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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).

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).

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 $(*)$.

[EDIT 3. This edit consists in the addition of the present paragraph.] This is the definition of $y_m$ when $m$ is an $n$-tuple of positive integers. If $m$ is an $n$-tuple of non-negative integers with at least two nonzero coordinates, then we define $y_m$ by ignoring the zero coordinates. For instance, if $m=(0,1,1,1)$, we consider that the starting indeterminates are $X_2,X_3,X_4$, and we define $y_m$ as a in the case $n=3$. In the question below, we consider the $y_m$ for all $n$-tuples of non-negative integers with at least two nonzero coordinates.

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 1.

(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.)

EDIT 2. The above identity corresponds to the case $n=3,m=(1,1,1)$. For $n=3,m=(2,1,1)$ we get $$ \frac{1}{(a-b)^2}\ \frac{1}{a-c}\ +\ \frac{1}{a-b}\ \frac{1}{(a-c)^2} $$ $$-\frac{1}{(a-b)^2}\ \frac{1}{b-c}\ -\ \frac{1}{(a-c)^2}\ \frac{1}{c-b}=0. $$ In the question I tried to explain where these identities come from. They are more and more messy to write down explicitly, but their origin is in the elementary notion of partial fraction decomposition. For instance, the identity of the previous edit comes from the partial fraction decomposition of $$ f(x):=\frac{1}{(x-a)(x-b)(x-c)}\quad, $$ whereas the identity of this edit comes from the partial fraction decomposition of $$ f(x):=\frac{1}{(x-a)^2(x-b)(x-c)}\quad. $$ In both cases, the recipe goes as follows: Write the partial fraction decomposition of $f$, multiply through by the denominator of $f$, and compare the coefficients of $x^{d-1}$, where $d$ is the degree of the denominator.

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).

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.)

EDIT 1.

(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.)

EDIT 2. The above identity corresponds to the case $n=3,m=(1,1,1)$. For $n=3,m=(2,1,1)$ we get $$ \frac{1}{(a-b)^2}\ \frac{1}{a-c}\ +\ \frac{1}{a-b}\ \frac{1}{(a-c)^2} $$ $$-\frac{1}{(a-b)^2}\ \frac{1}{b-c}\ -\ \frac{1}{(a-c)^2}\ \frac{1}{c-b}=0. $$ In the question I tried to explain where these identities come from. They are more and more messy to write down explicitly, but their origin is in the elementary notion of partial fraction decomposition. For instance, the identity of the previous edit comes from the partial fraction decomposition of $$ f(x):=\frac{1}{(x-a)(x-b)(x-c)}\quad, $$ whereas the identity of this edit comes from the partial fraction decomposition of $$ f(x):=\frac{1}{(x-a)^2(x-b)(x-c)}\quad. $$ In both cases, the recipe goes as follows: Write the partial fraction decomposition of $f$, multiply through by the denominator of $f$, and compare the coefficients of $x^{d-1}$, where $d$ is the degree of the denominator.

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 $(*)$.

[EDIT 3. This edit consists in the addition of the present paragraph.] This is the definition of $y_m$ when $m$ is an $n$-tuple of positive integers. If $m$ is an $n$-tuple of non-negative integers with at least two nonzero coordinates, then we define $y_m$ by ignoring the zero coordinates. For instance, if $m=(0,1,1,1)$, we consider that the starting indeterminates are $X_2,X_3,X_4$, and we define $y_m$ as a in the case $n=3$. In the question below, we consider the $y_m$ for all $n$-tuples of non-negative integers with at least two nonzero coordinates.

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 1.

(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.)

EDIT 2. The above identity corresponds to the case $n=3,m=(1,1,1)$. For $n=3,m=(2,1,1)$ we get $$ \frac{1}{(a-b)^2}\ \frac{1}{a-c}\ +\ \frac{1}{a-b}\ \frac{1}{(a-c)^2} $$ $$-\frac{1}{(a-b)^2}\ \frac{1}{b-c}\ -\ \frac{1}{(a-c)^2}\ \frac{1}{c-b}=0. $$ In the question I tried to explain where these identities come from. They are more and more messy to write down explicitly, but their origin is in the elementary notion of partial fraction decomposition. For instance, the identity of the previous edit comes from the partial fraction decomposition of $$ f(x):=\frac{1}{(x-a)(x-b)(x-c)}\quad, $$ whereas the identity of this edit comes from the partial fraction decomposition of $$ f(x):=\frac{1}{(x-a)^2(x-b)(x-c)}\quad. $$ In both cases, the recipe goes as follows: Write the partial fraction decomposition of $f$, multiply through by the denominator of $f$, and compare the coefficients of $x^{d-1}$, where $d$ is the degree of the denominator.