The Stacks project

Proof. Proof in case (7). Let $K$, resp. $L$ be the fraction field of $A$, resp. $B$. By Algebra, Lemma 10.30.7 we see that the unique point of $\mathop{\mathrm{Spec}}(B)$ which maps to the generic point $(0) \in \mathop{\mathrm{Spec}}(A)$ is $(0) \in \mathop{\mathrm{Spec}}(B)$. We conclude that $B \otimes _ A K$ is a ring with a unique prime ideal whose residue field is $L$ (in fact it is equal to $L$ but we do not need this). Choose $b \in J$ nonzero. Then $b$ maps to a unit of $L$. Hence $b$ maps to a unit of $B \otimes _ A K$ (Algebra, Lemma 10.19.2). Since $B \otimes _ A K = \mathop{\mathrm{colim}}\nolimits _{f \in A \setminus \{ 0\} } B_ f$ we see that $b$ maps to a unit of $B_ f$ for some $f \in A$ nonzero. This means that $b b' = f^ n$ for some $b' \in B$ and $n \geq 1$. Thus $f^ n \in A \cap J$ as desired.

In the rest of the proof, we show that each of the other assumptions imply (7). Under assumptions (1) – (5), the ring map $A \to B$ is flat and hence $A \to B$ is injective (since flat local homomorphisms are faithfully flat by Algebra, Lemma 10.39.17). Hence the generic point of $\mathop{\mathrm{Spec}}(B)$ maps to the generic point of $\mathop{\mathrm{Spec}}(A)$. Now, if there are no nontrivial specializations between points of fibres of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$, then of course this generic point of $\mathop{\mathrm{Spec}}(B)$ has to be the unique point mapping to the generic point of $\mathop{\mathrm{Spec}}(A)$. So (6) implies (7). Finally, to finish we show that in cases (1) – (5) there are no nontrivial specializations between the points of fibres of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$. Namely, see Algebra, Lemma 10.36.20 for the integral case, Algebra, Definition 10.122.3 for the quasi-finite case, and use that unramified and étale ring maps are quasi-finite (Algebra, Lemmas 10.151.6 and 10.143.6). $\square$