Difference between revisions of "Schonemann's criterion"

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.

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@@ Line 2: / Line 2: @@
 * <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) \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.

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Difference between revisions of "Schonemann's criterion"

Revision as of 12:25, 14 June 2021

Proof