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.

  • 1
    $\begingroup$ Your proof by contraposition is sound. (You've applied it for both $\Bbb Z/n\Bbb Z$ generally and quartics in $\Bbb F_p$ in particular.) Note that unless you're allowing $p$ to be arbitrary powers of primes instead of just primes, $\Bbb F_p\cong\Bbb Z/p\Bbb Z$ is just a special case of the other scenario. I don't see anything special about $n=p$ prime, at least in the kernels or root implications. $\endgroup$ – anon Oct 6 '12 at 2:35
  • $\begingroup$ Yah I know that $\mathbb{F}_p\cong\mathbb{Z}_p$, I was just criticized previously for using the latter notation when I'm looking at it as a field instead of a group. I'm confused by your mention of quartics however, it's for polynomials of all degree correct? $\endgroup$ – Thoth Oct 6 '12 at 2:41
  • $\begingroup$ Yes. I only mentioned it because you mentioned it. Jeesh! $\endgroup$ – anon Oct 6 '12 at 2:43
  • $\begingroup$ Sorry if I came off as curt, I wasn't intending to be. Ok so what you're saying is that there is nothing special about $n$ being prime, and whether $n$ is prime or not, any level of irreducibility over $\mathbb{Z}_n$ must be mimicked over $\mathbb{Z}$, (assuming the degree of the poly doesn't drop under the homomorphism). $\endgroup$ – Thoth Oct 6 '12 at 3:03
  • $\begingroup$ Indeed, the argument even works when the degree of the polynomial does drop due to the characteristic of the codomain. Contraposition is a powerful logical tool. $\endgroup$ – anon Oct 6 '12 at 3:06

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.

  • $\begingroup$ Ok cool, so it's for reasons of efficiency that we choose $p$ prime, good to see that I'm not going crazy here. $\endgroup$ – Thoth Oct 6 '12 at 2:50

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


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