Prime/Maximal Ideals in $\mathbb{Z}[\sqrt d]$

Question

Let $d \in \mathbb{Z}$ be a square free integer. $R=\mathbb{Z}[\sqrt d]$ =$\{a+b\sqrt d | a,b \in \mathbb{Z} \}$. Overall, I'm trying to show that every prime ideal $P \subset R$ is a maximal ideal.

So far I showed that $I \subset R$ is finitely generated. $I=\{(x,s+y \sqrt d)\}$ And now I'm trying to show that R/P, for some prime ideal is a finite ring with no zero divisors. From there it would follow that R/P is a field and any prime ideal is maximal.

I know R could also be written as $R= \mathbb{Z}[x]/(x^2-d)$. How could I show that R/P is a quotient of $\mathbb{Z}/n\mathbb{Z}[x]/(x^2-d)$?

Answer 1

The zero ideal is prime, but not maximal. If $P$ is a nonzero prime ideal then $R/P$ is finite, and an integral domain. All finite integral domains are fields...

Answer 2

In $\mathbb{Z}[\sqrt{d}]$ you can show that $0 \ne a+b \sqrt{d} \in P \implies (a+b \sqrt{d})(a-b \sqrt{d})= a^2-b^2 d \ne 0 \in P$ and $\mathbb{Z}[\sqrt{d}]/(n) = \{ u+v \sqrt{d}, (u,v) \in \mathbb{Z}/n\mathbb{Z}\}$ is a finite ring for any integer $n \ne 0$.

Thus $\mathbb{Z}[\sqrt{d}]/P = \mathbb{Z}[\sqrt{d}]/(n)/P$ is a finite ring, for any proper ideal $P$.

In general in $\mathcal{O}_K$, you'll show directly that $\mathcal{O}_K/(\alpha)\cong \mathbb{Z}^N / A \mathbb{Z}^N$ is finite, where $A \in \mathbb{Z}^{N \times N}$ is the (inversible) matrix representing the multiplication by $A$, or that $\alpha \in P \implies N_{K/\mathbb{Q}}(\alpha)\in P$, where $|N_{K/\mathbb{Q}}(\alpha)| = |\det(A)|$.