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 ![$f(x)\in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/0/a/0/0a04571f5290305c8b91f732a34451f2457ed370.png)
is monic![$g(x), h(x)\in \mathbb{Z}[x]$](//latex.artofproblemsolving.com/f/1/9/f196024102f0a214d55af2c62687c352a7d81851.png)
, a prime 
and an integer 
such that 
is an irreducible polynomial in 
and does not divide 
![$g(x), h(x)\in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/f/1/9/f196024102f0a214d55af2c62687c352a7d81851.png)
, 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 ![$p(x), q(x)\in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/a/b/f/abf0874c6ad3fe0dd5ea11c1485d502e8e6ec0a1.png)
. Since 
, we get 
, so 
. Therefore, we have 
and 
for some 
and 
. Therefore,
![\[g(x)^n+ph(x)=f(x)=p(x)q(x)=g(x)^n+p(P_1(x)g(x)^{n-r}+Q_1(x)g(x)^r+pP_1(x)Q_1(x))\]](http://latex.artofproblemsolving.com/9/7/8/9782d795c71550a21fe845076d586335ec7ff9f1.png)
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.
