Proving that the multiplication on a formal derivation ring is associative

Question

Let $R$ be a ring , $\delta$ a derivation of $R$, and $x$ an indeterminate. Let $R[x,\delta]$ be for the formal differential operator ring over $R$ (so for $a \in R$, $xa = ax + \delta(a)$). Multiplication on $R[x,\delta]$ is defined by \begin{align*} \left(\sum_i a_ix^i\right)\left(\sum_j b_jx^j\right) = \sum_k \left( \sum_{l=0}^k \sum_{i\ge l} {i \choose l} a_i \delta^{i-l}(b_{k-l}) \right) x^k. \end{align*} I want to show that this operation is associative. My work is as follows (this will get messy...): \begin{align} &\left [\left(\sum_i a_ix^i\right) \left(\sum_j b_jx^j\right)\right] \left(\sum_k c_kx^k\right) \\ & = \left (\sum_m \left( \sum_{l=0}^m \sum_{i\ge l} {i \choose l} a_i \delta^{i-l}(b_{m-l}) \right) x^m \right) \left(\sum_k c_kx^k\right) \\ & = \sum_n \left ( \sum_{r=0}^n \sum_{s \ge r} {s \choose r} \left( \sum_{l=0}^s \sum_{i\ge l} {i \choose l} a_i \delta^{i-l}(b_{s-l}) \right) \delta^{s-r}(c_{n-r}) \right ) x^n \\ & = \sum_n \left ( \sum_{r=0}^n \sum_{s \ge r} \sum_{l=0}^s \sum_{i\ge l} {s \choose r} {i \choose l} a_i \delta^{i-l}(b_{s-l}) \delta^{s-r}(c_{n-r}) \right ) x^n. \end{align} On the other hand, we have (and note that my choice of summation indices aren't great...): \begin{align*} &\left(\sum_i a_ix^i\right) \left [ \left(\sum_j b_jx^j\right) \left(\sum_k c_kx^k\right) \right] \\ & = \left(\sum_i a_ix^i\right) \left (\sum_m \left( \sum_{l=0}^m \sum_{k\ge l} {k \choose l} b_k \delta^{k-l}(c_{m-l}) \right) x^m \right) \\ & = \sum_n \left( \sum_{r=0}^n \sum_{s \ge r} {s \choose r} a_s \delta^{s-r} \left( \sum_{l=0}^{n-r} \sum_{k \ge l} {k \choose l}b_k \delta^{k-l}(c_{n-r-l}) \right)\right ) x^n \\ & = \sum_n \left( \sum_{r=0}^n \sum_{s \ge r} \sum_{l=0}^{n-r} \sum_{k \ge l} {s \choose r} {k \choose l} a_s \delta^{s-r} \left(b_k \delta^{k-l}(c_{n-r-l}) \right)\right ) x^n \\ & = \sum_n \left( \sum_{r=0}^n \sum_{s \ge r} \sum_{l=0}^{n-r} \sum_{k \ge l} \sum_{t=0}^{s-r}{s \choose r} {k \choose l} {s-r \choose t}a_s \delta^{s-r-t}(b_k) \delta^{t+k-l}(c_{n-r-l})\right ) x^n. \end{align*} That's pretty ugly. I now want to simplify the right hand side and relabel my indices to obtain the earlier ugly expression. Since I have an extra summation, I think I need to use some binomial coefficient identity, but can't seem to find one that fits this problem. Can anyone provide help?

Answer 1

I've sorted this out now, so I'll post my solution.

It's sufficient to consider products of monomials. We have \begin{align*} \Big((ax^r)(bx^s)\Big)(cx^t) &= \left(a \left(\sum_{i=0}^r {r \choose i} \delta^{r-i}(b)x^i \right) x^s\right) cx^t \\ &= \sum_{i=0}^r {r \choose i} a \delta^{r-i}(b) x^i \Big (x^{s} c \Big) x^t \\ &= \sum_{i=0}^r {r \choose i} a \delta^{r-i}(b) x^i \left (\sum_{j=0}^s {s \choose j} \delta^{s-j}(c) x^j \right) x^t \\ &= \sum_{i=0}^r \sum_{j=0}^s {r \choose i}{s \choose j} a \delta^{r-i}(b) \Big( x^i \delta^{s-j}(c) \Big) x^{j+t} \\ &= \sum_{i=0}^r \sum_{j=0}^s {r \choose i}{s \choose j} a \delta^{r-i}(b) \left(\sum_{k=0}^i {i \choose k} \delta^{i-k+s-j}(c) x^k \right) x^{j+t} \\ &= \sum_{i=0}^r \sum_{j=0}^s \sum_{k=0}^i {r \choose i}{s \choose j} {i \choose k} a \delta^{r-i}(b) \delta^{i-k+s-j}(c) x^{k+j+t}\,. \end{align*} On the other hand, \begin{align*} (ax^r)\Big((bx^s)(cx^t)\Big) &= (ax^r)\left(b \left(\sum_{j=0}^s {s \choose j} \delta^{s-j}(c)x^j \right) x^t \right) \\ &= \sum_{j=0}^s {s \choose j} a \Big(x^r b \Big) \delta^{s-j}(c)x^{j+t} \\ &= \sum_{j=0}^s {s \choose j} a \left( \sum_{i=0}^r {r \choose i} \delta^{r-i}(b) x^i \right) \delta^{s-j}(c)x^{j+t} \\ &= \sum_{i=0}^r \sum_{j=0}^s {r \choose i} {s \choose j} a \delta^{r-i}(b) \Big( x^i \delta^{s-j}(c) \Big) x^{j+t} \\ &= \sum_{i=0}^r \sum_{j=0}^s {r \choose i} {s \choose j} a \delta^{r-i}(b) \left( \sum_{k=0}^i {i \choose k} \delta^{i-k+s-j}(c) x^k \right) x^{j+t} \\ &= \sum_{i=0}^r \sum_{j=0}^s \sum_{k=0}^i {r \choose i}{s \choose j} {i \choose k} a \delta^{r-i}(b) \delta^{i-k+s-j}(c) x^{k+j+t}\,, \end{align*} and so \begin{align*} \Big((ax^r)(bx^s)\Big)(cx^t) = (ax^r)\Big((bx^s)(cx^t)\Big) \end{align*} as required.