Polynomials in Z modulus 9

  • Thread starter jeffreydk
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  • #1
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I'm trying to figure out how to prove that every polynomial in [itex]\mathbb{Z}_9[/itex] can be written as the product of two polynomials of positive degree (except for the constant polynomials [3] and [6]). This basically is just showing that the only possible irreducible polynomials in [itex]\mathbb{Z}_9[/itex] are the constant ones, [3] and [6], and that all the other constant polynomials can be written as the product of polynomials with degrees greater than 0, kind of like how [1] can be written as,

[tex]([3]x+[1])([6]x+[1])=[0]x^2+[3]x+[6]x+[1]=[1][/tex]

but I'm a bit lost on how to show it all, because it's not a field so the theorems I've been studying regarding irreducibility in polynomials don't apply to such a situation. Thanks, any help is greatly appreciated.
 

Answers and Replies

  • #2
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If ##f(x)\in \mathbb{Z}_9[x]## is irreducible, and we have a decomposition ##f(x)=f(x)\cdot 1=f(x)(3x+1)(6x+1)## then either ##f(x)(3x+1)\in \mathbb{Z}_9## or ##f(x)(6x+1)\in \mathbb{Z}_9##. In both cases we get ##3f(x) =3a## for an ##a\in \mathbb{Z}_9\,.## Thus ##3(f(x)-a) = 0## and ##9\,|\,3(f(x)-a)## which means ##3\,|\,(f(x)-a)## and ##f(x)=3n+a## for some ##n\in \mathbb{Z}##. Hence ##f(x)## is constant, i.e. all irreducible polynomials are constant.
 

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