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The following fact is stated (but not proved) in Chapter 0 of Noll's Finite Dimensional Spaces (it's 07.21):

If $R$ is a commutative ring with unity and $r_1, ..., r_n$ elements of $R$ then $$ n!\text{ }r_1\cdots r_n=\sum_{k=1}^{n}(-1)^{n-k}\sum_{1\le i_1<\cdots<i_k\le n}(r_{i_1}+\cdots+r_{i_k})^n $$

Now as it happens I (think I) can see a proof of this.

Start by showing that it holds when setting $r_{1}=X_{1}, ..., r_n=X_{n}$ in $\mathbb{Z}[X_{1}, ...,X_{n}]$. To do this, first show that the RHS is zero at $X_{i}=0$ so is divided by $X_{i}$, so is divided by the product $X_{1}\cdots X_{n}$ (by UFD-ness). Thus by considering degrees the RHS is a constant times this product. Setting $X_{i}=1$ then reveals this constant and gives the polynomial identity.

Then use the free-ness of $\mathbb{Z}[X_{1}, ...,X_{n}]$ to map these elements onto any $r_i$ in any commutative ring with 1, to deduce that the identity holds there.

However seeing as how this fact is stated without proof in the introductory chapter to a not particularly advanced textbook (a textbook that does not have knowledge of abstract algebra as a prerequisite), and appears in the company of some trivial facts like the binomial theorem, I was wondering if there was a more elementary proof of the identity?

user26857
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Blunka
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1 Answers1

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While I like your proof, here is one that does not use the fact that $\mathbb{Z}[X_1, \ldots, X_n]$ is a UFD.

As you pointed out, this identity holds "universally" by considering the elements $r_i$ as formal variables. Doing this, it is easy to see that in the right hand side of the equation, the coefficient in front of $r_1\cdots r_n$ is $n!$. What remains to be shown is that the coefficient in front of every other monomial is zero.

To do this, let's modify the notations a bit. Let $I=\{1,\ldots n\}$. Then the right hand side of the equation can be written as $$ \sum_{k=1}^{n}(-1)^{n-k}\sum_{\stackrel{J\subseteq I}{|J|=k}} (\sum_{\alpha\in J}r_\alpha)^n. $$

Now fix a monomial $r_{i_1}^{a_1}\cdots r_{i_l}^{a_l}$ with $\sum_{j=1}^{l}a_j = n$. Fix $K=\{i_1, \ldots, i_l\}$. The only terms in the above sum contributing to the coefficient in front of this monomial are those terms corresponding to a subset $J\subseteq I$ containing $K$. Moreover, given such a subset $J$ containing $K$, the coefficient in front of $r_{i_1}^{a_1}\cdots r_{i_l}^{a_l}$ in $(\sum_{\alpha\in J}r_\alpha)^n$ is

$$\binom{n}{a_1}\binom{n-a_1}{a_2}\cdots \binom{n-a_1-\ldots-a_{l-1}}{a_l}. $$

As such, it does not depend on $J$. Finally, the number of subsets $J$ containing $K$ and such that $|J|=k$ is $\binom{n-l}{k-l}$ whenever $l\leq k \leq n$.

Therefore the coefficient in front of $r_{i_1}^{a_1}\cdots r_{i_k}^{a_k}$ in the sum is $$ \sum_{k=l}^{n}(-1)^{n-k} \binom{n-l}{k-l} \binom{n}{a_1}\binom{n-a_1}{a_2}\cdots \binom{n-a_1-\ldots-a_{l-1}}{a_l} $$ which is in turn equal to $$ \binom{n}{a_1}\binom{n-a_1}{a_2}\cdots \binom{n-a_1-\ldots-a_{l-1}}{a_l} \sum_{k=l}^{n}(-1)^{n-k} \binom{n-l}{k-l} $$ and, changing indices, to $$ \binom{n}{a_1}\binom{n-a_1}{a_2}\cdots \binom{n-a_1-\ldots-a_{l-1}}{a_l} \sum_{k=0}^{n-l}(-1)^{n-l-k} \binom{n-l}{k}. $$

The alternating sum of the first $n-l$ binary coefficients is zero if $n-l>0$; thus the above expression is zero if $l\neq n$. This finishes the proof.

Pierre-Guy Plamondon
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