Present $\frac{1}{x^p-x}$ as a sum of simple fractions in $\mathbb{Z}_p$

Question

I've done some (maybe incorrect) algebra, which has led me to a strange equality: $$\frac{1}{x^p-x}$$ $$\frac{1}{(1+(x+p-1))^p+(p-1)x}$$ $$\frac{1}{1+p(x+p-1)+\cdots+p(x+p-1)^{p-1}+(x+p-1)^p+(p-1)x}$$ $p(x+p-1)+\cdots+p(x+p-1)$ are $0$ in $\mathbb{Z}_p$. $$\frac{1}{(x-1)^p +1 -x}$$ $$\frac{1}{(x-1)^p -(x-1)}$$

Which gave me an idea to use induction on $x$ (I don't know if this idea is legal), but I'm struggling to find the base.

I also see that $\frac{1}{x^p-x}=\frac{1}{x(x^{p-1}-1)}$, but that doesn't seem to lead anywhere.

The answer in the book is $-\sum_{a=0}^{p-1}\frac{1}{x-a}$.

Hints or full answers, I will be very grateful for any help.

Thank you.

Answer 1

You cannot use induction on $x$, since in $\mathbb Z_p[x]$, $x$ does not actually represent a number, let along an integer; it is merely a placeholder, or an algebraic object. You can only induct over things like integers or natural numbers, so this is out of the question.

Here is a more beneficial approach. Because $x^p-x$ equals $0$ for all $x\in\mathbb Z_p$, you may factor it as $x(x-1)...(x-p+1),$ which suggests (if you are familiar with partial fractions) that the decomposition is of the form $$\frac{a_0}{x}+\frac{a_1}{x-1}+...+\frac{a_{p-1}}{x-p+1}$$ But by symmetry (or, by substituting $x\to x+1$), you can see that all of these coefficients must be equal. So instead, you have a decomposition in the form $$a_0\bigg(\frac{1}{x}+\frac{1}{x-1}+...+\frac{1}{x-p+1}\bigg)$$ In which case all you have to do is show that this constant $a_0$ is equal to $-1$. Does this help?

Answer 2

I shall find the partial fraction decomposition of $$f(x):=\frac{1}{x^q-x}$$ over $\mathbb{F}_q$, where $q=p^r$ for some prime natural number $p$ and for some positive integer $r$. In particular, when $r=1$, $\mathbb{F}_q=\mathbb{F}_p=\mathbb{Z}_p$, which is what the OP asks for.

Note that $x^q-x\in\mathbb{F}_q[x]$ factors into linear factors $\prod\limits_{t\in\mathbb{F}_q}\,(x-t)$, with each linear factor occurring with multiplicity $1$. Therefore, the partial fraction decomposition of $f(x)$ is a sum of simple fractions $$f(x)=\sum_{t\in\mathbb{F}_q}\,\frac{s_t}{x-t}$$ for some $s_t\in\mathbb{F}_q$ for each $t\in\mathbb{F}_q$.

Now, $g_\tau(x):=\dfrac{1}{(x-\tau)\,f(x)}$ is a polynomial in $\mathbb{F}_q[x]$ for each $\tau\in\mathbb{F}_q$. It can be easily seen that $g_\tau(\tau)$ is the (first) derivative of $x^q-x$ evaluated at $x:=\tau$, which is $-1$. From $$\frac{1}{g_\tau(x)}=s_\tau+\sum_{t\in\mathbb{F}_q\setminus\{\tau\}}\,\frac{s_t(x-\tau)}{x-t}\text{ for all }\tau\in\mathbb{F}_q\,,$$ we evaluate this expression at $x:=\tau$ to get $$s_\tau=-1\text{ for every }\tau\in\mathbb{F}_q\,.$$ This shows that $$\frac{1}{x^q-x}=f(x)=-\sum_{t\in\mathbb{F}_q}\,\frac{1}{x-t}\,.$$