Checking whether $x^2-5$ is prime but not maximal I want to find an example of a prime ideal that is not maximal. I thought about $x^2-5$. We know that $Z[\sqrt{-5}]\cong Z[x]/(x^2-5)$ is an integral domain, therefore is $x^2-5$ prime. However, I would need to show now that $Z[\sqrt{-5}]$ is not a field but I don't know where to go from here. Is this a right approach to show that $x^2-5$ is a prime but not maximal? If so, how would finish the justification? Or should I consider finding another example?
 A: This is a good approach. You could carry on by showing that $(x^2 - 5)$ is the kernel of the homomorphism from $\mathbb{Z}[x]$ to $\mathbb{R}$ that maps $x$ to $\sqrt{5}$. Then show that the image of this homomorphism (which is isomorphic to $\mathbb{Z}[x]/(x^2-5)$) is the subring $A$ say of $\mathbb{R}$ comprising all numbers of the form $m + n\sqrt{5}$ where $m, n \in \mathbb{Z}$. You can then show that $A$ is not a field, e.g., by showing that $2$ is not invertible in $A$ (which is most easily done using the norm on $A$, see https://en.wikipedia.org/wiki/Quadratic_integer).
A: You can prove it is not maximal by finding a prime $p$ such that $x^2-5$  is irreducible modulo $p$. Thus we'll have 
$$(x^2-5)\varsubsetneq(p,x^2-5)$$
and this last one  is indeed a maximal ideal since $\mathbf Z[x](p,x^2-5)\simeq\mathbf F_p[x]/(x^2-5)\simeq\mathbf F_{p^2}$.
For instance $x^2-5$ is irreducible modulo $3$, since the only squares mod. $3$ are $0,1$.
A: The direction you are trying is possible. However, it is important to notice that the zero ideal is prime in every integral domain, and hence...
