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 $1< e\_i <p$ 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.



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


