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 $\mathfrak p = \mathfrak r \cap R$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).