Difference between revisions of "Ostrowski's criterion"

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Ostrowski's Criterion states that:
 
Ostrowski's Criterion states that:
  
Left <math>f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in \mathbb{Z}[x]</math>. If <math>a_0</math> is a prime and
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Let <math>f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in \mathbb{Z}[x]</math>. If <math>a_0</math> is a prime and
 
<cmath>|a_0|>|a_n|+|a_{n-1}|+\cdots+|a_1|</cmath>
 
<cmath>|a_0|>|a_n|+|a_{n-1}|+\cdots+|a_1|</cmath>
 
then <math>f(x)</math> is irreducible.
 
then <math>f(x)</math> is irreducible.

Latest revision as of 11:23, 15 June 2021

Ostrowski's Criterion states that:

Let $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in \mathbb{Z}[x]$. If $a_0$ is a prime and \[|a_0|>|a_n|+|a_{n-1}|+\cdots+|a_1|\] then $f(x)$ is irreducible.

Proof

Let $\phi$ be a root of $f(x)$. If $|\phi|\leq 1$, then \[|a_0|=|a_1\phi+\cdots+a_n\phi^n|\leq |a_1|+\cdots+|a_n|\] a contradiction. Therefore, $|\phi|>1$.

Suppose $f(x)=g(x)h(x)$. Since $f(0)=a_0$, one of $g(0)$ and $h(0)$ is 1. WLOG, assume $g(0)=1$. Then, let $g_n$ be the leading coefficient of $g(x)$. If $\phi_1,\phi_2,\cdots,\phi_r$ are the roots of $g(x)$, then $1<|\phi_1\phi_2\cdots \phi_n|=\frac{1}{|b|}\leq 1$. This is a contradiction, so $f(x)$ is irreducible.

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