Vandermonde identity in a ring

Question

Let $R$ be a commutative $\mathbb{Q}$-algebra. For $r \in R$ and $n \in \mathbb{N}$ we can define the binomial coefficient $\binom{r}{n}$ as usual by $\binom{r}{0}=1$ and $\binom{r}{n+1}=\frac{r-n}{n+1} \cdot \binom{r}{n}$. I would like to prove the following identity for $r,s \in R$ and $n \in \mathbb{N}$:

$$\binom{r+s}{n}=\sum_{p+q=n} \binom{r}{p} \cdot \binom{s}{q}$$

For $r,s \in \mathbb{N}$ this is known as the Vandermonde identity. For $R=\mathbb{C}$ it is known as the Chu–Vandermonde identity. In general, let's just call it the Vandermonde identity in $R$.

One can prove the identity for general $R$ by observing that both sides are polynomials in $r,s$, so that it suffices to consider $R=\mathbb{Q}[x,y]$ with $r=x$, $s=y$. Then the polynomials agree on $\mathbb{N} \times \mathbb{N}$ by the usual Vandermonde identity. Hence they are equal (using that $\mathbb{N} \times \mathbb{N} \subseteq \mathbb{A}^2$ is Zariski dense). See also Darij Grinberg's notes on $\lambda$-rings, Theorem 3.2.

Question. Is there a more direct and algebraic proof of the Vandermonde identity in a given commutative $\mathbb{Q}$-algebra $R$?

I am looking for a proof which works directly for $R$, without any reduction arguments. There should be some proof which just consists of simple algebraic manipulations. I have tried induction on $n$, but didn't succeed. Note that the claim is equivalent to

$$\prod_{k=0}^{n-1} (r+s-k)=\sum_{p=0}^{n} \binom{n}{p} \cdot \prod_{i=0}^{p-1} (r-i) \cdot \prod_{j=0}^{n-p-1} (s-j).$$

I would also be very happy with a proof following the principle of categorification: For $r,n$ try to find some $R$-module (or some complex of $R$-modules?) $\Lambda^n(r)$, construct (via universal properties?) an isomorphism $\Lambda^n(r+s) \cong \oplus_{p+q=n} \Lambda^p(r) \otimes \Lambda^q(s)$ and evaluate this using some rank function. I have no idea what $\Lambda^n(r)$ should be, but for $r \in \mathbb{N}$ it should be the usual exterior power $\Lambda^n(R^r)$.

Answer 1

Here's one way to do the induction step for the equivalent reformulation: $$\begin{align*} (r+s-n)\prod_{k=0}^{n-1} (r+s-k) & =\sum_{p=0}^{n} \binom{n}{p} \left([r-p]+[s-(n-p)]\right)\cdot \prod_{i=0}^{p-1} (r-i) \cdot \prod_{j=0}^{n-p-1} (s-j) \\ & =\sum_{p=0}^{n} \binom{n}{p} \cdot \prod_{i=0}^{\color{red}{p}} (r-i) \cdot \prod_{j=0}^{n-p-1} (s-j) \\&\phantom{=}+ \sum_{p=0}^{n} \binom{n}{p} \cdot \prod_{i=0}^{p-1} (r-i) \cdot \prod_{j=0}^{\color{red}{n-p}} (s-j) \\ &=\binom{n}{n} \cdot \prod_{i=0}^{n} (r-i) + \sum_{p=\color{red}{1}}^{\color{red}{n}} \binom{n}{\color{red}{p-1}} \cdot \prod_{i=0}^{\color{red}{p-1}} (r-i) \cdot \prod_{j=0}^{\color{red}{n-p}} (s-j) \\&\phantom{=}+ \binom{n}{0}\prod_{j=0}^{n} (s-j) +\sum_{p=1}^{n} \binom{n}{p} \cdot \prod_{i=0}^{p-1} (r-i) \cdot \prod_{j=0}^{n-p} (s-j) \\ &=\binom{n+1}{n+1} \cdot \prod_{i=0}^{n} (r-i) \\ &\phantom{=}+ \sum_{p=1}^{n} \left[\binom{n}{p-1}+\binom{n}{p}\right] \cdot \prod_{i=0}^{p-1} (r-i) \cdot \prod_{j=0}^{n\color{red}{+1}-p\color{red}{-1}} (s-j) \\&\phantom{=}+ \binom{n+1}{0}\prod_{j=0}^{n} (s-j) \\ &= \sum_{p=0}^{n+1} \binom{n+1}{p} \cdot \prod_{i=0}^{p-1} (r-i) \cdot \prod_{j=0}^{n+1-p-1} (s-j). \end{align*}$$