The Stacks project

Ah, I see that in the definition of "field" given in the next section does explicitly rule out $1=0$. Apologies for not spotting this.

The definition of a field appears to have a minor error in it. A field is a (commutative) ring in which $1\neq0$ and all nonzero elements are invertible. The definition given here allows for $1=0$. (One can avoid this by defining a field to be a ring whose nonzero elements form a group under multiplication, or by defining a field as a ring with exactly two ideals.)

The first $K,L$ should be $L,M$.

The second $\mathcal{C}'$ should be $\mathcal{C}$.

In the proof of (3) from (2), * for the case $m=0$, $B_i$ should rather be the identity of $\mathcal{O}_{U}^n$ and $t$ should be the same as $s$, and * "but the first column of" should be "but the last column of".

I believe it would help the reader to change the section title to "Separable algebraic extensions".

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.

(Lest this comment does not compile well in your device, you can read its plain text here.)

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 $\sqrt{I}$."