Difference between revisions of "Schonemann's criterion"

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.

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

@@ Line 1: / Line 1: @@
-For a polynomial <math>q</math> denote by <math>q^*</math> the residue of <math>q</math> modulo <math>p</math>.
+If <math>f(x)\in \mathbb{Z}[x]</math>
-Suppose the following conditions hold:
+* <math>f(x)</math> is monic
-* <math>k=f^n+pg</math> with <math>n\geq 1</math>, <math>p</math> prime, and <math>f,g\in \mathbb{Z}[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>\text{deg}(f^n)>\text{deg}(g)</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>
-* <math>k</math> is primitive.
+then <math>f(x)</math> is irreducible.
-* <math>f^*</math> is irreducible in <math>\mathbb{F}_p[X]</math>.
-* <math>f^*</math> does not divide <math>g^*</math>.
+==Proof==
-Then <math>k</math> is irreducible in <math>\mathbb{Q}[X]</math>.
+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>
+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.
+{{stub}}
 See also [[Eisenstein's criterion]].

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Difference between revisions of "Schonemann's criterion"

Latest revision as of 16:03, 31 October 2023

Proof