Difference between revisions of "Schonemann's criterion"
(A Proof of Schonemann's) |
m (Fixed several typos where f's needed to be g's.) |
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* <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> | + | * <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> |
then <math>f(x)</math> is irreducible. | then <math>f(x)</math> is irreducible. | ||
==Proof== | ==Proof== | ||
− | We know that <math>f(x)</math> is monic, so deg <math>f=n</math> deg<math>g</math> and 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 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, |
− | <cmath>g(x)^n+ph(x)=f(x)=p(x)q(x)=g(x)^n+p(P_1(x) | + | <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) | + | 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}} | {{stub}} | ||
See also [[Eisenstein's criterion]]. | See also [[Eisenstein's criterion]]. |
Revision as of 23:10, 25 November 2019
If
is monic
, a prime
and an integer
such that
is an irreducible polynomial in
and does not divide
then is irreducible.
Proof
We know that is monic, so deg
deg
and we may assume that
is monic. Assume
, where
. Since
, we get
, so
. Therefore, we have
and
for some
and
. Therefore,
This means that
, which means that
, a contradiction. This means that
is irreducible.
This article is a stub. Help us out by expanding it.
See also Eisenstein's criterion.