Let consider the integer of $\mathbb{Q}(\sqrt{d})$, where $d$ is square free, i.e. if $d=4k+3$ or $d=4k+2$ it is $\mathcal{O}=\mathbb{Z}+\mathbb{Z}\sqrt{d}$. If $d=4k+1$, it is $\mathcal{O}=\mathbb{Z}+\mathbb{Z}\frac{1+\sqrt{d}}{2}$.

If $\mathcal{P}$ is a non-zero prime ideal of $\mathcal{O}$, then we have $\mathcal{P}\cap\mathbb{Z}=p\mathbb{Z}$ for some prime number $p$. What I want to prove is $\mathcal{O}/\mathcal{P}$ is a field, actually an at-most-two-degree extension of $\mathbb{F}\_p$.

If we pick $[x]\neq[0]\in\mathcal{O}/\mathcal{P}$, we need to show $x\notin \mathcal{P}$, there exists $y\in \mathcal{O}$, such that $[yx]=1 \text{ mod } p$. 

Further, if we consider the norm of $x=a+b\sqrt{d}$, how can we deduce $n(x)=\bar{x}x$ and $p$ are coprime? If we can deduce this property then $y:=c(a-b\sqrt{d})$ can be an inverse of $x$.