Lemma 1

Let $\rho : A \to B$ be a ring homomorphism, and let $p$ be a prime ideal of A. Consider the following two properties: (i) the canonical homomorphism $\tilde{\rho}$ of $A_p$ into $A_p \otimes_A B$ is bijective; (ii) there exists a single prime ideal $r$ of B lying over $p$ and the canonical homomorphism $\rho_p$ of $A_p$ into $B_r$ is bijective. One has (i) $\Rightarrow$ (ii). If $p$ is minimal, or if B is integral over A, one has (i) $\Leftrightarrow$ (ii) $^1$.

1 This lemma remains valid when the ring A is not noetherian.

The ring $A_p \otimes_A B$ is identified with the ring of fractions $S^{-1}B$ of $B$ defined by the multiplicative subset $S = \rho(A - p)$ of $B$. The prime ideals of $S^{-1}B$ are therefore the $S^{-1}q$, where $q$ is a prime ideal of $B$ such that $\rho^{-1}(q) \subset p$; if $q$ is such an ideal, $(S^{-1}B)_{S^{-1}q}$ is identified with $B_q$ (II, § 2, No. 5, prop. 11).

If condition (i) is satisfied, there exists (V, § 2, No. 1, Lemma 1) a unique prime ideal $r$ of $B$ such that $\rho^{-1}(r) = p$. Moreover, $B_r$ is identified with the ring of fractions $(S^{-1}B)_{S^{-1}r}$, hence also with $(A_p)_s$, where $s$ is the inverse image of $S^{-1}r$ by the isomorphism $\tilde{\rho}: A_p \to S^{-1}B$; now $\tilde{\rho}^{-1}(S^{-1}r) = (A - p)^{-1}p = pA_p$, whence (ii).

Conversely, suppose that (ii) is satisfied, and let $r$ be the unique prime ideal of $B$ lying over $p$. Since $(S^{-1}B)_{S^{-1}r}$ is identified with $B_r$, it is enough to prove that $S^{-1}B$ is local with maximal ideal $S^{-1}r$, that is to say, that every prime ideal $q$ of $B$ such that $\rho^{-1}(q) \subset p$ is contained in $r$. If $p$ is minimal, one has $\rho^{-1}(q) = p$, hence $q = r$. If $B$ is integral over $A$, there exists, by V, § 2, no. 1, cor. 2 to th. 1, a prime ideal $r'$ of $B$ such that $q \subset r'$ and $\rho^{-1}(r') = p$; necessarily one has $r' = r$, hence $q \subset r$.