Let $k$ be an arbitrary field such that $\mathrm{char}(k) = p> 0$ . We will assume that $0\not\in\mathbb{N}$ .

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 $0\leq n\leq m$ 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. $\square$

``$\Rightarrow$ '' 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 $c_0^{\mathrm{gcd}(n,p^m)}\in k$ , but $\mathrm{gcd}(n,p^m)=p^s$ 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 $n<p^m$ because we assumed that both $\mathrm{deg}(V(X))>0$ and $\mathrm{deg}(U(X))>0$ ), which completes the proof of the first part. $\square$

Now let $n\in\mathbb{N}$ be a maximal natural number such that $n\leq m$ 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. $\square$