Prove for any nonzero polynomial $f\in \mathbb{Z}_p$ these statements are equivalent

Question

• f is a divisor of $x^{p^n} -x$ for some $n \geq 1$
• f has no multiple roots in the closure $\bar{\mathbb{Z}_p}$ of $\mathbb{Z}_p$

I think I have the forward direction:

Computing the formal derivative of $x^{p^n} -x$ yields $p^nx^{p^{n-1}} - 1 = -1 \neq 0$ because $x^{p^{n-1}} = 0$ so the derivative of the polynomial is constant and therefore has no roots and a criteria for something having multiple roots is that its also a root of the derivative so f has no multiple roots.

I have no idea how to do the backwards direction. There are a bunch of equivalent statements about algebraic closures in my notes:

1. F is algebraically closed
2. the only irreducible polynomials are of degree 1
3. Every polynomial is a product of first degree polynomials

among others that I don't think are useful here.

I can't see how any of this is useful nor do I have any other thoughts on how to proceed.

Answer 1

You don’t need to know anything about algebraic closures here. The statement that $f$ has no multiple roots in the algebraic closure of $\Bbb F_p$ is just a way of saying that no matter what finite field extension of $\Bbb F_p$ you look at, there are no multiple roots of $f$ there.

Notice that if $\psi$ is a simple root of $f$ in some extension $k\supset\Bbb F_p$, then $\psi$ will be a simple root in every bigger field $k'\supset k$.

In other words, if the roots are simple in the algebraic closure, there is some finite field extension $K\supset\Bbb F_p$ containing all roots of $f$; call the set of all such roots $\Psi$. Then $$f(x)=\prod_{\psi\in\Psi}(x-\psi)\in K[x]\,,$$ and notice that no factor $x-\psi$ has an exponent $e>1$, because you’re assuming that the roots are not multiple.

Say that the above field $K$ has $p^n$ elements… I’m sure you see the rest of the argument.

Answer 2

Since totoro finds my comment unsatisfying, I write a detailed proof here.

Let $K$ be the splitting field of $f$. It is well known that $K$ has $p^n$ elements for some $n$.

Since $K^{\times}$ is an abelian group of order $p^n-1$, for all $a \in K^{\times}$ you have $$a^{p^n-1}=1$$ or, equivalently, $$a^{p^n}-a=0$$ Notice that this holds for $a=0$ too. This means that all elements of $K$ (and in particular the roots of $f$) are roots of $x^{p^n}-x$.

This concludes the proof.