Lemma 15.107.1. Let $A$ and $B$ be domains and let $A \to B$ be a ring map. Assume $A \to B$ has additionally at least one of the following properties

1. it is the localization of an étale ring map,

2. it is flat and the localization of an unramified ring map,

3. it is flat and the localization of a quasi-finite ring map,

4. it is flat and the localization of an integral ring map,

5. it is flat and there are no nontrivial specializations between points of fibres of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$,

6. $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ maps the generic point to the generic point and there are no nontrivial specializations between points of fibres, or

7. exactly one point of $\mathop{\mathrm{Spec}}(B)$ is mapped to the generic point of $\mathop{\mathrm{Spec}}(A)$.

Then $A \cap J$ is nonzero for every nonzero ideal $J$ of $B$.

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). $\square$

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