The Stacks project

The mentioned chain of morphisms can also have all its arrows reversed for m_i and m_i' to be equivalent (by definition of generated equivalence relations)

The mentioned chain of morphisms can also have all its arrows reversed for m_i and m_{i}' to be equivalent (by definition of generated equivalence relations)

I think we should use $\pi_x$ instead of $\pi_X$ here.

We may replace $LG: DF^+(\mathcal{A}) \to DF^+(\mathcal{B})$ to $LG: DF^-(\mathcal{A}) \to DF^-(\mathcal{B})$.

We may replace $DF^-(\mathcal{A}) \cong K^+(\mathcal{P})$ to $DF^-(\mathcal{A}) \cong K^-(\mathcal{P})$.

Small typo: "map finite separable extensions" should be "map of finite separable extensions" I think.

This is a good question, thank you! The answer is that it is constructed in the proof. But let me explain a more general canonical map.

I claim: given an abelian category $\mathcal{A}$ with countable products and enough injectives, an inverse system $\ldots \to K_3^\bullet \to K_2^\bullet \to K_1^\bullet$ of bounded below complexes of $\mathcal{A}$, an (unbounded) complex $K^\bullet$, and maps of complexes $K^\bullet \to K_n^\bullet$ compatible with the transition maps in the inverse system, there is a canonical map $\gamma : K^\bullet \to R\lim K_n^\bullet$ in $D(\mathcal{A})$.

To construct $\gamma$ we argue as in the proof of tag 13.34.5. Namely, we find an inverse system $\ldots \to I_3^\bullet \to I_2^\bullet \to I_1^\bullet$ of bounded below complexes of injectives with termwise surjective transition maps and quasi-isomorphisms $K_n^\bullet \to I_n^\bullet$ compatible with transition maps. Then $R\lim K_n$ is represented by the complex $I^\bullet$ where $I^p = \lim I_n^p$ as in the proof of 13.34.5. The map $\gamma$ is defined to be the one corresponding to the obvious map of complexes $K^\bullet \to I^\bullet$. Of course we have to show that this map is independent of choices...

If the transition maps in the system $\ldots \to I_3^\bullet \to I_2^\bullet \to I_1^\bullet$ are not termwise surjective, we can still get a unique map easily. Namely, then the Rlim is represented by $C[-1]$ where $C$ is the cone on $\prod I_n^\bullet \to \prod I_n^\bullet$, see references given in proof of 13.34.5. But now you can use that $K^\bullet \to \prod I_n^\bullet \to \prod I_n^\bullet$ is zero on the nose, to get a canonical map into $C[-1]$, see Lemma 13.9.3 and its proof. The non-uniqueness that your are pointing out corresponds, in the proof of Lemma 13.9.3, to the choice of the homotopy witnessing the vanishing of $g \circ f$; however, since in our case $g \circ f = 0$ it makes sense to take the $0$ homotopy and this does indeed give a well defined map into the cone. Again, one has to show that the map $\gamma$ so obtained does not depend on the choice of the inverse system $I_n^\bullet$ and the quasi-isomorphisms $K_n^\bullet \to I_n^\bullet$... but this does not present any problems (as far as I can see at the moment).

The next time I go through all the comments I will make the corresponding edits. Please feel free to complain if you think this doesn't work.

Can you explain what is the canonical map $K^{\bullet}\rightarrow R \lim \tau_{\geq -n} K^{\bullet}$ in tag 13.34.5? I imagine that it comes from the exact sequence $\mathrm{Hom}(K^{\bullet},R \lim \tau_{\geq -n} K^{\bullet})\rightarrow \mathrm{Hom}(K^{\bullet}, \prod \tau_{\geq -n} K^{\bullet})\rightarrow \mathrm{Hom}(K^{\bullet}, \prod \tau_{\geq -n} K^{\bullet})$, since the canonical map $K^{\bullet}\rightarrow \prod \tau_{\geq -n} K^{\bullet}$ in the second term will be mapped to $0$ in the third term. But I can not get a unique map from $K^{\bullet}$ to $R \lim \tau_{\geq -n} K^{\bullet}$.

I learned another example from Gillman and Jerison's book 'Rings of continuous functions', Chapter 14: the ring of real valued continuous functions $C({\beta}R^{+}-R^{+})$, where $R^{+}$ is the set of posivtive real numbers. Indeed, for every $C(X)$ on a completely regular space $X$, $\rm Spec$ $(C(X)/P)$ forms an (infinite) chain when $P$ is a proper prime ideal. And ${\beta}R^{+}-R^{+}$ is a compact connected F-space, F-space means every maximal ideal in the ring contains only one minimal prime ideal. So its local rings are domains but it isn't a domain.

There is a typo at the end. It should be $\mathcal{G} \to g_*g^*\mathcal{G}$.

In statement (3) : W is dense in every fibre of X→S

The reference to Conrad's paper has a broken link.

Look, there is a typo in the proof of the first lemma and it should be $A' \otimes_{R'} R = A$ and not the other way around. I will fix this the next time I go through all the comments.

About the annihilator of $f$ what you say is correct. Here is an often used abuse of language: if $C$ is a ring with a square zero ideal $J$, then $J$ has a unique structure of a module over $C/J$ which "agrees" with the $C$-module structure on $J$ (this means exactly what you think it means). This works more generally for any $C$-module $M$ such that $JM = 0$.

Thus when we say in the statement of the first lemma: "$I$ is the annihilator of $f$ in $R$" this is the module structure on $\text{Ker}(R' \to R)$ we are using.

I might be missing something on the relations between $R$ and $R'$ in 23.7.1-23.7.3. With the condition that $I$ is the annihlator in $R$ of $f\in R'$ and $A'\otimes_R R'$, it seems that it is assumed that $R'$ is an $R$-algebra, i.e., $R\rightarrow R'$. But the usage of the results in the proof of 23.7.4 does not suggest $R\rightarrow R'$.

It seems that $I$ is the image of $Ann_{R'}(f)$ in $R$, am I correct? With $A=A'\otimes_{R}R'$ replaced by $A=A'\otimes_{R'}R$, the statements and the proof might make sense (with a few places of $R$-linear to replaced by $R'$-linear).

In "Interlude I", we should write "Choose a local equation $f\in\mathfrak{m}_y\subset \mathscr{O}_{Y,y}$" instead of "Choose a local equation $f\in\mathfrak{m}_y=\mathscr{O}_{Y,y}$". (Please delete the comment above)

In the commutative diagram, the first $\sigma_{\geq j}$ in the top left term should be replaced by $\sigma_{\geq i}$.

@#8281: where?

It seems aa though in (2) => (1) you don't actually use the colimit aasumption, right? That is to say it is enough to assume we have a cocone X for which the map in (2) is bijective.

The reason I'm asking is that in lemma 13.14.6, you spend the last part of the proof proving (in a purely formal manner) the colimit assumption on the object denoted C in that lemma, and it seems as though that may be unecessary.

I think we haven't proved that $\mathcal{K}_X(U)$ equals the total quotient ring of $\mathcal{O}_X(U)$ for any affine open $U ⊂ X$, we use it directely.

I think the reduction to the affine case is a bit confusing as stated, specifically the fact T is closed. Mainly, if X is not q.c you may end up taking an union of infinitely many closed sets in X and then it's not clear why you would get a closed set.

A better way to phrase the reduction to the non affine case is to note the construction given agrees on passing to principal opens in X, and so it glues together tp give the desired closed set T.