8
votes
2answers
186 views

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 ...