In the excellent expository papers of Keith Conrad, I stuck at a proof of a proposition in the Finite Fields.
Proposition.
Let $\pi(X)\in \Bbb Z_p[X]$ be an irreducible polynomial of $\Bbb Z_p[X]$ of degree $\deg \pi(X)=d$ and $K/\Bbb Z_p$ an extension of $\Bbb Z_p$, in which $\pi(X)$ has a root $\alpha\in K$. Then:
- $\pi(X)\in \Bbb Z_p[X]$ has roots $\alpha,\alpha^{p},\alpha^{p^2},\dots,\alpha^{p^{d-1}} \in K$ and all these $d$ roots are distinct.
- In particular, $\alpha^{p^i}=\alpha^{p^j} \iff i\equiv j\bmod d,$ for any $i,j\in \Bbb N$.
Proof.
By a Lemma, $\pi(X^{p^m})=\pi(X)^{p^m}$ and by hypothesis $\pi(\alpha)=0$. So, $\pi(\alpha^{p^m})=\pi(\alpha)^{p^m} \iff \pi(\alpha^{p^{p^m}})=0$. So, $\alpha,\alpha^{p},\alpha^{p^2},\dots,\alpha^{p^{d-1}}\in K$ are roots of $\pi(X)\in \Bbb Z_p[X]$.
We observe that once we reach the $p^d$-th power of $\alpha$, that is $\alpha^{p^d}$, we have cycled back to start. Indeed,
\begin{align*}
d|d \quad & \iff \quad \pi(X) | X^{p^d}-X \\
& \iff \quad X^{p^d}-X=\pi(X)q(X),\ q(X)\in \Bbb Z_p[X] \\
& \iff \quad X^{p^d}=X+\pi(X)q(X),\ q(X)\in \Bbb Z_p[X] \\
&\implies \quad \alpha^{p^d}=\alpha + \pi(\alpha)q(\alpha) \\
& \iff \quad \alpha^{p^d}= \alpha,
\end{align*}
by substituting $\alpha$ for $X$ and since $\pi(\alpha)=0$.
Now, we want to show that $\alpha^{p^i}=\alpha^{p^j} \iff i\equiv j\bmod d$, where $i,j\in \Bbb N$. We assume without loss of generality that $i\leq j\iff j=i+k$, for some $k\in \Bbb N$. Then,
\begin{align*}
\alpha^{p^i}=\alpha^{p^j} \quad \iff \quad & \alpha^{p^i}=(\alpha^{p^k})^{p^i} \\
\quad \iff \quad & (\alpha^{p^k})^{p^i}-\alpha^{p^i}=0 \\
\quad \iff \quad & (\alpha^{p^k}-\alpha)^{p^i}=0 \\
\quad \iff \quad & \alpha^{p^k}-\alpha=0.
\end{align*}
This paper continues by claiming that
\begin{align*}
\alpha^{p^k}=\alpha \iff \pi(X)|X^{p^k}-X, \text{ inside } \Bbb Z_p[X]
\tag{*}
\end{align*}
Questions.
I can not see why the equivalence (*) is true (at least the $\Longrightarrow$ part).
Could you please give me a hand at this point?
Also, could you please explain why this equivalence reassures us that all these roots are distinct?
Best Answer
We know that $\alpha$ is a root of $\pi(x)$. Write $f(x) = x^{p^k}-x$.
Firstly, if $\pi(x) | f(x)$, then we may write $$f(x) = \pi(x) g(x)$$ for some $g(x) \in \mathbb{F}_p[x]$. Then we have $\alpha^{p^k} - \alpha = \pi(\alpha) g(\alpha) = 0$
Conversely, if $\alpha^{p^k} - \alpha = 0$, then $\alpha$ is a root of $f(x)$. But $\pi(x)$ is the minimal polynomial for $\alpha$ (up to multiplication by a unit to ensure that it is monic), hence $\pi(x)$ divides $x^{p^k} - x$.
This is because we can apply the division algorithm so that $f(x) = h(x) \pi(x) + r(x) $ where $r(x) $ has degree $ < \deg(\pi(x))$. Then $$0 = f(\alpha) = h(\alpha) \pi(\alpha) + r(\alpha) = r(\alpha)$$ But $r(x)$ has degree less than $\deg(\pi(x))$, hence $r(x) = 0$.
Notice that there is nothing special about the polynomial $f$.