Suppose we have a field $F$ of characteristic $p$ and a degree $n$
irreducible polynomial $f \in F[x]$.
\[
f(x) = \sum\limits_{i = 0}^n a_i x^i \text{, where } a_i \in F
\,\forall i = 0,1,\dots,n.
\]
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
\[
\beta = \sum\limits_{i = 0}^{n - 1} b_i \alpha^i.
\]
Compute the $p$-th power of $\beta$.
\[
\beta^p = \left(\sum\limits_{i = 0}^{n - 1} b_i \alpha^i \right)^p =
\sum\limits_{k_0,\dots,k_{n
- 1}} \frac{p!}{\prod\limits_{i = 0}^{n - 1} k_i!} \prod\limits_{i =
- 0}^{n - 1} (b_i
\alpha^i)^{k_i},
\]
\begin{equation}
\text{ where } 0 \le k_i \le p \,\forall i = 0,\dots,n \text{ and }
\sum\limits_{i = 0}^{n - 1} k_i = p. \label{eq:cond}
\end{equation}
Case 1: $\exists k_j = p$, then $k_i = 0 \,\forall i \ne j$.
\[
\frac{p!}{\prod\limits_{i = 0}^{n - 1} k_i!} \prod\limits_{i = 0}^{n -
1} (b_i \alpha^i)^{k_i} = 1 \cdot (b_j \alpha^j)^{k_j} = b_j
\alpha^{pj}
\]
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,
\[
\frac{p!}{\prod\limits_{i = 0}^{n - 1} k_i!} = 0 \text{ in } \Z_p
\]
Since one only has $n$ choices of $k_0,\dots,k_{n - 1}$ which
satisfy \eqref{eq:cond}, we conclude that
\begin{equation}
\beta^p = \left(\sum\limits_{i = 0}^{n - 1} b_i \alpha^i \right)^p =
\sum\limits_{i = 0}^{n - 1} b_i^p \alpha^{pi} \label{eq:powp}
\end{equation}
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
\[
(f(\alpha))^p = \left(\sum\limits_{i = 0}^n a_i \alpha^i \right)^p =
\sum\limits_{i = 0}^n a_i \alpha^{pi} = f(\alpha^p) = 0
\]
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.
