Difference between revisions of "Ostrowski's criterion"
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Ostrowski's Criterion states that: | Ostrowski's Criterion states that: | ||
− | + | 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. | ||
− | Proof | + | ==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> | <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>. | a contradiction. Therefore, <math>|\phi|>1</math>. |
Latest revision as of 11:23, 15 June 2021
Ostrowski's Criterion states that:
Let . If
is a prime and
then
is irreducible.
Proof
Let be a root of
. If
, then
a contradiction. Therefore,
.
Suppose . Since
, one of
and
is 1. WLOG, assume
. Then, let
be the leading coefficient of
. If
are the roots of
, then
. This is a contradiction, so
is irreducible.
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