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.