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Difference between revisions of "Schonemann's criterion"
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==Proof==
 
==Proof==
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We know that <math>f(x)</math> is monic, so deg <math>f=n</math> deg<math>g</math> and we may assume that <math>g(x)</math> is monic.  Assume <math>f(x)=p(x)q(x)</math>, where <math>p(x), q(x)\in \mathbb{Z}[x]</math>. Since <math>f(x)=p(x)q(x) \pmod{p}</math>, we get <math>\overline{F}=f(x) \pmod{p}</math>, so <math>g(x)^n \pmod{p}=\overline{P}\cdot \overline{Q}</math>. Therefore, we have <math>p(x)=g(x)^r+pP_1(x)</math> and <math>q(x)=g(x)^{n-r}+pQ_1(x)</math> for some <math>P_1(x)</math> and <math>Q_1(x)</math>. Therefore,
+
We know that <math>f(x)</math> is monic, so <math>\deg f=n\deg g</math> and we may assume that <math>g(x)</math> is monic.  Assume <math>f(x)=p(x)q(x)</math>, where <math>p(x), q(x)\in \mathbb{Z}[x]</math>. Since <math>f(x)=p(x)q(x) \pmod{p}</math>, we get <math>\overline{F}=f(x) \pmod{p}</math>, so <math>g(x)^n \pmod{p}=\overline{P}\cdot \overline{Q}</math>. Therefore, we have <math>p(x)=g(x)^r+pP_1(x)</math> and <math>q(x)=g(x)^{n-r}+pQ_1(x)</math> for some <math>P_1(x)</math> and <math>Q_1(x)</math>. Therefore,
 
<cmath>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))</cmath>
 
<cmath>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))</cmath>
 
This means that <math>h(x)=P_1(x)g(x)^{n-r}+Q_1(x)g(x)^r+pP_1(x)Q_1(x)</math>, which means that <math>g(x)\pmod{p}\vert h(x)\pmod{p}</math>, a contradiction. This means that <math>f(x)</math> is irreducible.
 
This means that <math>h(x)=P_1(x)g(x)^{n-r}+Q_1(x)g(x)^r+pP_1(x)Q_1(x)</math>, which means that <math>g(x)\pmod{p}\vert h(x)\pmod{p}</math>, a contradiction. This means that <math>f(x)</math> is irreducible.

Latest revision as of 16:03, 31 October 2023

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.

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See also Eisenstein's criterion.

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