Lying over for integral ring maps

Let $\bar{A}:=A\mathrm{/}\ker f$, let $\pi :A\to \bar{A}$ be the quotient map, and let $\bar{\mathfrak{p}}:=\mathfrak{p}\mathrm{/}\ker f$. By [L6], prime ideals of $\bar{A}$ correspond to prime ideals of $A$ containing $\ker f$, so $\bar{\mathfrak{p}}$ is prime. The map $f$ factors through an injective integral map $\bar{f}:\bar{A}\hookrightarrow B$, so it is enough to find a prime of $B$ contracting to $\bar{\mathfrak{p}}$ and then pull it back through [L6].

Replace $A$ by $\bar{A}$ and write again $\mathfrak{p}$ for the chosen prime. Set $S:=A\setminus \mathfrak{p}$. By [L5], $A\mathrm{/}\mathfrak{p}$ is a domain, so $S$ contains no zero element of $A$; since $A\hookrightarrow B$ is injective, $0\mathrm{\notin }S$ inside $B$ as well. Hence [L9] shows that $S^{-1}B$ is nonzero. By [L2], the map $S^{-1}A\to S^{-1}B$ is integral.

By [L7], prime ideals of $A_{\mathfrak{p}}=S^{-1}A$ correspond to prime ideals of $A$ contained in $\mathfrak{p}$. Therefore every prime of $A_{\mathfrak{p}}$ lies inside $\mathfrak{p}{A}_{\mathfrak{p}}$, so $\mathfrak{p}{A}_{\mathfrak{p}}$ is the unique maximal ideal of $A_{\mathfrak{p}}$. By [L8], the nonzero ring $S^{-1}B$ has a maximal ideal $\mathfrak{n}$. Then [L3] implies that $\mathfrak{n}\cap A_{\mathfrak{p}}=\mathfrak{p}{A}_{\mathfrak{p}}$.

By [L7], the prime ideal $\mathfrak{n}$ of $S^{-1}B$ is $S^{-1}\mathfrak{q}$ for a unique prime ideal $\mathfrak{q}$ of $B$ disjoint from $S$, and its contraction to $A$ is the contraction of $\mathfrak{p}{A}_{\mathfrak{p}}$, namely $\mathfrak{p}$. Therefore $\mathfrak{q}$ lies over $\mathfrak{p}$. Pulling back through step 1.1 gives a prime of $B$ lying over the original prime of $A$.