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I'm considering a polynomial $f(x)\in\mathbb{Z}[x]$ and looking at it both in $\mathbb{Z}_n[x]$ ($n$ composite) and in $\mathbb{F}_p[x]$, to gather information about possible roots and irreducibility in $\mathbb{Z}[x]$.

I thought I understood the situation well but it is clear I do not. At first I thought that considering roots/irreducibility in $\mathbb{Z}_n[x]$ gave you nothing. But if there's a homomorphism $\phi:\mathbb{Z}[x]\rightarrow\mathbb{Z}_n[x]$, then if $\alpha$ is a root of $f(x)\in\mathbb{Z}[x]$, then we have for $f(x) = f_0 + f_1x + ... + f_tx^t$:

$0=\phi(0)=\phi(f_0 + f_1\alpha + ... + f_t\alpha^t)=[f_0]_n+[f_1]_n[\alpha]_n + ...+[f_t]_n[\alpha]_n^t$

And thus by the contrapositive, if $f(x)$ has no root in $\mathbb{Z}_n[x]$ then it has no root in $\mathbb{Z}[x]$. Or does the fact that $\mathbb{Z}_n[x]$ is not an integral domain somehow make this not follow? I suspect it does, but I'm having trouble seeing exactly how and why.

As a somewhat related question. How does the homomorphism $\phi:\mathbb{Z}[x]\rightarrow\mathbb{Z}_n[x]$ differ from the homomorphism $\psi:\mathbb{Z}[x]\rightarrow\mathbb{F}_p[x]$? Since these are both surjective homs I thought it would be in the kernel, but the kernels don't seem to be qualitatively different. The theorem in my professor's notes says that irreducibility in $\mathbb{F}_p[x]$ implies irreducibility in $\mathbb{Z}[x]$, so there must be something special about $p$ prime.

Also what about if you have say a 4th degree polynomial which factors into irreducible quadratics in $\mathbb{F}_p[x]$, then it's not irreducible in $\mathbb{F}_p[x]$ but it doesn't have any roots in there, can we say it then has no roots in $\mathbb{Z}[x]$?

As you can see I'm harboring a number of related confusions about what's going on here, if anyone could help enlighten me it would be much appreciated, thanks.

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    lol, we were talking about roots, but we were also talking about reducibility more generally I thought.2012-10-06

2 Answers 2

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Your argument that if the polynomial $P(x)$ with integer coefficients has a root in $\mathbb{Z}$ then it has a root in $\mathbb{Z}_n$ is certainly correct. However, if $p$ is a prime divisor of $n$, it is in general "harder" for $P(x)$ to have a root in $\mathbb{Z}_p$ than to have a root in $\mathbb{Z}_n$. So if we are trying to show that $P(x)$ has no integer roots, we might as well confine attention to prime $n$.

For irreducibility, the same considerations apply.

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    Ok cool, so it's for reasons of efficiency that we choose $p$ prime, good to see that I'm not going crazy here.2012-10-06
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Factorization theory is more complicated in non-domains, e.g. $\rm\:x = (3+2x)(2-3x)\in \Bbb Z_6[x].\:$ Basic notions such as associate and irreducible bifurcate into a few inequivalent notions, e.g. see

When are Associates Unit Multiples?
D.D. Anderson, M. Axtell, S.J. Forman, and Joe Stickles.
Rocky Mountain J. Math. Volume 34, Number 3 (2004), 811-828.

Factorization in Commutative Rings with Zero-divisors.
D.D. Anderson, Silvia Valdes-Leon.
Rocky Mountain J. Math. Volume 28, Number 2 (1996), 439-480