Difference between revisions of "Schonemann's criterion"
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==Proof== | ==Proof== | ||
− | We know that <math>f(x)</math> is monic, so | + | 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 17:03, 31 October 2023
If
is monic
, a prime
and an integer
such that
is a irreducible polynomial in
and does not divide
then is irreducible.
Proof
We know that is monic, so
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.