The Stacks project

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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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Krull intersection requires the local ring to be noetherian. See: https://stacks.math.columbia.edu/tag/00IP

As you observed at the end of the proof of Lemma 24.4, one can join Lemmas 24.3 and 24.4 in a unique result proving that TFAE: $$\begin{enumerate} \item $v\circ u$ ($u\circ v$ resp.) is fully faithful; \item $u$ ($v$ resp.) is fully faithfull; \item $\eta:id\Rightarrow v\circ u$ ($\varepsilon:u\circ v\Rightarrow id$ resp.) is a natural isomorphism; \item there exists a natural isomorphism $id\Rightarrow v\circ u$ ($u\circ v\Rightarrow id$ resp.). \end{enumerate}$$

Lemma 05A3 (2) is also immediate from https://stacks.math.columbia.edu/tag/059M and that restriction of scalars along $R \to S$ commutes with arbitrary products.

I think Lemma 06WX (2) needs the assumption that f is flat, since $f^{-1}$ is a functor to $f^{-1}O_Y$ modules and tensoring to get $f^\star$ is only exact when f is flat. For instance if f is the inclusion of a closed point in $A^1$ then $O_X$ is injective but $f_*O_X$ is not.

I think Lemma 06WX (2) needs the assumption that f is flat, since $f^{-1}$ is a functor to $f^{-1}O_Y$ modules and tensoring to get $f^\star$ is only exact when f is flat. For instance if f is the inclusion of a closed point in $A^1$ then $O_X$ is injective but $f_*O_X$ is not.

In the definition of "purely inseparable extension", it is written that "these extensions only show up in positive characteristic". Hence, using the definitions in this section, it would follow that radicial morphisms also only show up in positive characteristic, which is not what we want, right? Perhaps one should change the definition of "purely inseparable extension" to include the characteristic zero case, by saying that $K/F$ is purely inseparable if each $\alpha \in K - F$ is purely inseparable over $F$.

From Street Art to Canvas

In the vibrant tapestry of contemporary art, the evolution of street art has transcended its urban origins, finding a prominent place within the hallowed halls of galleries and auction houses. This metamorphosis from ephemeral graffiti on the wall to coveted canvas pieces illustrates not only a shift in artistic expression but also a burgeoning recognition of street art as a legitimate form of fine art.

Graffiti on The Wall Understanding Street Art

Street art, often characterized by its bold colors and provocative themes, serves as a powerful medium for social commentary and cultural reflection. It challenges the boundaries of traditional art forms, inviting viewers to engage with the narratives woven into the urban landscape. As society increasingly embraces this genre, understanding its nuances becomes essential for both artists and collectors alike.

Learning to Spray Painting

For those aspiring to delve into the world of street art, mastering the technique of spray painting is paramount. This dynamic form of expression requires not only technical skill but also an innate understanding of color theory and composition. Workshops and tutorials abound, providing budding artists with the tools necessary to transform their visions into tangible works of art.

Drawing to Spray Painting

Transitioning from drawing to spray painting can be both exhilarating and daunting. Artists often find that their foundational skills in sketching enhance their ability to manipulate spray cans, allowing for intricate designs and bold statements. This synergy between traditional drawing techniques and modern spray painting opens new avenues for creativity, enabling artists to explore their unique styles.

Instigating the ‘Shift’ to Bankable Art

As street art garners recognition within the mainstream art market, a palpable shift occurs—one that transforms once-underground artists into bankable commodities. This transition not only elevates the status of street art but also prompts discussions about authenticity, ownership, and the commercialization of what was once deemed transgressive. As we navigate this evolving landscape, it becomes imperative to celebrate the roots of street art while acknowledging its potential as a lucrative investment.

Perhaps one should also mention that the isomorphisms $H^i(G,M)\to H^i_{cont}(G,M)$ have been known also for $G$ locally profinite by \ref{https://arxiv.org/pdf/2106.04473}. (This should also follow from the two facts: (i) the functor taking $M$ to its associated condensed abelian group $\underline{M}$ is exact for discrete $M$, (ii) condensed group cohomology agrees with continous group cohomology for $G$ locally profinite and $M$ discrete (or more generally if $\underline{M}$ is solid). See the notes \ref{https://janschuetz.perso.math.cnrs.fr/skripte/homology_profinite.pdf}.)

If $R$ is a normal domain, and $S\subset R$ is a multiplicative set containing $0$, then $S^{-1}R$ is the zero ring, which is not even an integral domain. I suppose the zero ring is, strictly speaking, a normal ring, since it is vacuously true that the localizations at prime ideals are normal domains, but I wonder if this was really intended.

Why are there finitely many maximals containing $x$? This seems to assume that $R$ is Noetherian -- it is equivalent to say that $R/xR$ has finitely many minimal primes, and I think this already requires the Noetherian hypothesis, which is only stated later.

Typo: distinghuised

I presume that (ii) should be "equivalence relation" or, equivalently, "monomorphism". But I don't understand how (iii) is supposed to be interpreted at all. If the condition holds for all points u, then it's of course fine. But if it holds for all but one orbit, it is not fine. Take for example Hironaka's example of a proper smooth algebraic space $X$ of dimension 3 which is not a scheme. It can be obtained as the quotient of a free involution on a proper scheme $U$. There is a closed orbit $\lbrace u_1, u_2 \rbrace$ in $U$ which has no affine neighborhood and the complement of $\lbrace u_1, u_2 \rbrace$ is quasi-projective. So $X\smallsetminus \lbrace x\rbrace$ is a scheme where $x$ corresponds to $\lbrace u_1, u_2 \rbrace$. Any dense open affine in $X\smallsetminus \lbrace x\rbrace$ pulls-back to a dense $R$-stable affine open in $U$. Does dense have another meaning in this context?

@10056: per the conventions, all rings in the Stacks Project are commutative with 1.