4
$\begingroup$

I'm reading Bhatt and Scholze's, "Prisms and prismatic cohomology", and I have a few questions about Lemma 3.1.

  1. Why can they choose a finite number of elements generating $A$ such that $IA[1/g_i]$ is principal? I don't understand how this follows from the fact that $I$ is locally principal.
  2. What do they mean by the Zariski localization of a ring along a closed subset of its spectrum? From Def. 2.1.6 in Bhatt and Scholze's, "The pro-étale cohomology for schemes", I understand that the localization of $Spec(A)$ along $V(I)$ should be something like $$\cup_{\mathfrak p \in V(I)} Spec(A_{\mathfrak p}),$$ which can be seen as a subset of $Spec(A)$. On the other hand, seeing Def. 2.1.12, ibid., regarding Zariski localizations of spectral spaces, I imagine it should be something like $$\sqcup_{\mathfrak p \in V(I)} Spec(A_{\mathfrak p})$$ endowed with the natural map to $Spec(A)$.
    And so the Zariski localization of $A$ along $V(I)$ should be something like $$\prod_{\mathfrak p \in V(I)} A_{\mathfrak p},$$ but I don't know how this fits Def. 2.2.1, ibid., regarding Zariski localizations of rings.
$\endgroup$

2 Answers 2

3
$\begingroup$

Regarding question 2:

My interpretation of the phrase (which fits the proof, i.e. makes the proof of the Lemma work) is the following:

Given a ring $B$ and an ideal $I \subseteq B$, the localization of $B$ along $V(I)$ is $S^{-1}B$ where $S=B \setminus \bigcup V(I)=B \setminus \bigcup_{\mathfrak{p}\in V(I)} \mathfrak{p}$ (note that $S$ is multiplicatively closed).

This is, firstly, ind-Zariski localization of the ring $B$ as per the mentioned Definition 2.2.1.

Secondly, it does what is claimed in the proof of Lemma 3.1, namely $IS^{-1}B$ is contained in the radical:

Indeed, primes of $S^{-1}B$ correspond to primes $\mathfrak{p}$ of $B$ that are contained in $\bigcup V(I)$. Given such a prime, the inclusion $\mathfrak{p}+I \subseteq \bigcup V(I)(=\bigcup_{I \subseteq \mathfrak{q} \in \mathrm{Spec}\,B}\mathfrak{q})$ shows that $\mathfrak{p}+I$ does not contain $1$, so there is a maximal ideal $\mathfrak{m}$ of $B$ with $\mathfrak{p}, I \subseteq \mathfrak{m}$. This shows that every prime ideal of $S^{-1}B$ is contained in a maximal ideal coming from $V(I)$, so in particular, every maximal ideal of $S^{-1}B$ contains $I$.

In fact, this argument shows that $\mathrm{Spec}\,S^{-1}B=\bigcup_{\mathfrak{p}\in V(I)}\mathrm{Spec}\,A_{\mathfrak{p}}$ as you mention.

$\endgroup$
1
  • $\begingroup$ It is worthwhile to mention that there is a neater (depends on taste) description of $S$, that is, $S=1+I$ with the usual assumption of Axiom of Choice. See Atiyah-Macdonald, Exercise 3.7. $\endgroup$
    – user20948
    Mar 20, 2020 at 17:32
4
$\begingroup$

Let me answer question 1. The short answer is: "this is essentially by definition of locally prinicipal, and the facts that ${\rm Spec}A$ is quasi-compact, and principal opens form a basis for its topology".

The extended version of an answer is: Let $\mathscr{I}$ denote the ideal sheaf on ${\rm Spec} A$ corresponding to $I$. The fact that $I$ is locally principal means precisely that that locally on ${\rm Spec} A$, $\mathscr{I}$ is generated by one global section. THus there is an open covering ${\rm Spec} A = \bigcup_{i \in I} U_i$ such that $\mathscr{I}|_{U_i}$ is generated -- as an $\mathcal{O}_{U_i}$-module -- by one global section. Note that this property remains true after restricting to an open subset of $U_i$. Thus, and as the principal open subsets form a basis for topology of ${\rm Spec} A$, we can refine our covering to one by principal opens $D(g_i) \cong {\rm Spec}A[g_i^{-1}] \subseteq {\rm Spec} A$. Moreover, as ${\rm Spec} A$ is quasi-compact, we can pick a finite subcovering and so we have finitely many $g_i$'s such that ${\rm Spec} A = \bigcup_i D(g_i)$, which is equivalent to the claim that the unit ideal of $A$ is generated by the $g_i$'s. Now $\mathscr{I}|_{D(g_i)}$ is the ideal sheaf on $D(g_i) = {\rm Spec}A[g_i^{-1}]$ corresponding to the $A[g_i^{-1}]$-module $IA[g_i^{-1}]$, and and it is generated by one global section by construction. This implies that $IA[g_i^{-1}] = \Gamma(D(g_i), \mathscr{I})$ is generated (as $A[g_i^{-1}]$-module) by one element.

$\endgroup$
1
  • $\begingroup$ Thank you for the detailed explanation! This is clear now. $\endgroup$
    – Yiiwa
    Nov 29, 2019 at 11:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.