Suppose we have a field $F$ of characteristic $p$ and a degree $n$ irreducible polynomial $f \in F[x]$.

One can find a root $\alpha \notin F$ of $f$ in a field extension $E$
of $F$ by Kronecker’s Theorem. Then $f(\alpha) = 0$. **How about the
other $n - 1$ roots of $f$?**

Half a month ago, I *couldn’t* find out the answer directly — I used
the fact that all finite fields of order $p^n$ were isomorphic to
$\F_{p^n}$, which was the collection of roots of $x^{p^n} - x$ in
$\Z_p$ in $\overline{\Z_p}$. However, this makes use of *too many*
abstract facts.

Yesterday night, by computing the $p$-th power of an element $\beta$ in $E$, I finally know the direct way of finding the other $n - 1$ roots of $f$ in $E$. Let $b_0,\dots,b_n \in F$ such that

Compute the $p$-th power of $\beta$.

Case 1: $\exists k_j = p$, then $k_i = 0 \,\forall i \ne j$.

Note that one can prove that $\forall b \in F, b^p = b$ by induction.

Case 2: $k_i \ne p \,\forall i = 0,\dots,n$. Since $p$ is a prime,

Since one only has $n$ choices of $k_0,\dots,k_{n - 1}$ which satisfy \eqref{eq:cond}, we conclude that

Since $\alpha$ is a root of $f$ in $E$, $(f(\alpha))^p = 0$. Replacing “$n - 1$” by “$n$” in the derivation of \eqref{eq:powp}, one gets

Therefore, *without* learning induction, one can sense that
$f(\alpha^{p^m}) = 0 \,\forall m \in \N$. That’s not the end. Since
the degree of $f$ is $n$, we expect to that the number of roots of $f$
is *finite*. Since we expect $n$ roots of $f$, we hope that
$\alpha^{p^m}$ will repeat itself for sufficiently large $m$. This
hope comes true due to Lagrange’s Theorem — $|F^\times| = p^n -
1$, so $\alpha^{p^n - 1} = 1$.

~~
Unluckily, I’ve just found out that $f$ ~~*has* some root $\alpha’$
which *doesn’t* hit either one of
$\alpha,\alpha^p,\alpha^{p^2},\dots,\alpha^{p^{n - 1}}$. For example,
if one sets $f(x) := x^{p^2} - x \in \Z_p[x]$ and let $\alpha$ be a
root of $f$ in $\overline{\Z_p}$, then it’s trivial that
$\alpha^{p^2} = \alpha$, thus the collection of the $p^m$-th power of
$\alpha$ is just $\left\{ \alpha,\alpha^p \right\}$. (i.e.
$\left\{ \alpha^{p^m} \mid m \in \N \right\} = \left\{
\alpha,\alpha^p \right\}$) Nevertheless, $f$ should have $p^2$ roots
in $\overline{\Z_p}$.

Hence, I *didn’t* succeed in answering **the above bolded question**,
but I still learn something about the roots of an irreducible
polynomial in an algebraic extension.

(Edited on MAR 28, 2015)

With the results from perfect fields, we *immediately* know that $f$
has $n$ different roots.