Why is the (-1)-th coefficient of $f^n f'$ equal to 0, without dividing by $n+1$?

Let $R$ be a commutative ring, and $n$ be a nonnegative integer. Let $f\in R\left[t,t^{-1}\right]$ be a Laurent polynomial in one variable $t$ over $R$ (this means a formal $R$-linear combination of terms of the form $f^m$ with $m\in\mathbb Z$ such that only finitely many of these terms have nonzero coefficients).

We let $f^{\prime}$ denote the derivative of $f$ with respect to $t$, defined in the formal way: $f^{\prime} = \sum\limits_{m\in \mathbb Z} mf_m t^{m-1}$, where $f_m$ is the coefficient of $f$ before $t^m$.

Theorem: The coefficient of the Laurent polynomial $f^n f^{\prime}$ before $t^{-1}$ is zero.

Let me sketch the standard proof of this theorem, to show what I want to avoid:

First of all, it is very easy to verify the Theorem in the case $n=0$. Hence, we can apply the Theorem to $f^{n+1}$ and $0$ instead of $f$ and $n$, and conclude that the coefficient of the Laurent polynomial $\left(f^{n+1}\right)^0 \left(f^{n+1}\right)^{\prime}$ before $t^{-1}$ is zero. Since

$\left(f^{n+1}\right)^0 \left(f^{n+1}\right)^{\prime} = \left(f^{n+1}\right)^{\prime} = \left(n+1\right)f^n f^{\prime}$ (by the Leibniz identity or the chain rule, as you wish),

this yields that the coefficient of the Laurent polynomial $\left(n+1\right)f^n f^{\prime}$ before $t^{-1}$ is zero.

Now, if $n+1$ is not a zero-divisor in $R$, then this immediately yields that the coefficient of the Laurent polynomial $f^n f^{\prime}$ before $t^{-1}$ is zero. Thus, the Theorem is proven in the case when $n+1$ is not a zero-divisor in $R$. In particular, the Theorem is proven in the case when $R$ is a polynomial ring over $\mathbb Z$. But since the statement of the Theorem (for a given value of $n$ and for a given up-degree and low-degree of the Laurent polynomial $f$) is a polynomial identity in the coefficients of $f$, proving it when $R$ is a polynomial ring over $\mathbb Z$ automatically entails that it holds for any commutative ring $R$ (by an elementary fact which has many names, among them "principle of permanence of identities"). Thus, the Theorem is proven.

Question: Is there a (not too long or ugly) proof of the Theorem which avoids the use of the principle of permanence of identities? For $n=0$ and $n=1$, the Theorem can be shown by "expanding" the polynomial, but this seems to become messy for higher $n$'s. I believe there should be some smart induction-over-$n$ argument (maybe through generalization of the Theorem).

-

I don't think it becomes that messy to do explicitly. The coefficient you want is given by $$\sum_{i_1+i_2+\dots+i_n+i_{n+1}=0}\left(i_{n+1}\prod_k f_{i_k}\right) \, ;$$ the $k$th index on this sum comes from the $k$th factor of $f$ in the original product, while the $(n+1)$st index comes from the derivative factor.

Now, combine all the terms that contain the same combination $\prod_k f_{i_k}$, permuted somehow. Each $i_k$ will show up as a coefficient exactly $n!$ times, so the condition $\sum i_k = 0$ means that each of these combined terms vanishes.

-
No to exactly $n!$; if some of the $i_j$ are equal, then it will be only $\dfrac{n!}{\text{number of permutations in }S_n\text{ which stabilize }\left(i_1,i_2,...,i_n\right)}$ times. One can salvage your argument, but this is somewhat messy... – darij grinberg May 11 '12 at 12:38
Oops, you're right, but I don't think the fix has to be that horrible. Instead of combining terms to the maximum extent possible, fix some $n+1$-cyclic $G$ in $S_{n+1}$ and only combine terms that are identified by $G$. Then your stabilizer is considerably neater... – Micah May 11 '12 at 15:45

If your ring is an integral domain, remember that the characteristic of the ring must be a prime number, and that in that case, by the binomial theorem applied inductively, $$(a_1 + \dots + a_k)^p = a_1^p + a_2^p + \dots + a_k^p,$$ (this is known as the freshman's dream, which is a common mistake in rings of characterstic $0$ made by starters in algebra, but is true in characteristic $p$) which means that proving the theorem for $n = 0, \dots, p-1$ is sufficient, because for $n = p$ a prime number, we get \begin{align*} f(x)^p (f'(x)) & = \left(\sum_{n \in \mathbb Z} f_n x^n \right)^p \left( \sum_{m \in \mathbb Z} m f_m x^{m-1} \right) \\ & = \left(\sum_{n \in \mathbb Z} f_n^p x^{np} \right) \left( \sum_{m \in \mathbb Z} m f_m x^{m-1} \right) \\ & = \sum_{n, m \in \mathbb Z} f_n^p f_m m x^{np+m-1} \end{align*} The coefficients before $x^{-1}$ are those with an exponent of the form $np+m-1 = -1$, but in characteristic $p$, this means $m = 0$. You can readily see that the coefficients of all such terms ($f_n^p f_m m$) will indeed vanish.

I don't know if you get so easily past the $p^{th}$ powers ($n = 2p, 3p, \dots$), but this is an idea to get around it. In rings which are not integral domains I have no idea what to do.

Hope that helps,

-