Lying over theorem

Remarks

One localizes at $\mathfrak{p}$: the ring $B_{\mathfrak{p}}$ is integral over the local ring $A_{\mathfrak{p}}$ and is not zero, since $A_{\mathfrak{p}} \subseteq B_{\mathfrak{p}}$. Take a maximal ideal $\mathfrak{n}$ of $B_{\mathfrak{p}}$. Then $B_{\mathfrak{p}}/\mathfrak{n}$ is a field integral over $A_{\mathfrak{p}}/(\mathfrak{n} \cap A_{\mathfrak{p}})$, so the latter is a field by integral extensions of fields, and $\mathfrak{n} \cap A_{\mathfrak{p}}$ is the maximal ideal $\mathfrak{p} A_{\mathfrak{p}}$. The contraction of $\mathfrak{n}$ to $B$ is the wanted $\mathfrak{q}$.

Example: for $\Z \subseteq \Z[i]$ and $\mathfrak{p} = 5\Z$, the prime $\mathfrak{q} = (2 + i)$ satisfies $\mathfrak{q} \cap \Z = 5\Z$, since $5 = (2 + i)(2 - i)$.

Integrality is needed: for $\Z \subseteq \Q$, the only prime ideal of $\Q$ is $0$, so no prime of $\Q$ lies over $2\Z$. Injectivity is needed too: for the integral map $\Z \to \Z/2\Z$, no prime lies over $3\Z$.

Going deeper

For a student

Lying over is the first of the Cohen–Seidenberg theorems; going-up, incomparability and going-down build on it. In number theory it says that each prime number $p$ is divisible by some prime ideal of the ring of integers of a number field. Together with going-up and incomparability it gives $\dim A = \dim B$ for an integral extension.