The Stacks project

Maybe this is pedantic, but do you need some argument on why $k(\alpha)$ are separable over $k$ (i.e. that every element of $k(\alpha)$ is separable over $k$), and similarly for $k(\beta)$?

For example, one could use Tags 9.12.11 and 9.12.4.

Alternative way to say it (but maybe worse exposition): if there were an element $x \in k(\alpha)$ which is not separable over $k$, we would get an injection $k(x) \otimes_k \overline{k} \rightarrow k(\alpha) \otimes_k \overline{k}$ from a non-reduced ring into a reduced ring, hence a contradiction.

In the proof the last two $\mathcal{K}_1$ should be $\mathcal{K}_2$ and $\mathcal{K}_3$

$\mathcal{G}$ after the second last $\to$ should be $f^*\mathcal{G}$ instead.

Here's a possibly useful lemma which I thought was already on here. I also looked in the tor dimension section but couldn't find it. Apologies if I missed this or if there is a mistake below.

Let $P$ be perfect of tor-amplitude in $[a,b]$. Then $P^\vee$ has tor-amplitude in $[-b,-a]$. Indeed, if $M$ is an $A$-module then $P^\vee\otimes^L M=R\mathrm{Hom}(P,M)$ and the RHS can be represented by a complex $E^\bullet$ with $E^i$ non-zero for $i\in [a,b]$. Relabelling we see that $\mathrm{Ext}^i(P,M)$ is non-zero for $i\in [-b,-a]$ and so we conclude.

typo: "if $f\colon Y \to X$ be a morphism of schemes"

I don't know if it's any worth, but one can generalize 13.25.1 by replacing in its statement “let $F:\mathcal{A}\to\mathcal{B}$ be an additive functor into an abelian category” by “let $F:K^+(\mathcal{A})\to\mathcal{D}$ be an exact functor into a triangulated category” and by replacing $D^+(\mathcal{B})$ by $\mathcal{D}$ in the statement's commutative triangle. The proof is the same.

In the big diagram, the vertical arrows should be downward-pointing.

Hi it seems there is no content about the genral Cartier diviaors, rather than the effective case on Stacks Project?

As in other proofs, we are using AB5 to get that $T_\beta(M^\bullet)\to T_\alpha(M^\bullet)$ is injective for a limit ordinal $\alpha\neq 0$ and $\beta<\alpha$, by means of the lemma in Comment #9497.

To assure the existence of $\alpha$ with $\operatorname{cf}\alpha>\max\{\kappa,|U|\}$ one could invoke Sets, Proposition 3.7.2.

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For those who care, here's the proof that $j_{M^\bullet}$ is a qis (as it is hinted the last sentence of the proof).

Definition. Extending the terminology introduced in Derived Categories, Remark 13.4.4, if $\mathcal{A}$ is an abelian category, a triangle $A\to B\to C\to A[1]$ in $K(\mathcal{A})$ or in $D(\mathcal{A})$ is said to be $H^*$-special if its induced long sequence in cohomology is exact.

We know that all distinguished triangles in $K(\mathcal{A})$ and in $D(\mathcal{A})$ are $H^*$-special. The converse isn't true in general: Let $\varepsilon_i\in\{1,-1\}$ be any signs satisfying $\varepsilon_1\varepsilon_2\varepsilon_3=-1$. If $A\xrightarrow{a}B\xrightarrow{b}C\xrightarrow{c} A[1]$ is a d.t. in $K(\mathcal{A})$ or in $D(\mathcal{A})$, then the anti-distinguished triangle $A\xrightarrow{\varepsilon_1 a}B\xrightarrow{\varepsilon_2 b}C\xrightarrow{\varepsilon_3 c} A[1]$ is again $H^*$-special but not necessarily distinguished. Note that if $(f,g,h)$ is a morphism of $H^*$-special triangles and at least two among $f,g,h$ are quasi-isomorphisms, then so is the third.

Now, consider the following diagram in $D(\mathcal{A})$: $$\require{AMScd} \begin{CD} J^\bullet(M^\bullet)@>>> C(u_{M^\bullet})@>-p>>M^\bullet[1]@>u_{M^\bullet}>>J^\bullet(M^\bullet)[1]\\ @|@VVqV@|@|\\ J^\bullet(M^\bullet)@>\smash{v_{M^\bullet}}>>Q^\bullet(M^\bullet)@>\smash{-pq^{-1}}>>M^\bullet[1]@>\smash{u_{M^\bullet}}>>J^\bullet(M^\bullet)[1]\\ @|@|@VVj_{M^\bullet}[1]V@|\\ J^\bullet(M^\bullet)@>\smash{v_{M^\bullet}}>>Q^\bullet(M^\bullet)@>\smash i>>C(v_{M^\bullet})@>\smash{(0,1)}>>J^\bullet(M^\bullet)[1] \end{CD}$$ where $j_{M^\bullet}=<script type="math/tex; mode=display">\begin{pmatrix}0\\u_{M^\bullet}\end{pmatrix}$ , $p=(0,1)$ and $q=(v_{M^\bullet},0)$ is a quasi-isomorphism as it is argued in Derived Categories, Section 13.12, before Lemma 13.12.1. In the diagram, the top row is a distinguished triangle. Since the three upper squares commute and $q$ is an iso in $D(\mathcal{A})$, the middle row is a distinguished triangle too (in $D(\mathcal{A})$). On the other hand, the bottom row is anti-distinguished, so it is $H^*$-special. Thus, to see that $j_{M^\bullet}[1]$ is a qis, it suffices to see that the three lower squares commute. Clearly the left and right lower squares commute. Commutativity of the middle square amounts to $i\circ q=-j_{M^\bullet}[1]\circ p$. It suffices to verify this equality in $K(\mathcal{A})$. This is the content of the following

Lemma. Let $A\xrightarrow{a} B\xrightarrow{b} C$ be morphisms in some additive category. Then the morphisms $<script type="math/tex; mode=display">\begin{pmatrix}b&0\\0&0\end{pmatrix}$, $$\begin{pmatrix}0&0\\0&-a[1]\end{pmatrix}$$ :C(a)\rightrightarrows C(b) are chain homotopic.

Proof. A homotopy is given by $$h^n=\begin{pmatrix}0&0\\1_{B^n}&0\end{pmatrix}:C(a)^n=B^n\oplus A^{n+1}\to C^{n-1}\oplus B^n=C(b)^{n-1}.\quad\square$$

There is a small typo: $Y'$ should be $V'/R'$.

When $G$ is profinite and $M$ is discrete with a continuous $G$-action, the $H^i_{cont}(G, M)$ as defined on this page form a universal cohomological $\delta$-functor, and therefore, the canonical map $H^i(G, M) \to H^i_{cont}(G, M)$ is an isomorphism. A reference for this — though probably not the original one, and certainly not a very recent one — is Stephen S. Shatz' book Profinite groups, Arithmetic, and Geometry, Section II.§2; specifically, one needs to combine Proposition 5 and Theorem 8. (Note that Shatz writes $H^i$ for what this page denotes by $H^i_{cont}$.)

Let me also weigh in on comments 7013–7014 above. Gille & Szamuely define $H^i_{\mathrm{cont}}(G, M) := \varinjlim H^i(G/U, M^U)$ where $U$ ranges over the open normal subgroups of $G$. (I use the notation $H^i$ on the right-hand side because the way these groups are defined by Gille & Szamuely agrees with 59.57.2.(2).) On the other hand, in Shatz' book, it is proved (see Corollary 1 of Theorem 7 in Section II.§2) that $H^i_{cont}(G, M) = \varinjlim H^i_{cont}(G/U, M^U)$ where $U$ ranges over the open normal subgroups of $G$. But each $G/U$ is discrete — even finite — so we can identify $H^i_{cont}(G/U, M^U) = H^i(G/U, M^U)$ for all $U$; therefore, Gille–Szamuely's definition is equivalent to the one on this page, after all.

What I think Remark 4.2.4 from Gille & Szamuely's book actually tells us is that, if $G_{\mathrm{disc}}$ denotes $G$ equipped with the discrete topology, as opposed to its given profinite topology, then $H^i(G_{\mathrm{disc}}, M) \ne H^i_{cont}(G, M)$ in general.

Also in the penultimate line, it's strange to take $S/IS$ since $I$ is not inside $S$ and its image in $S$ is zero. I think what it means is that the map $R/I\to S$ is an injective ring map, therefore by lemma \ref{https://stacks.math.columbia.edu/tag/00FK}, the corresponding spectra map $\operatorname{Spec}(S)\to \operatorname{Spec}(R/I)$ hits all minimal prime ideals of $R/I$. A minimal prime ideal of $R$ will contain the nilpotent elements of $R$, therefore also $I$ by (1) thus corresponding to a minimal prime ideal of $R/I$ which in turn will have a preimage in $\operatorname{Spec}(S)$.

The equality $\sqrt{I}=\cap_{q\in S}R\cap q$ is better to be written as "the preimage of the nilradical $\sqrt{(0)}$ of $S$ equals $\sqrt{I}$."

Proof of lemma 29.39.1, in the explaination of condition (1), $\Epsilon$ should be a quasi-coherent $\mathcal{O}_X$ module instead of a $\mathcal{O}_S$ module?

I guess $S'$ is not the ring generated by these elements, but rather the $R$-subalgebra

In the proof, in “then $M^\bullet\to M^\bullet(M^\bullet)$ is a functor” one could write '$\mapsto$' instead to make it agree with the style displayed in the statement.

When the proof says “the cokernel of $j_{M^\bullet}$ is isomorphic to the direct sum of the cokernels of the maps $K^{\bullet}_i\to L^\bullet_i$ hence acyclic” we are using that $\mathcal{A}$ is AB5. Specifically, we are using firstly that AB5$\Rightarrow$AB4 (as it was pointed out in #9401) and secondly that in an AB4 category a direct sum of acyclic complexes is acyclic (actually, this is equivalent to the AB4 condition, see Proposition 1 here).

Correction to #9512: it suffices to consider some cardinal $\kappa$ greater than $\prod_{i\in\mathbb{Z}}|M^i|$ and such that there is some other cardinal strictly in-between.

Logic issues in the construction of the transfinite sequence of subcomplexes $K^\bullet_\alpha$. Cases (1) and (3) are independent of previous steps, but it seems (2) is not: it requires to make a choice since $N_\alpha^\bullet$ is not unique. Instead of transfinite recursion on $\alpha$, maybe what one actually needs is the generalized axiom of dependent choices? Let $\kappa$ be an aleph. This axiom says:

$\mathsf{DC}_\kappa$ [1, Sect. 8.1]. Let $S$ be a nonempty set and let $R$ be a binary relation such that for every $\alpha<\kappa$ and every $\alpha$-sequence $s=\left\langle x_{\xi}: \xi<\alpha\right\rangle$ of elements of $S$ there exists $y \in S$ such that $sRy$. Then there is a function $f: \kappa \rightarrow S$ such that for every $\alpha<\kappa,(f \mid \alpha) R f(\alpha)$.

(One has $\mathsf{AC}\Leftrightarrow\forall\kappa,\mathsf{DC}_\kappa$ [1, Theorem 8.1].)

The axioms $\mathsf{DC}_\kappa$ alone do not provide a transfinite sequence indexed on the class of all ordinals, but rather a transfinite sequence of length equal to an aleph. However, for our proof it suffices to consider some cardinal $\kappa$ greater than $\prod_{i\in\mathbb{Z}}|M^i|$ (as it is hinted in #9511, 4).

A disclaimer: All set theory I know I have learnt it in the last month, from scratch. I hadn't ever been into any kind of foundations of math, logical or set-theoretic ones.

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References

1. T. Jech, The Axiom of Choice

1. In the statement, to match the style used in Lemma 19.12.6 and in Derived Categories, Section 13.31, one could display “K-injective” instead of “$K$-injective.”

2. In the construction of the subcomplexes $K^\bullet_\alpha$, in case (3), $\alpha$ is a limit ordinal, we are inadvertently using that $\mathcal{A}$ has AB5 to get that the filtered colimit of subobjects is again a subobject (maybe it's worth mentioning AB5?).

3. When the proof says “note that the transition maps in the system are surjective” how is this fact being used? Lemma 12.31.8 does not require the transition maps being surjective as a hypothesis.

4. In “it is clear that $M^\bullet=K^\bullet_\alpha$ for a suitably large ordinal $\alpha$” one could add “(take an ordinal $\alpha$ greater than $\prod_{i\in\mathbb{Z}}|M^i|$).”

5. For-posterity details for those who care: The reason we get $$\tag{1}\label{seq} \operatorname{Hom}_{K(\mathcal{A})}(K_\beta^{\bullet}, I^{\bullet}) \leftarrow \operatorname{Hom}_{K(\mathcal{A})}(K_\alpha^{\bullet}, I^{\bullet}) \leftarrow \operatorname{Hom}_{K(\mathcal{A})}(N_{\dot{\alpha}}^{\bullet}, I^{\bullet})$$ exact is because $\operatorname{Hom}^\bullet(-,I^\bullet)$ is exact (this is equivalent to $I^j$ being injective for all $j$), so a s.e.s. of cochain complexes $S$ is sent to another s.e.s. of cochain complexes $\operatorname{Hom}^\bullet(S,I^\bullet)$. Application of Homology, Lemma 12.13.12 to $\operatorname{Hom}^\bullet(S,I^\bullet)$ gives exactness of \eqref{seq} (Homology, Equation 15.71.0.1).