Lemma 10.30.5. Let $R \subset S$ be an injective ring map. Then $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ hits all the minimal primes.

Proof. Let $\mathfrak p \subset R$ be a minimal prime. In this case $R_{\mathfrak p}$ has a unique prime ideal. Hence it suffices to show that $S_{\mathfrak p}$ is not zero. And this follows from the fact that localization is exact, see Proposition 10.9.12. $\square$

Comment #4211 by Aaron on

I think it should be: $\text{Spec}(S)\to\text{Spec}(R)$ hits all the minimal primes.

Comment #6009 by Ivan on

could you please explain why it suffices to show that $S_p$ is not zero and we get a $q \in Spec(S)$ such that $q \cap R=p$? in additon, here $S_p$ is a ring?

Comment #6010 by Fan on

See the section on localisation (00CM)

Comment #6012 by Ivan on

but 00CM doesn't mention about it. even if $0 \rightarrow R_p \rightarrow S_p$ is exact and $S_p$ is non-zero, is it possible that $(p)S_p=S_p$? why $S_p$ is non-zero and then there exist $q \in Spec(S)$ such that $q \cap R = p$?

Comment #6014 by Ivan on

I think I got it, thanks

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