[proofplan] We localise the integral extension $A \subset B$ at $\mathfrak{p}$ to reduce the problem to a local ring. The localised extension $A_\mathfrak{p} \subset B_\mathfrak{p}$ is still integral. Any maximal ideal of $B_\mathfrak{p}$ contracts to the unique maximal ideal of $A_\mathfrak{p}$ (by preservation of maximality under integral extensions), and its preimage in $B$ under the localisation map is a prime ideal lying over $\mathfrak{p}$. [/proofplan] [step:Localise the extension at $\mathfrak{p}$ and verify integrality is preserved] Let $S := A \setminus \mathfrak{p}$, which is a multiplicative subset of $A$. Form the localisations $A_\mathfrak{p} := S^{-1}A$ and $B_\mathfrak{p} := S^{-1}B$. The inclusion $A \hookrightarrow B$ induces an inclusion $A_\mathfrak{p} \hookrightarrow B_\mathfrak{p}$ (since the localisation of an injective map at a multiplicative set disjoint from the kernel remains injective; here $A \hookrightarrow B$ is injective by hypothesis). By [Stability Under Quotients and Localisation](/theorems/2867), part (1b), the extension $A_\mathfrak{p} \subset B_\mathfrak{p}$ is integral. The localised ring $B_\mathfrak{p}$ is nonzero: since $1 \in B$ and $1/1 \neq 0/1$ in $B_\mathfrak{p}$ (as $S$ consists of non-zero-divisors in $A$, and $A \hookrightarrow B$ is injective), $B_\mathfrak{p}$ contains the nonzero element $1/1$. [guided] The strategy is to move from the global extension $A \subset B$ to a local one where the target prime $\mathfrak{p}$ becomes the unique maximal ideal. Set $S = A \setminus \mathfrak{p}$ and localise both rings at $S$. We must verify two things. First, that $A_\mathfrak{p} \hookrightarrow B_\mathfrak{p}$ is still an injection. The inclusion $A \hookrightarrow B$ is injective, and localisation preserves injectivity (since [Localization Is Exact](/theorems/2847)), so the induced map $S^{-1}A \to S^{-1}B$ is injective. Second, that $B_\mathfrak{p}$ is integral over $A_\mathfrak{p}$. This is exactly [Stability Under Quotients and Localisation](/theorems/2867), part (1b): for any multiplicative set $S \subset A$, if $B$ is integral over $A$, then $S^{-1}B$ is integral over $S^{-1}A$. Finally, $B_\mathfrak{p} \neq 0$ because $1/1 \neq 0/1$ in $B_\mathfrak{p}$. (If $1/1 = 0/1$, then $s \cdot 1 = 0$ for some $s \in S \subset A \subset B$, so $s = 0$ in $B$; but $s \in A \setminus \mathfrak{p}$ is nonzero in $A$ and hence in $B$ since $A \hookrightarrow B$ is injective.) [/guided] [/step] [step:Choose a maximal ideal of $B_\mathfrak{p}$ and show it contracts to $\mathfrak{p}A_\mathfrak{p}$] Since $B_\mathfrak{p} \neq 0$, by Zorn's lemma $B_\mathfrak{p}$ has a maximal ideal $\mathfrak{m}$. In particular, $\mathfrak{m}$ is a prime ideal of $B_\mathfrak{p}$. By [Maximality Is Preserved Under Integral Extensions](/theorems/2869), applied to the integral extension $A_\mathfrak{p} \subset B_\mathfrak{p}$: $\mathfrak{m}$ is maximal in $B_\mathfrak{p}$ if and only if $\mathfrak{m} \cap A_\mathfrak{p}$ is maximal in $A_\mathfrak{p}$. Therefore $\mathfrak{m} \cap A_\mathfrak{p}$ is a maximal ideal of $A_\mathfrak{p}$. The ring $A_\mathfrak{p}$ is local with unique maximal ideal $\mathfrak{p}A_\mathfrak{p}$. Hence $\mathfrak{m} \cap A_\mathfrak{p} = \mathfrak{p}A_\mathfrak{p}$. [guided] Since $B_\mathfrak{p}$ is a nonzero ring, Zorn's lemma guarantees the existence of a maximal ideal $\mathfrak{m}$ of $B_\mathfrak{p}$. We need to determine what $\mathfrak{m}$ contracts to in $A_\mathfrak{p}$. The key tool is [Maximality Is Preserved Under Integral Extensions](/theorems/2869): for an integral extension $A_\mathfrak{p} \subset B_\mathfrak{p}$ and a prime ideal $\mathfrak{m}$ of $B_\mathfrak{p}$, the ideal $\mathfrak{m}$ is maximal in $B_\mathfrak{p}$ if and only if $\mathfrak{m} \cap A_\mathfrak{p}$ is maximal in $A_\mathfrak{p}$. Since $\mathfrak{m}$ is maximal by construction, we conclude $\mathfrak{m} \cap A_\mathfrak{p}$ is maximal in $A_\mathfrak{p}$. But $A_\mathfrak{p}$ is a local ring: its unique maximal ideal is $\mathfrak{p}A_\mathfrak{p} = \{a/s : a \in \mathfrak{p}, s \in S\}$. Since $A_\mathfrak{p}$ has only one maximal ideal, we must have $\mathfrak{m} \cap A_\mathfrak{p} = \mathfrak{p}A_\mathfrak{p}$. [/guided] [/step] [step:Contract $\mathfrak{m}$ back to $B$ and verify that the resulting prime lies over $\mathfrak{p}$] Let $\iota: B \to B_\mathfrak{p}$ denote the canonical localisation map $b \mapsto b/1$. Define $\mathfrak{q} := \iota^{-1}(\mathfrak{m}) = \mathfrak{m} \cap B$ (identifying $B$ with its image in $B_\mathfrak{p}$). Then $\mathfrak{q}$ is a prime ideal of $B$, being the preimage of a prime ideal under a ring homomorphism. We verify $\mathfrak{q} \cap A = \mathfrak{p}$. Consider the commutative diagram of localisation maps: \begin{align*} A &\hookrightarrow B \\ \downarrow &\quad\quad \downarrow \\ A_\mathfrak{p} &\hookrightarrow B_\mathfrak{p} \end{align*} where the vertical arrows are the canonical localisation maps $a \mapsto a/1$ and $b \mapsto b/1$. For any $a \in A$: \begin{align*} a \in \mathfrak{q} \cap A &\iff a \in B \text{ and } a/1 \in \mathfrak{m} \\ &\iff a/1 \in \mathfrak{m} \cap A_\mathfrak{p} = \mathfrak{p}A_\mathfrak{p} \\ &\iff a \in \mathfrak{p}. \end{align*} The last equivalence holds because the contraction of $\mathfrak{p}A_\mathfrak{p}$ to $A$ under the localisation map $A \to A_\mathfrak{p}$ is $\mathfrak{p}$ (by [Prime Ideals Under Localization](/theorems/2927), since $\mathfrak{p} \cap S = \varnothing$). Therefore $\mathfrak{q} \cap A = \mathfrak{p}$. [guided] Define $\mathfrak{q} := \mathfrak{m} \cap B$, the preimage of $\mathfrak{m}$ under $\iota: B \to B_\mathfrak{p}$. Since $\mathfrak{m}$ is prime in $B_\mathfrak{p}$ and $\iota$ is a ring homomorphism, $\mathfrak{q} = \iota^{-1}(\mathfrak{m})$ is prime in $B$. It remains to show $\mathfrak{q} \cap A = \mathfrak{p}$. We use the commutativity of the localisation diagram: an element $a \in A$ lies in $\mathfrak{q} \cap A$ if and only if $a/1 \in \mathfrak{m}$, which happens if and only if $a/1 \in \mathfrak{m} \cap A_\mathfrak{p} = \mathfrak{p}A_\mathfrak{p}$. Now we use a standard fact about localisation at a prime: the contraction of the extended ideal $\mathfrak{p}A_\mathfrak{p}$ back to $A$ is $\mathfrak{p}$ itself. This follows from [Prime Ideals Under Localization](/theorems/2927): $\mathfrak{p}$ is a prime ideal disjoint from $S = A \setminus \mathfrak{p}$, so it is a contracted ideal, meaning $(\mathfrak{p}A_\mathfrak{p}) \cap A = \mathfrak{p}$. Therefore $a/1 \in \mathfrak{p}A_\mathfrak{p}$ if and only if $a \in \mathfrak{p}$, giving $\mathfrak{q} \cap A = \mathfrak{p}$. [/guided] [/step]