The Stacks project

In the proof of Krull-Akizuki the beggining seems to be circular, or not well-written: To begin we may assume that $L$ is the fraction field of $A$ by replacing $L$ by the fraction field of $A$ if necessary.

I think $(B,I,\delta)$ should be an $A$-algebra. In the sentense "...and let $\Omega_{D'}^i$ resp. $\Omega_{D'}^i$ be ...", the former $\Omega_{D'}^i$ should be replaced by $\Omega_D^i$.

In Lemma 50.15.2, it maybe worth mentioning that there's another short exact sequence of complexes: 0 $\rightarrow$ $\Omega_{Y/S}^{\bullet}$ $\rightarrow$ $i^{*}\Omega_{X/S}^{\bullet}(log\ Y)$ $\xrightarrow{Res}$ $\Omega_{Y/S}^{\bullet}[-1]$ $\rightarrow$ 0, where $i: Y\rightarrow X$ is the closed immersion. Moreover, it's locally split (locally Res has a section $\omega\mapsto\omega dlogf$) which is dependent of choices of local equation $f$ of $Y$.

Sorry, I do not notice that your convention is start from 0.

In the 2nd paragraph of the proof, W should be $\text{pr}_2(…)$?

In the phrase "The second map is quasi-compact as it is the base change of $f$...", I believe $f$ should be replaced with $g\circ f$.

There's a typo at the third proof of part (1). You should add one more bracket on the right, it should be $\operatorname{Hom}_R (M, \Gamma(X, \mathcal{Hom}_{\mathscr{O}_X} (\widetilde{N}, \mathcal{F})))$.

There's a typo at the third proof of part (1). You should add one more bracket on the right, it should be $\operatorname{Hom}R (M, \Gamma(X, \mathcal{Hom}{\mathscr{O}_X} (\widetilde{N}, \mathcal{F}))).

There are two typos in the proof where $g \circ f$ is $f \circ g$.

Small grammatical issue, and unclear phrasing. This should probably say something like:

In a preadditive category $\mathcal{A}$, we call an object that is both final and initial (as in Lemma 12.3.2 above) a zero object and denote it by $0$.

Should the $Y'$ in Lemma 0ABP be actually $X'$? Also, should one interchage the roles of X and Y in Lemma 0BB0?

Is there a typo on line -3 of the example? I believe $P\to X^{\gamma}$ should be $P\to X^g$.

The definition of a spectral map seems to be exactly the same as that of quasi-compact map from 005A.

Isn't it easier to use that M is a direct summand of a finite free module?

I want to remark that this definition of “right acyclic for $F$” might not be equivalent to Lipman's definition of “right-$F$-acyclic” [L, 2.2.5 and second paragraph of Sect. 2.7]. Whereas Lipman's definition always implies Definition 13.15.3.3, the converse might not be always true; although the converse holds in $K^+(\mathcal{A})$ with the extra assumption that $\mathcal{A}$ “has enough right acyclics for $F$ (in the sense of Definition 13.15.3.3).” All of this is explained in [GH, Remark 1.17].

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[GH] —, A Gospel Harmony of Derived Functors https://eliasguisado.wordpress.com/work/

[L] J. Lipman. “Notes on Derived Functors and Grothendieck Duality”. In: Foundations of Grothendieck Duality for Diagrams of Schemes. Lecture Notes in Mathematics. Springer-Verlag, 2009

For the interested reader, I wrote the generalization of this result to arbitrary triangulated categories and I merged it with the list of equivalent definitions of K-injective complex from Lipman [L, 2.3.8], including the proof (I claim no originality over any of this). See [GH, Prop. 2.4].

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[GH] —, A Gospel Harmony of Derived Functors https://eliasguisado.wordpress.com/work/

[L] J. Lipman. “Notes on Derived Functors and Grothendieck Duality”. In: Foundations of Grothendieck Duality for Diagrams of Schemes. Lecture Notes in Mathematics. Springer-Verlag, 2009

If anyone is interested, I generalized all results from this section to arbitrary triangulated categories [GH, Sect. 3]; they particularize now to K-injective resolutions in $K(\mathcal{A})$ (in $K^+(\mathcal{A})$, if $\mathcal{A}$ has enough injectives, a K-injective resolution is the same as a resolution by a bounded below complex of injectives [GH, Prop. 2.9]). Maybe this is what they meant in #9044 when they said that this section was "a bit obsolete"?

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[GH] —, A Gospel Harmony of Derived Functors https://eliasguisado.wordpress.com/work/

Does this hold more generally when $f$ is only locally equi-dimensional (the proof may be different)? This proposition seems to be called Chevalley's theorem according to https://people.kth.se/~dary/thesis/thesis-paperIV.pdf Cor 6.3 and (EGAIV, Theorem 14.4.1). Maybe it is also related to https://stacks.math.columbia.edu/tag/0GIQ.

I don't know if this is relevant, but maybe one could stress out the fact that in (2) the choice for the isomorphism $F'\circ[1]\cong[1]\circ F'$ is uniquely determined by the isomorphism $F\circ[1]\cong [1]\circ F$. Specifically, if $(F,\xi):\mathcal{D}\to\mathcal{D}'$ is a triangulated functor sending all morphism in $S$ to isos in $\mathcal{D}'$ then there is a unique triangulated functor $(F',\xi'):S^{-1}\mathcal{D}\to\mathcal{D}'$ factoring $(F,\xi)$ through $(Q,\zeta)$ (where $\zeta:Q\circ[1]=[1]\circ Q$ is the identity natural transformation).

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