The Stacks project

Comments 1 to 20 out of 8857 in reverse chronological order.

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

On Joe Lamond left comment #9536 on Section 9.1 in Fields

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


On Joe Lamond left comment #9535 on Section 9.1 in Fields

The definition of a field appears to have a minor error in it. A field is a (commutative) ring in which and all nonzero elements are invertible. The definition given here allows for . (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.)


On nkym left comment #9534 on Lemma 21.35.1 in Cohomology on Sites

The first should be .


On nkym left comment #9533 on Lemma 18.39.4 in Modules on Sites

The second should be .


On nkym left comment #9532 on Lemma 18.28.14 in Modules on Sites

In the proof of (3) from (2), * for the case , should rather be the identity of and should be the same as , and * "but the first column of" should be "but the last column of".


On Laurent Moret-Bailly left comment #9531 on Section 9.12 in Fields

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


On Anonymous left comment #9530 on Lemma 9.12.13 in Fields

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

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 which is not separable over , we would get an injection from a non-reduced ring into a reduced ring, hence a contradiction.


On nkym left comment #9529 on Lemma 21.17.7 in Cohomology on Sites

In the proof the last two should be and


On nkym left comment #9528 on Remark 18.27.3 in Modules on Sites

after the second last should be instead.


On Shubhankar left comment #9527 on Section 15.74 in More on Algebra

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 be perfect of tor-amplitude in . Then has tor-amplitude in . Indeed, if is an -module then and the RHS can be represented by a complex with non-zero for . Relabelling we see that is non-zero for and so we conclude.


On Andrea Panontin left comment #9526 on Section 50.2 in de Rham Cohomology

typo: "if be a morphism of schemes"


On left comment #9525 on Section 13.25 in Derived Categories

I don't know if it's any worth, but one can generalize 13.25.1 by replacing in its statement “let be an additive functor into an abelian category” by “let be an exact functor into a triangulated category” and by replacing by in the statement's commutative triangle. The proof is the same.


On left comment #9524 on Lemma 13.14.6 in Derived Categories

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


On Riv left comment #9523 on Section 31.13 in Divisors

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


On left comment #9522 on Theorem 19.12.6 in Injectives

As in other proofs, we are using AB5 to get that is injective for a limit ordinal and , by means of the lemma in Comment #9497.

To assure the existence of with one could invoke Sets, Proposition 3.7.2.


On left comment #9521 on Lemma 19.12.5 in Injectives

(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 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 is an abelian category, a triangle in or in is said to be -special if its induced long sequence in cohomology is exact.

We know that all distinguished triangles in and in are -special. The converse isn't true in general: Let be any signs satisfying . If is a d.t. in or in , then the anti-distinguished triangle is again -special but not necessarily distinguished. Note that if is a morphism of -special triangles and at least two among are quasi-isomorphisms, then so is the third.

Now, consider the following diagram in : where , and 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 is an iso in , the middle row is a distinguished triangle too (in ). On the other hand, the bottom row is anti-distinguished, so it is -special. Thus, to see that 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 . It suffices to verify this equality in . This is the content of the following

Lemma. Let be morphisms in some additive category. Then the morphisms ,:C(a)\rightrightarrows C(b) are chain homotopic.

Proof. A homotopy is given by


On Gyujin Oh left comment #9520 on Lemma 81.6.2 in Pushouts of Algebraic Spaces

There is a small typo: should be .


On Giancarlo Castellano left comment #9519 on Section 59.58 in Étale Cohomology

When is profinite and is discrete with a continuous -action, the as defined on this page form a universal cohomological -functor, and therefore, the canonical map 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 for what this page denotes by .)

Let me also weigh in on comments 7013–7014 above. Gille & Szamuely define where ranges over the open normal subgroups of . (I use the notation 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 where ranges over the open normal subgroups of . But each is discrete — even finite — so we can identify for all ; 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 denotes equipped with the discrete topology, as opposed to its given profinite topology, then in general.


On Goodluckthere left comment #9518 on Lemma 10.30.6 in Commutative Algebra

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


On Goodluckthere left comment #9517 on Lemma 10.30.6 in Commutative Algebra

The equality is better to be written as "the preimage of the nilradical of equals ."