5 minor change

In view of Mariano Suárez-Alvarez's answer I posted this see how badly phrased my question on Mathematics Stack Exchange (link)was, and decided to rewrite it. The drawback is that some comments of Martin Brandenburg are now incomprehensible, but got no answer so farI 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 $\ge2$. \ge3$. For$1\le i\neq i < j\le n$put \varepsilon:K[(Y_{ij})_{1\le i\neq j\le n}]\to K[(x_{ij})_{1\le i < j\le n}],\quad varepsilon:K[(Y_{ij})]\to K[(x_{ij})],\quad Y_{ij}\mapsto x_{ij}.$$Obviously I contains$$for$1\le i < j\le n$. But Is$I$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$S(i)$finitely generated? If it isthe , can one give an explicit finite 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 generators? Note that$y_m$is in$I$. Indeed, in view of this Mathematics Stack Exchange answer, if$P(T)$is defined bywhere$T$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(*)a_{ik}=(-1)^k\sum_{u\in S(i,k)}\\prod_{j\neq i}\ frac{1}{a-b}\ \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 frac{1}{a-c}+\frac{1}{b-a}\ \sum_{j\neq i}frac{1}{b-c}+\frac{1}{c-a}\ \u_j=k.Then the claim follows from the fact shows 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$m$I$ is an $n$-tuple of non-negative integers with at least two nonzerocoordinates, 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 $K[(x_{ij})_{1\le i < j\le n}]$, K[(x_{ij})]$, viewed as a$K[(Y_{ij})_{1\le i\neq j\le n}]$-module, K[(Y_{ij})]$-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 question had been posted before on which $X_i$ acts by $0$. Mathematics Stack Exchange (By the way, I'll be happy to remove this tag if it is indeed inappropriate.link)

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}$$-\frac{1}{(a-b)^2}\ \frac{1}{b-c}\ -\ \frac{1}{(a-c)^2}\ \frac{1}{c-b}=0.In the question I tried to explain where these identities come from. They are more and more messy to write down explicitly, but their origin is in the elementary notion of partial fraction decomposition. For instance, the identity of the previous edit comes from the partial fraction decomposition of whereas the identity of this edit comes from the partial fraction decomposition of In both cases, the recipe goes as follows: Write the partial fraction decomposition of f, multiply through by the denominator of f, and compare the coefficients of x^{d-1}, where d is the degree of the denominator. 4 edit clearly indicated I posted this question on Mathematics Stack Exchange (link), but got no answer so far. Let K be a commutative ring, and let X_1,\dots,X_n be indeterminates. Here n is an integer \ge2. For 1\le i\neq j\le n put$$ 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. $$In both cases, the recipe goes as follows: Write the partial fraction decomposition of f, multiply through by the denominator of f, and compare the coefficients of x^{d-1}, where d is the degree of the denominator. 3 edit clearly indicated EDIT 1. EDIT 2. The above identity corresponds to the case n=3,m=(1,1,1). For n=3,m=(2,1,1) we get \frac{1}{(a-b)^2}\ \frac{1}{a-c}\ +\ \frac{1}{a-b}\ \frac{1}{(a-c)^2}$$-\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 whereas the identity of this edit comes from the partial fraction decomposition of 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.

2 edit clearly indicated
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