special reducible polynomials over a field with positive characteristic

Let $k$ be an arbitrary field such that $\mathrm{char}(k)=p>0$. We will assume that $0\notin \mathbb{N}$.

Proposition. Let $m\in \mathbb{N}$. Then for any $a\in k$ the polynomial $W(X)=X^{p^m}-a$ is reducible if and only if there exist $c\in k$ and $n\in \mathbb{N}$ such that $c^{p^n}=a$. Moreover the factorization of $W(X)$ is given by the formula

$$W(X)={(X^{p^{m-n}}-c)}^{p^n},$$

where $n$ is a maximal natural number such that $0\le n\le m$ and $a=c^{p^n}$ for some $c\in k$.

Proof. “$\Leftarrow$” Assume that $a=c^{p^n}$ for some $c\in k$ and $n\in \mathbb{N}$. It is well known that if $\mathrm{char}(k)=p>0$ and $t\in \mathbb{N}$ then for any $x,y\in k$ we have ${(x+y)}^{p^t}=x^{p^t}+y^{p^t}$. Therefore

$$W(X)=X^{p^m}-a=X^{p^m}-c^{p^n}={(X^{p^{m-1}})}^p-{(c^{p^{n-1}})}^p={(X^{p^{m-1}}-c^{p^{n-1}})}^p={(V(X))}^p.$$

Note that $p^m>\mathrm{deg}(V(X))=p^{m-1}>0$ and therefore $W(X)$ is reducible. $\mathrm{\square }$

“$\Rightarrow$” Assume that $W(X)$ is reducible. Therefore there exist $V(X),U(X)\in k[X]$ such that $W(X)=V(X)\cdot U(X)$ and both $\mathrm{deg}(V(X))>0$ and $\mathrm{deg}(U(X))>0$.

Recall that there exists an algebraically closed field $\overline{k}$ such that $k$ is a subfield of $\overline{k}$ (generally it is true for any field). Therefore there exists $c_0\in \overline{k}$ such that $c_0^{p^m}=a$ and thus we have:

$$W(X)=X^{p^m}-a=X^{p^m}-c_0^{p^m}={(X-c_0)}^{p^m}$$

in $\overline{k}[X]$. Now $V(X)\cdot U(X)=W(X)={(X-c_0)}^{p^m}$ and since $\overline{k}[X]$ is a unique factorization domain then for $n=\mathrm{deg}(V(X))>0$ we have:

$$V(X)={(X-c_0)}^n.$$

But $V(X)\in k[X]$ (the factorization was assumed to be over $k$) and therefore $c_0^n\in k$. It is easy to see that since $c_0^n\in k$ and $c_0^{p^m}\in k$ then $c_0^{\gcd (n,p^m)}\in k$, but $\gcd (n,p^m)=p^s$ for some $s\in \mathbb{N}$. Thus if we put $c=c_0^{p^s}$ we gain that $c^{p^{m-s}}=a$. But $m>s$ (since because we assumed that both $\mathrm{deg}(V(X))>0$ and $\mathrm{deg}(U(X))>0$), which completes the proof of the first part. $\mathrm{\square }$

Now let $n\in \mathbb{N}$ be a maximal natural number such that $n\le m$ and $a=c^{p^n}$ for some $c\in k$. Then we have

$$W(X)={(X^{p^{m-n}}-c)}^{p^n}.$$

Note that the polynomial $X^{p^{m-n}}-c$ is irreducible. Indeed, assume that $X^{p^{m-n}}-c$ is reducible. Then (due to first part of the proposition) $c=u^{p^k}$ for some $k\in \mathbb{N}$ and $u\in k$. But then $a={(u^{p^k})}^{p^n}=u^{p^{n+k}}$. Contradiction, since $n+k>n$ and $n$ was assumed to be maximal. $\mathrm{\square }$