X_1,X_2,X_3,X_4,Y_{1,2},Y_{1,3},Y_{1,4},Y_{2,3},Y_{2,4},Y_{3,4}$$ for $n=4$ we find the Groebner basis $$\begin{array}{l} Y_{2,3} Y_{2,4}+Y_{3,4} Y_{2,4}-Y_{2,3} Y_{3,4} \\\\ Y_{1,3} Y_{1,4}+Y_{3,4} Y_{1,4}-Y_{1,3} Y_{3,4} \\\\ Y_{1,2} Y_{1,4}+Y_{2,4} Y_{1,4}-Y_{1,2} Y_{2,4} \\\\ Y_{1,2} Y_{1,3}+Y_{2,3} Y_{1,3}-Y_{1,2} Y_{2,3} \\\\ X_3 Y_{3,4}-X_4 Y_{3,4}-1 \\\\ X_2 Y_{2,4}-X_4 Y_{2,4}-1 \\\\ X_2 Y_{2,3}-X_3 Y_{2,3}-1 \\\\ X_1 Y_{1,4}-X_4 Y_{1,4}-1 \\\\ X_1 Y_{1,3}-X_3 Y_{1,3}-1 \\\\ X_1 Y_{1,2}-X_2 Y_{1,2}-1 \end{array}$$ The same pattern is seen for all $n$. It is very easy to see that all these polynomials are in $((X_i-X_j)Y_{i,j}-1:1\leq i&lt;j\leq n)$, and it should not be difficult to show that they are a Groebner basis in general. I expect checking that the above claim $(\star)$ can actually be proved without much pain.

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These are the "same" relations that occur in Arnold's presentation of the de Rham cohomology of the complement of the braid arrangement or, equivalently, the cohomology of the pure braid group with real coefficients. The only difference is that Arnold's generators anticommute and square to zero, while yours commute. – Mariano Suárez-Alvarez Feb 3 '12 at 18:38
Dear Mariano: Thanks a lot! Your answer is very concise, but I'm sure it contains a lot of maths. I'll try to digest and assimilate it. – Pierre-Yves Gaillard Feb 3 '12 at 18:44
@Mariano: in fact, Arnold's generators satisfy X_{ij}=X_{ji}, so the difference is more subtle. However, there is an even analogue of Arnold's relation (still with square zero though) for antisymmetric generators, it is described in a paper by Olivier Mathieu called "The symplectic operad" from a collection of articles for Gelfand's 80th birthday. The relations are precisely these ones. Now one just has to figure out what parts of this observation are coincidental and what are not. – Vladimir Dotsenko Feb 3 '12 at 20:54
I wrote throughout the $Y_{i,j}$ with the convention that $Y_{i,j}=-Y_{j,i}$, too. If Pierre-Ives wants to work without making that identification, then one should add the relations $Y_{i,j}+Y_{j,i}$ along with the quadratic relations above to generate the ideal. – Mariano Suárez-Alvarez Feb 3 '12 at 21:21
Yes, I understand that - and that's partly because I am making my comment. In Mathieu's setting, the generators are in a sense dual to Poisson brackets \{x_i,x_j\} and hence are naturally antisymmetric. So it matches the story very well (unlike the Arnold's story, where there is symmetry, not skew-symmetry in i and j), except for the square zero condition which is absent here. – Vladimir Dotsenko Feb 3 '12 at 21:43

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