Schonemann's criterion

Revision as of 15:46, 14 August 2018 by Naman12 (talk | contribs) (A Proof of Schonemann's)

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 $f(x)=g(x)^n+ph(x)$
  • $f(x) \pmod{p}$ is an 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 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, \[g(x)^n+ph(x)=f(x)=p(x)q(x)=g(x)^n+p(P_1(x)f(x)^{n-r}+Q_1(x)f(x)^r+pP_1(x)Q_1(x))\] This means that $h(x)=P_1(x)f(x)^{n-r}+Q_1(x)f(x)^r+pP_1(x)Q_1(x)$, which means that $f(x)\pmod{p}\vert h(x)\pmod{p}$, a contradiction. This means that $f(x)$ is irreducible.

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See also Eisenstein's criterion.