Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 Theorem to $f^{n+1}$ and $0$ instead of $f$ and $n$, and conclude that the coefficient of the Laurent polynomial $f^0 \left(f^{n+1}\right)^{\prime}$ before $t^{-1}$ is zero. Since

$f^0 \left(f^{n+1}\right)^{\prime} = \left(f^{n+1}\right)^{\prime} = \left(n+1\right)f^n f^{\prime}$ (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).

share|improve this question
add comment

2 Answers 2

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.

share|improve this answer
    
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
add comment

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,

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.