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Let $g(z) = a\prod_{i=1}^N (z-\lambda_i) \in \mathbb{Q}[z]$ be square-free. At each root $\lambda_i \in \mathbb{C}$, let $r_i$ denote the residue $\mathrm{Res}_{\lambda_i} 1/g(z)$. Let $I_g(z)$ denote the unique function of degree less than $N$ such that $I_g(\lambda_i) = r_i$ for all $i$. Note that $I_{ag}(z) = aI_{g}(z)$, so we may take $g(z) \in \mathbb{Z}[t]$ if so desired. For some basic examples, I've computed $$I_g(z) = \frac{1}{5}(2z-1) \qquad \text{for} \qquad g(z) = z^2-z-1$$ $$I_g(z) = \frac{1}{22}(2z-1)(z-3) \qquad \text{for} \qquad g(z) = z^3 -z^2+z+1$$

Has anyone seen reference to such a construction in the literature? My main questions are the following:

Q1: Do we have $I_g(z) \in \mathbb{Q}[z]$ for $g \in \mathbb{Q}[z]$?

Q2: Where $\Delta(g)$ denotes the discriminant of $g$, do we have $I_g(z) \in \frac{1}{\Delta(g)} \mathbb{Z}[z]$ for $g \in \mathbb{Z}[z]$? (A strengthening of Q1.)

Final note: the generalizations for $g$ not necessarily square-free are not difficult: take $I_g(z)$ be the unique function of degree less than $\deg g$ such that $I_g(\lambda_i)=r_i$, attained with multiplicity at least that of $\mathrm{ord}_{\lambda_i}g(z)$. We may also consider generalizations for polynomials with coefficients in a general number field.

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up vote 1 down vote accepted

Q1: Yes. Note that the residue at $\lambda_i$ is given by $$\lim_{z \to \lambda_i} \frac{z - \lambda_i}{g(z)} = \lim_{z \to \lambda_i} \frac{1}{g'(z)} = \frac{1}{g'(\lambda_i)}$$

by l'Hopital's. Since $g$ is assumed to be squarefree, we have $\gcd(g, g') = 1$, so by Bezout's lemma applied to $\mathbb{Q}[z]$ there exist polynomials $a, b \in \mathbb{Q}[z]$ such that $b$ has degree less than $\deg g$ and $$ag + b g' = 1.$$

It follows that $\frac{1}{g'(\lambda_i)} = b(\lambda_i)$ for all $i$, so in fact $b = I_g$ and in particular $I_g \in \mathbb{Q}[z]$. (Abstractly $q$ is the multiplicative inverse of $g'$ in $\mathbb{Q}[z]/g(z)$.)

Q2: The statement that should be true is that there exist $A, B \in \mathbb{Z}[z]$ such that $$Ag + B g' = \Delta(g)$$

from which the conclusion follows, but I don't know how to prove it. Morally it should follow by some $p$-adic argument.

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I should also mention that from anon's deleted answer you can prove Q2. Anon uses Lagrange interpolation and although there was an error a correct use of Lagrange interpolation gives the result after applying the fundamental theorem of symmetric functions, but it's messy. – Qiaochu Yuan Mar 22 '12 at 18:09

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