Incomparability theorem

Remarks

Let $\mathfrak{p} = \mathfrak{q} \cap A$ and localize at $\mathfrak{p}$. Then $B_{\mathfrak{p}}/\mathfrak{q} B_{\mathfrak{p}}$ is a domain integral over the field $A_{\mathfrak{p}}/\mathfrak{p} A_{\mathfrak{p}}$, so it is a field by integral extensions of fields. Hence $\mathfrak{q} B_{\mathfrak{p}}$ is maximal, and the larger prime $\mathfrak{q}' B_{\mathfrak{p}}$ must equal it, which gives $\mathfrak{q} = \mathfrak{q}'$.

A consequence: a prime $\mathfrak{q}$ of $B$ is maximal if and only if $\mathfrak{q} \cap A$ is maximal in $A$.

Example: in $\Z \subseteq \Z[i]$, the primes $0 \subsetneq (2 + i)$ contract to $0$ and $5\Z$, which differ, as the theorem demands.

Integrality is needed: for $k[x] \subseteq k[x, y]$, the primes $0 \subsetneq (y)$ both contract to $0$.

Going deeper

For a student

Incomparability says that chains of primes cannot get longer when one passes to an integral extension. With going-up it gives $\dim B = \dim A$. It is also the reason that the maximal ideals of $B$ are exactly the primes over maximal ideals of $A$.