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The Stacks project

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. $\square$


Comments (7)

Comment #4211 by Aaron on

I think it should be: hits all the minimal primes.

Comment #6009 by Ivan on

could you please explain why it suffices to show that is not zero and we get a such that ? in additon, here 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 is exact and is non-zero, is it possible that ? why is non-zero and then there exist such that ?

Comment #6014 by Ivan on

I think I got it, thanks

Comment #10079 by Junyan Xu on

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 is injective and ) of the following more general going-up result:

If is a homomorphism of commutative semirings and is an ideal in , then for any prime ideal , there exists a prime ideal such that . If is minimal over , then of course .

This result directly follows from the following fact: if is a submonoid of a commutative semiring and is an ideal disjoint from it, then there exists a maximal such ideal containing , 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 is disjoint from an ideal (or subset) iff .

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

In https://stacks.math.columbia.edu/tag/00HY, one can directly show the image is equal to the zero locus of . Every prime ideal in clearly contains , and on the other hand, if a prime ideal contains then there is a prime such that by the general result, so because and is closed under specialization.


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