The Stacks project

$S[t]$ should be used instead of $S[x]$ as in the previous lemma and it should be mentioned that $A=R[It]$.

$S[t]$ should be used instead of $S[x]$ as in the previous lemma and it should be mentioned that $A=R[It]$.

Just to be clear, for Lemma 0B5I, this simple tensor is a priori an elment of $F(D_+(f))\otimes O_X(-nd)$?

Just to be clear, for Lemma 0B5I, this simple tensor is a priori an elment of F(D_+(f))\otimes O_X(-nd)?

The term "on the nose" is used here for the first time (according to the search function) and is used later on as well a few times. From context I gather it roughly means actual equality (instead of up to isomorphism)? I think it would be helpful for this notion to be defined somewhere concretely (or defined on-the-nose so to speak)

The term "on the nose" is used here for the first time (according to the search function) and is used later on as well a few times. From context I gather it roughly means actual equality (instead of up to isomorphism)? I think it would be helpful for this notion to be defined somewhere concretely (or defined on-the-nose so to speak)

In Lemma 01KO, (2) why do we bother saying that "$U\cap V$ is a finite union of affine opens in $X$", instead of saying that "$U\cap V$ is quasi-compact"? Is this to compare with Lemma 01KP, (1)(a) where we require $U\cap V$ to be affine?

Also personally I think it is more clear and parallel to rewrite Lemma 01KP in the form of Lemma 01KO, i.e. 3 things are equivelent : $f$ being separated is equivalent to for any pair of affine opens such that ... , and is equivalent to there is an affine open covering such that ... This is because in this form Lemma 01KO and Lemma 01KP would easily imply that being separated and quasi-separated is local on the target (which is also missing in this tag)

In Remark 0816, "Moreover, if $X\to Y$ and $Z\to Y$ are morphisms $\mathcal{C}$", do you want to say "are morphisms in $\mathcal{C}$"?

In the statement, $(\mathcal{C},\mathcal{O})$ should be a ringed site.

In line with the proof of Lemma 02VX, which references Lemmas 01S1 and 01U9, a stronger statement holds: A surjective and flat morphism is a universal epimorphism in the category of schemes.

$\kappa \subset N''/N'$ should be $\kappa \subset N'/N''$

In the paragraph before Lemma 51.22.1, "we denote $d = d(S)=...$" all the $d$ should probably be $q$.

The composition of $(f,g):(U,b,L)→(U′,b′,L′)$ with $(f′,g′):(U′,b′,L′)→(U′′,b′′,L′′)$ is probably given by $(f'∘f,g∘f∗(g′)),$ i.e., the order of $f$ and $f'$ should be changed

Typos : the third paragraph $\mathcal{E}$ is miswritten as $\mathcal{L}$. The last paragraph, third line from below "cup product" (missing space).

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Am I right that there is a typo in the sentence And then we may choose $\mathfrak{q}_{-2}\subset$ $\mathfrak{q}_{e-1}$ lying over $\mathfrak{p}_{e-2}$?

If I understand it correctly, then it should be $\mathfrak{q}_{−2}\subset \mathfrak{q}_{−1}$ , since by the going down property, one defines a chain of prime ideals $\mathfrak{q}_i$ lying over the chain of prime ideals $\mathfrak{p}_i$ in the sense that the preimage of $\mathfrak{q}_i$ under the map $R\to S$ equals $\mathfrak{p}_i$.

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.

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