Ostrowski's criterion: Difference between revisions

Revision as of 15:30, 14 August 2018

Ostrowski's Criterion states that:

Left $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in \mathbb{Z}[x]$ . 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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@@ Line 5: / Line 5: @@
 then <math>f(x)</math> is irreducible.
-Proof: Let <math>\phi</math> be a root of <math>f(x)</math>. If <math>|\phi|\leq 1</math>, then
+==Proof==
+Let <math>\phi</math> be a root of <math>f(x)</math>. If <math>|\phi|\leq 1</math>, then
 <cmath>|a_0|=|a_1\phi+\cdots+a_n\phi^n|\leq |a_1|+\cdots+|a_n|</cmath>
 a contradiction. Therefore, <math>|\phi|>1</math>.

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Ostrowski's criterion: Difference between revisions

Revision as of 15:30, 14 August 2018

Proof