The Stacks project

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

Thanks and fixed here.

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?

See the section on localisation (00CM)

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 $q \cap R = p$?

I think I got it, thanks

Let me present a proof that does not rely on localizations or quotients (so it applies to semirings as well). This lemma is a special case (where $f$ is injective and $I=0$) of the following more general going-up result:

If $f : R\to S$ is a homomorphism of commutative semirings and $I$ is an ideal in $S$, then for any prime ideal $P \supseteq f^{-1}(I)$, there exists a prime ideal $Q \supseteq I$ such that $P\supseteq f^{-1}(Q)$. If $P$ is minimal over $f^{-1}(I)$, then of course $P=f^{-1}(Q)$.

This result directly follows from the following fact: if $M$ is a submonoid of a commutative semiring and $I$ is an ideal disjoint from it, then there exists a maximal such ideal containing $I$, and any such ideal is prime. A formalized proof is available at https://leanprover-community.github.io/mathlib4_docs/Mathlib/RingTheory/Ideal/Maximal.html#Ideal.isPrime_of_maximally_disjoint. To obtain the result, just notice that the submonoid $M := f(R\setminus P)$ is disjoint from an ideal (or subset) $I\subseteq S$ iff $P\supseteq f^{-1}(I)$.

The implication (1) ⇒ (2) in https://stacks.math.columbia.edu/tag/00FL also follows from this general result by taking $I=0$, so $f^{-1}(I)$ is the kernel of $f$. (1) says that the kernel is contained in the nilradical of $R$, but every prime contains the nilradical, so a minimal prime over the nilradical is the same as a minimal prime of $R$.

In https://stacks.math.columbia.edu/tag/00HY, one can directly show the image $T\subseteq \operatorname{Spec} R$ is equal to the zero locus of $f^{-1}(0)$. Every prime ideal in $T$ clearly contains $f^{-1}(0)$, and on the other hand, if a prime ideal $P$ contains $f^{-1}(0)$ then there is a prime $Q$ such that $P\supseteq f^{-1}(Q)$ by the general result, so $P\in T$ because $f^{-1}(Q)\in T$ and $T$ is closed under specialization.

Going to leave as is.