Proving $(x^{p}-x)^{a} \equiv 0\pmod{p^{a}}$ as a step to another problem?

I have a problem where I have to use that $$a^n\equiv a \pmod p$$ to show that there exist non-zero polynomials modulo $$n$$ that are the zero function. I have tested some examples and figured that for the examples I tested, it seems that for $$m\equiv p_1^{a_1}p_2^{a_2}\dots p_n^{a_n}$$, the polynomial:

$$(x^{p_1}-x)^{a_1}(x^{p_2}-x)^{a_2}\dots(x^{p_n}-x)^{a_n} \tag{\star}$$

Is the zero function but I can't figure out why, I also tested some variations such as:

$$(x^{p_1^{a_1}}-x)^{a_1}(x^{p_2^{a_2}}-x)^{a_2}\dots(x^{p_n^{a_n}}-x)^{a_n}$$

But noticed they fail. I thought about writing $$(\star)$$ as:

$$\begin{eqnarray*} {(x^{p_1}-x)^{a_1}(x^{p_2}-x)^{a_2}\dots(x^{p_n}-x)^{a_n} }&\equiv &{0\pmod{p_1^{a_1}} } \\ {(x^{p_1}-x)^{a_1}(x^{p_2}-x)^{a_2}\dots(x^{p_n}-x)^{a_n} }&\equiv &{0\pmod{p_2^{a_2}}} \\ {\vdots\hspace{3cm}}&&{} \\ {(x^{p_1}-x)^{a_1}(x^{p_2}-x)^{a_2}\dots(x^{p_n}-x)^{a_n} }&\equiv &{0\pmod{p_n^{a_n}}} \end{eqnarray*}$$

And realized that if that is actually true, we must have:

$$\begin{eqnarray*} {(x^{p_1}-x)^{a_1} }&\equiv &{0\pmod{p_1^{a_1}} } \\ {(x^{p_2}-x)^{a_2} }&\equiv &{0\pmod{p_2^{a_2}}} \\ {\vdots\hspace{4mm}}&&{} \\ {(x^{p_n}-x)^{a_n} }&\equiv &{0\pmod{p_n^{a_n}}} \end{eqnarray*}$$

So it all boils down to proving that $$(x^{p}-x)^{a} \equiv 0\pmod{p^{a}}$$ for any $$p$$ prime and any positive integer $$a$$. But I have no clue on how to do that. Could you give me a hint?

1 Answer

Hint If $$b \in \mathbb Z$$ you know that $$b^p-b \equiv 0 \pmod{p}$$

This means that $$p|b^p-b$$

Deduce from here that $$p^a|(b^p-p)^a$$

P.S. You can also show that there exist non-zero polynomials modulo $$n$$ that are the zero function by a very simple counting: there are finitely many functions $$f: \mathbb Z/n \mathbb Z \to \mathbb Z/n \mathbb Z$$ but infinitely many polynomials. Therefore, there are two polynomials $$P \neq Q$$ with the same function. Then $$R:=P-Q$$ is your polynomial. You can even restrict the degree to $$\deg(R)\leq n+1$$

• Also not that the implicit problem is non-trivial: $\ker(\Bbb{Z}[x]\to \Bbb{Z}/p^a \Bbb{Z})$ is not $(p^a, (x^p-x)^a)$ for $a >1$ because $\Bbb{Z}[x]/(p^a,(x^p-x)^a)$ is a group with $(p^a)^{p^a}$ elements whereas not all functions $\Bbb{Z}/p^a \Bbb{Z}\to \Bbb{Z}/p^a \Bbb{Z}$ come from a polynomial, as $f(b+p)=f(b)\bmod p$. Dec 10, 2020 at 4:16