Determining whether $(x^2-3)$ is a maximal ideal in $\mathbb{Z}[X]$

Question

I've got a question regarding abstract algebra and prime/maximal ideals. I need to determine whether $(x^2-3)$ is a maximal or prime ideal in $\mathbb{Z}[X]$. I have not yet been introduced to irreducibility, so I cannot make use of theorems about that.

I'm not quite sure how to get started. I thought maybe I could use long division, and I found we can get polynomials $q(x),r(x)q(x),r(x)$ such that $p(x)=q(x)(x^2-3)+r(x)p(x)=q(x)(x^2-3)+r(x)$ and $r(x)=ax+b$. But now I'm stuck. And I'm fairly certain it's supposed to be prime, so finding zero-divisors in $\mathbb{Z}[X]/(x^2-3)$ for example didn't seem like a good idea either.

Could anybody help me out? Thanks in advance!

Answer 1

I would approach problems like this by using ring homomorphisms, and applying the first isomorphism theorem $$ R/\operatorname{ker}(f)\simeq \operatorname{Im}(f). $$ Recall that

• $I$ is a prime ideal of a commutative ring $R$ iff $R/I$ is an integral domain, and
• $I$ is a maximal ideal of a commutative ring $R$ iff $R/I$ is a field.

Extended hints:

1. Can you show that the mapping $f: p(x)\mapsto \overline{p(0)}$ is a surjective homomorphism from $\Bbb{Z}[x]$ to $\Bbb{Z}_3$? Looks like $x^2-3$ is in the kernel, but does it generate the kernel?
2. Can you show that the mapping $g: p(x)\mapsto p(\sqrt3)$ is a homomorphism from $\Bbb{Z}[x]$ to $\Bbb{R}$? Use long division to prove that this time $x^2-3$ generates the kernel (you will also need the fact that $\sqrt3$ is irrational). Can you show that the image of this ring homomorphism is an integral domain?

Answer 2

The maximal ideals of $\mathbf Z[X]$ are known: they're generated by $(p, f(X)$, where $p$ is a prime, and $f$ is a polynomial irreducible modulo $p$.