# 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 ${c}_{0}^{\mathrm{gcd}(n,{p}^{m})}\in k$, 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 $$ 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. $\mathrm{\square}$

Title | special reducible polynomials over a field with positive characteristic |
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Canonical name | SpecialReduciblePolynomialsOverAFieldWithPositiveCharacteristic |

Date of creation | 2013-03-22 18:31:05 |

Last modified on | 2013-03-22 18:31:05 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 10 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 13F07 |