# Residues of $\frac{1}{\prod_{i=1}^n (x-P_i)^{e_i}}$

This is a problem occurring in my research about deformations of $$\mathbb{Z}/p^n$$-covers over a ring of power series. Given an algebraically closed field $$k$$ of characteristic $$p>0$$, suppose $$1< e_i for $$i=1,2, \ldots, n$$ are integers ($$n \ge 2$$). Then I conjecture that there exists some distinct $$P_i$$'s in $$k$$ such that

$$\text{Res}_{P_i}(\frac{1}{\prod_{i=1}^n (x-P_i)^{e_i}})=0,$$

for all $$i$$ if any only if $$\sum_{i=1}^n e_i \ge p+n$$ and $$\sum_{i=1}^n e_i \not \equiv 1 \pmod p$$. $$\text{Res}_{P_i}$$ here stands for the Residue at $$P_i$$, i.e. the $$-1$$ coefficient of the Laurent series expansion of the function at $$x=P_i$$. The condition on the residues tells us whether the fraction can be a derivative of a rational function.

Here is a little bit more background about the question. I am studying whether there could be a deformation of an Artin-Schreier cover branched at one point with ramification jump $$\sum_{ i=1}^n e_i -1$$ to an Artin-Schreier cover branched at $$n$$ points $$\{Q_1, Q_2, \ldots, Q_n\}$$ with ramification jump $$e_i-1$$ at $$Q_i$$ over $$k[[t]]$$. Thus, $$e_i$$ is not congruent to $$1 \pmod p$$. The conjecture, if correct, will prove that a deformation as before exists if and only if $$\sum_{ i=1}^n e_i \ge p+n$$.

I am able to prove it for $$n=2$$ by explicitly calculating the residue or applying the Cartier operator to the fraction. For $$n\ge 3$$, the conjecture still holds in all the examples I've checked using Gröbner bases. I believe that Gröbner bases would work if one tries hard enough. I would love to hear other ideas!

I have just added one condition that $$\sum e_i \not \equiv 1 \pmod p$$. That actually makes it much more interesting since I believe this conjecture determines the deformations of a $$\mathbb{Z}/p$$-cover that branched at one point with ramification jump $$\sum_{ i=1}^n e_i -1$$. Hence, $$\sum_{ i=1}^n e_i -1$$ is not congruent to $$0$$ modulo $$p$$ by Artin-Schreier theory.

Below is my proof for the case $$n=2$$. One might assume that $$P_1=0$$. Consider the rational function

$$\omega=\frac{1}{x^{e_1}(x-Q)^{e_2}}=\frac{x^{p-e_1}(x-Q)^{p-e_2}}{ x^p(x-Q)^p}.$$

One might check that if $$f'=g$$ then $$(fh^p)'=gh^p$$ in characteristic $$p>0$$. Thus, $$\omega$$ is a derivative of some rational functions if and only if $$x^{p-e_1}(x-Q)^{p-e_2}$$ is. The later is a derivative if and only if all the $$kp-1$$ coefficients are equal to $$0$$. Suppose $$e_1+e_2 \ge p+2$$. Then $$2p-(e_1+e_2) \le p-2$$. Thus it is clearly a derivative since all the $$kp-1$$ coefficients are equal to $$0$$. Suppose $$e_1+e_2 < p+2$$. Then $$2p-(e_1+e_2) > p-2$$ and the $$p-1$$th coefficient is not zero when $$Q$$ is different from zero.

Update: Gjergji Zaimi gave a counter example where $$p=7, n=7$$ and all $$e_i=2$$. Another counter-example is $$p=7, n=4$$ with $$e_1=e_2=e_3=2, e_4=6$$. So my conjecture is false! My question right now is whether there is a sufficient condition on $$e_i$$'s for $$\frac{1}{\prod_{i=1}^n (x-P_i)^{e_i}}$$ to be a derivative of some rational functions.

• I don't think that's right if some $e_i=1$. – Felipe Voloch Sep 15 at 21:13
• @FelipeVoloch You are absolutely correct, $e_i$ should not be congruent to $1 \pmod p$. I will fix it right away. – Huy Quoc Dang Sep 15 at 22:17

If we restrict to the case when $e_1=e_2=\cdots=e_n=2$ the right condition is $\sum_{i=1}^n e_i=n+kp$ or $n+kp+1$ for some $k\geq 1$. This provides counterexamples to the stated conjecture (see below) and it shows that the right condition is more complicated than just one inequality.

Here is a proof of my claim: The residue in this case has a simple formula $$\operatorname{Res}_{P_i}\left(\frac{1}{\prod_{i=1}^n(x-P_i)^2}\right)=\frac{2}{\prod_{j\neq i}(P_i-P_j)^2}\left(\sum_{j\neq i}\frac{1}{P_j-P_i}\right)$$ from which we conclude that $$\operatorname{Res}_{P_i}=0 \iff \sum_{j\neq i}\frac{1}{P_j-P_i}=0.$$ Using the fact that $$\frac{d^2}{dx^2}\left(\prod_{i=1}^n (x-P_i)\right)=\frac{1}{2}\sum_{i=1}^n\left(\prod_{j\neq i}(x-P_j)\sum_{j\neq i}\frac{1}{x-P_j}\right)$$ we notice that $\sum_{j\neq k}\frac{1}{P_j-P_k}=0$ is equivalent to the second derivative of $\prod_{i=1}^n(x-P_i)$ vanishing at $P_k$. Since the degree of the second derivative is $n-2$, the only way it can vanish at $n$ distinct points is if it is equal to $0$. So we have proven that $$\operatorname{Res}_{P_i}=0 \quad \text{for all i}\iff \frac{d^2}{dx^2}\left(\prod_{i=1}^n (x-P_i)\right)=0$$ By looking at the leading coefficient we see that the degree can only be $0$ or $1\pmod{p}$. Or, in other words, $\sum_{i=1}^n (e_i-1)\in \{p,p+1,2p,2p+1,\dots\}$. To show that each of these degrees work you can take $P_i$'s to be the roots of polynomials $x^{kp}-x-1$ and $x^{kp+1}-1$, respectively (both polynomials have distinct roots, and their second derivatives vanish).

For an explicit counter example to your conjecture look at $p=5, n=7$ with all $e_i=2$. We have $\sum e_i=14\neq 1\pmod 5$. Yet there exists no choice of distinct $P_i$ for which $\operatorname{Res}_{P_i}=0$ for all $i$.

• Notice that this can also give instances where $\sum e_i=1\pmod{p}$ and it is possible to find distinct $P_i$'s satisfying the conditions of the problem. – Gjergji Zaimi Sep 20 at 6:58

Easy part: the proof that $\sum e_i\geqslant n+p$.

Assume that $\sum e_i<n+p$. The rational antiderivative should have the form $g/f$, where $f=\prod (x-P_i)^{e_i-1}$, $\deg g<\deg f=\sum (e_i-1)<p$. This may be seen from expanding $\prod (x-P_i)^{-e_i}$ as a sum of elementary fractions and integrating them all. We have $(g/f)'=(g'f-f'g)/f^2$, and if $\deg f=a$, $\deg g=b$, the degree of the numerator equals $a+b-1$, since the leading coefficients of $f'g$, $g'f$ do not cancel (here we use that $p$ can not divide $a-b$). Thus $1=\prod (x-P_i)^{e_i} (g/f)'$ has degree $(n+a)+(a+b-1)-2a=n+b-1>0$, a contradiction.

• That is a very smart argument. I haven’t thought about $p$ does not divide $a-b$. Thank you so much! – Huy Quoc Dang Sep 15 at 23:40
• Residues equal to zero is not the same as having a rational antiderivative in positive characteristic, e.g. $x^{p-1}$. But the conditions $\sum e_i < p+n, e_i>1$ imply that each $e_i < p-1$ so it's OK. – Felipe Voloch Sep 16 at 6:48
• @FelipeVoloch exactly, and I even tried to sketch the proof:) – Fedor Petrov Sep 16 at 7:01
• @FelipeVoloch You are right. One can extend the conjecture to the case $e_i$ large and change the condition to $\sum_{i=1}^n \overline{e}_i < p+n$ where $\overline{e}_i \equiv e_i \pmod p$ and $1 <\overline{e}_i \le p$. – Huy Quoc Dang Sep 16 at 7:11
• @Huy this is up to you, but I do not consider the question being answered. – Fedor Petrov Sep 16 at 16:22