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


*

*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?

*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?

A: 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 $p$.
