understanding the definition of Q lies over p

Question

If $A \subset B$ is an extension of Dedekind domains and $p \in A$ is a prime ideal , does $Q_i$ lie over $p$ if $p\cdot B=Q_1^{e_1}\cdot...\cdot Q_r^{e_r}$ for $Q_i\subset B $ prime ideals ?

Answer 1

Remember that, in Dedekind domains, every nonzero prime is maximal. As $$p ⊆ pB = Q_1^{e_1}·…·Q_r^{e_r} ⊆ Q_i,$$ we have $p ⊆ Q_i ∩ A$ for $i = 1, …, r$. As $p$ is maximal, $p = Q_i ∩ A$.