Can all polynomials in Z modulus 9 be written as the product of two polynomials?

[jeffreydk]

I'm trying to figure out how to prove that every polynomial in \mathbb{Z}_9 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 \mathbb{Z}_9 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,

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

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.

 

[fresh_42]

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.