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 $0\not\in\mathbb{N}$.

. 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 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. $\square$

Title special reducible polynomials over a field with positive characteristic SpecialReduciblePolynomialsOverAFieldWithPositiveCharacteristic 2013-03-22 18:31:05 2013-03-22 18:31:05 joking (16130) joking (16130) 10 joking (16130) Theorem msc 13F07