Lemma 10.30.7. Let R \to S be a ring map. If a minimal prime \mathfrak p \subset R is in the image of \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R), then it is the image of a minimal prime.
Proof. Say \mathfrak p = \mathfrak q \cap R. Then choose a minimal prime \mathfrak r \subset S with \mathfrak r \subset \mathfrak q, see Lemma 10.17.2. By minimality of \mathfrak p we see that \mathfrak p = \mathfrak r \cap R. \square
Comments (0)