Difference between revisions of "Schonemann's criterion"

m (Fixed several typos where f's needed to be g's.)
Line 2: Line 2:
 
* <math>f(x)</math> is monic
 
* <math>f(x)</math> is monic
 
* <math>g(x), h(x)\in \mathbb{Z}[x]</math>, a prime <math>p</math> and an integer <math>n</math> such that <math>f(x)=g(x)^n+ph(x)</math>
 
* <math>g(x), h(x)\in \mathbb{Z}[x]</math>, a prime <math>p</math> and an integer <math>n</math> such that <math>f(x)=g(x)^n+ph(x)</math>
* <math>g(x) \pmod{p}</math> is an irreducible polynomial in <math>\mathbb{F}_p</math> and does not divide <math>h(x) \pmod{p}</math>
+
* <math>g(x) \pmod{p}</math> is a irreducible polynomial in <math>\mathbb{F}_p</math> and does not divide <math>h(x) \pmod{p}</math>
 
then <math>f(x)</math> is irreducible.
 
then <math>f(x)</math> is irreducible.
  

Revision as of 12:25, 14 June 2021

If $f(x)\in \mathbb{Z}[x]$

  • $f(x)$ is monic
  • $g(x), h(x)\in \mathbb{Z}[x]$, a prime $p$ and an integer $n$ such that $f(x)=g(x)^n+ph(x)$
  • $g(x) \pmod{p}$ is a irreducible polynomial in $\mathbb{F}_p$ and does not divide $h(x) \pmod{p}$

then $f(x)$ is irreducible.

Proof

We know that $f(x)$ is monic, so deg $f=n$ deg$g$ and we may assume that $g(x)$ is monic. Assume $f(x)=p(x)q(x)$, where $p(x), q(x)\in \mathbb{Z}[x]$. Since $f(x)=p(x)q(x) \pmod{p}$, we get $\overline{F}=f(x) \pmod{p}$, so $g(x)^n \pmod{p}=\overline{P}\cdot \overline{Q}$. Therefore, we have $p(x)=g(x)^r+pP_1(x)$ and $q(x)=g(x)^{n-r}+pQ_1(x)$ for some $P_1(x)$ and $Q_1(x)$. Therefore, \[g(x)^n+ph(x)=f(x)=p(x)q(x)=g(x)^n+p(P_1(x)g(x)^{n-r}+Q_1(x)g(x)^r+pP_1(x)Q_1(x))\] This means that $h(x)=P_1(x)g(x)^{n-r}+Q_1(x)g(x)^r+pP_1(x)Q_1(x)$, which means that $g(x)\pmod{p}\vert h(x)\pmod{p}$, a contradiction. This means that $f(x)$ is irreducible.

This article is a stub. Help us out by expanding it.

See also Eisenstein's criterion.