All prime ideals in ring of integers of a bi-quadratic field I want to find all the prime ideals of the ring of integers of bi-quadratic fields. As I know, every prime ideal $\mathcal{P}$ of an algebraic number field $K = \mathbb{Q}(\alpha)$ lies over an ideal generated by a prime $p$ in $\mathbb{Z}$. And, if the prime $p$ in $\mathbb{Z}$ divides $[\mathcal{O}_K : \mathbb{Z}[\alpha]]$ then from Prime ideals of the ring of integers of an algebraic number field I got explicitly all primes $\mathcal{P}$ in $\mathcal{O}_K$ that lies over $p$. Now my first question is how can I find the other prime ideals $\mathcal{P’}$ in $\mathcal{O}_K$ ?
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And my second question is that for these type of prime ideals $\mathcal{P’}$ how the field $\mathcal{O}_{K}/ \mathcal{P’}$ looks like?(as I know, for other cases $\mathcal{O}_{K}/ \mathcal{P}$ is isomorphic to some finite field).
For this, I have to calculate first [$ \mathcal{O}_K : Z[\alpha] $] for any bi-quadratic field $ K = \mathbb{Q}(\alpha) $. How can I do this?
 A: Question: "I got explicitly all primes P in OK that lies over p. Now my first question is how can I find the other prime ideals P′ in OK? And my second question is that for these type of prime ideals P′ how the field OK/P′ looks like?(as I know, for other cases OK/P is isomorphic to some finite field)."
Answer: If $K$ is any number field and $A$ is the ring of integers in $K$, the inclusion map $\mathbb{Z} \subseteq A$ is integral. Any non-zero prime ideal $\mathfrak{m} \subseteq A$ is maximal and $\mathfrak{m} \cap \mathbb{Z}=(p)$ for some prime number $p$. The field extension
$$F1.\text{  }\mathbb{F}_p \subseteq A/\mathfrak{m}$$
is finite hence there is an integer $r \geq 1$ with $A/\mathfrak{m} \cong \mathbb{F}_{p^r}$. Note that since $(p):=\mathfrak{m} \cap \mathbb{Z}$ it follows there is an inclusion $(p) \subseteq \mathfrak{m}$ and a well defined injective map
$$\phi: \mathbb{F}_p \rightarrow A/\mathfrak{m}$$
Hence any prime in $A$ "lies over" a prime ideal in $\mathbb{Z}$ and the isomorphism $F1$ holds for all maximal ideals $\mathfrak{m} \subseteq A$ (the prime number $p$ depends on the ideal $\mathfrak{m}$).
