The Stacks project

Comments 1 to 20 out of 9145 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 #9994 on Lemma 26.11.1 in Schemes

Lucas, in the Stacks Project an irreducible topological space is nonempty by definition (see https://stacks.math.columbia.edu/tag/004U#:~:text=Let%20X%20be%20a%20topological,maximal%20irreducible%20subset%20of%20X.). One reason this convention makes sense is that an irreducible topological space can be defined as a topological space which cannot be expressed as a finite union of proper closed subspaces. But the empty space equals the empty union, hence it is not irreducible. A similar line of reasoning explains why the empty topological space is not connected.


On anonymous left comment #9993 on Definition 13.19.1 in Derived Categories

Typo: "an projective" is written instead of "a projective" four times in this definition.


On Doug Liu left comment #9992 on Lemma 37.4.1 in More on Morphisms

It seems that the proof works for not only "first order" but also "finite order" thickenings.


On Jonas left comment #9991 on Section 57.3 in Derived Categories of Varieties

In the last step of the proof of Lemma 57.3.3, how exactly is TR3 used to obtain an element of mapping to both and ?


On Rankeya left comment #9990 on Lemma 10.99.1 in Commutative Algebra

Is this proof using that is noetherian anywhere? I understand in applications it is likely that is also noetherian.


On left comment #9989 on Lemma 15.87.8 in More on Algebra

Here are the omitted details that show that is a quasi-isomorphism. (The following is an adaptation of this proof.)

Lemma. Let be an abelian category. Let be a cartesian square in . If is onto and is a quasi-isomorphism, then is a quasi-isomorphism.

Proof. Consider the sequence It is left exact by Homology, Lemma 12.5.11, and it is exact at for is onto. It gives rise to a distinguished triangle in . Since is an iso in , the lemma follows from the following result.

Lemma. Let be a triangulated category. Let be a distinguished triangle in . If is an isomorphism, then is an isomorphism.

Proof. Without loss of generality we may assume , since we have an isomorphism of distinguished triangles

\xymatrix{ A \ar@{->}[d]^{\mathrm{id}} \ar@{->}[r]^{{(f,g)}} & B\oplus C \ar@{->}[d]^{\mathrm{id}} \ar@{->}[r]^{{(h,k)}} & D \ar@{->}[d]^{k^{-1}} \ar@{->}[r] & A[1] \ar@{->}[d]^{\mathrm{id}} \ A \ar@{->}[r]^{{(f,g)}} & B\oplus C \ar@{->}[r]^{{(k^{-1}h,\mathrm{id})}} & C \ar@{->}[r] & A[1] }

But we also have an isomorphism of distinguished triangles

\xymatrix{ A \ar@{->}[d]^{\mathrm{id}} \ar@{->}[r]^{{(f,g)}} & B\oplus C \ar@{->}[d]^{} \ar@{->}[r]^{{(h,\mathrm{id})}} & C \ar@{->}[d]^{\mathrm{id}} \ar@{->}[r] & A[1] \ar@{->}[d]^{\mathrm{id}} \ A \ar@{->}[r]{{(f,0)}} & B\oplus C \ar@{->}[r]{{(0,\mathrm{id})}} & C \ar@{->}[r] & A[1] }

The matrix in the middle is invertible for its inverse is . The right square clearly commutes and the left square commutes as , for the top sequence is a complex. Let . Let . After inspecting the long exact sequence obtained by mapping through the bottom distinguished triangle from the last diagram, we deduce that is an isomorphism. By the Yoneda lemma, is an isomorphism.


On left comment #9988 on Lemma 15.59.2 in More on Algebra

The converse holds: Let be an acyclic complex. Then the quasi-isomorphism is mapped through the functor to a quasi-isomorphism .


On left comment #9987 on Section 15.86 in More on Algebra

Continuation to #9936: The functor is actually full [ref]. Also is conservative for any index category (use Lemma 3 from here).


On left comment #9986 on Section 13.22 in Derived Categories

Ignore #9982, it's wrong.


On Fabrice Orgogozo left comment #9985 on Section 91.2 in Deformation Theory

Two sentences before Lemma 08S6, *solutuion.


On left comment #9984 on Section 15.87 in More on Algebra

One could add the following result to this section.

Let be an inverse system of rings and let be a subset. Define to be the category of inverse systems of abelian groups such that each is given the structure of a -module and the transition maps is -linear for each with . This is an abelian category. We have a restriction functor , which is exact.

Lemma. Let be an infinite subset. The commutative diagram of functors

\xymatrix{ Mod(\mathbf{N},(A_n)) \ar@{->}[rd]^{\lim} \ar@{->}[dd] & \ & Mod(A) \ Mod(\mathbf{M},(A_n)) \ar@{->}[ru]_{\lim} & }

promotes to a commutative diagram of right derived functors

\xymatrix{ D(Mod(\mathbf{N},(A_n))) \ar@{->}[rd]^{R\lim} \ar@{->}[dd] & \ & D(Mod(A)) \ D(Mod(\mathbf{M},(A_n))) \ar@{->}[ru]_{R\lim} & }

Proof. Very similar to #9983.


On left comment #9983 on Section 15.86 in More on Algebra

I propose to add the following result to this section.

For each set , denote to the category , where is a poset and hence it may be viewed as a category. We have a restriction functor , which is exact.

Lemma. Let be an infinite subset. The commutative diagram of functors

\xymatrix{ Ab(\mathbf{N}) \ar@{->}[rd]^{\lim} \ar@{->}[dd] & \ & Ab \ Ab(\mathbf{M}) \ar@{->}[ru]_{\lim} & }

promotes to a commutative diagram of right derived functors

\xymatrix{ D(Ab(\mathbf{N})) \ar@{->}[rd]^{R\lim} \ar@{->}[dd] & \ & D(Ab) \ D(Ab(\mathbf{M})) \ar@{->}[ru]_{R\lim} & }

Proof. We will apply #9981. Let be a complex of inverse systems. By Lemma 15.86.1, there is a quasi-isomorphism , where is right -acyclic for each . We must check that is right -acyclic. It turn, by Lemma 15.86.1, it suffices to show that is right -acyclic for each .

More generally, we will show: if is right -acyclic, then is right -acyclic. By the proof of Lemma 15.86.1, every inverse system of abelian groups has a ML resolution (actually, a resolution by surjective inverse systems). Pick a ML resolution . Then, inside , is concentrated in degree .

Since is also a ML resolution (for is exact and preserves ML systems), we get is concentrated in degree . Hence, by Derived Categories, Lemma 13.16.4, is right -acyclic.


On left comment #9982 on Section 13.22 in Derived Categories

(The hypothesis in #9981 of being everywhere defined may be dropped since it is already implied by the rest.)


On left comment #9981 on Section 13.22 in Derived Categories

(I missed the title in the reference in #9980, it is J. Lipman, Notes on Derived Functors and Grothendieck Duality.)

This might be added as well. It is Görtz, Wedhorn, Algebraic Geometry II, Remark F.177.

Lemma. Let be additive functors between abelian categories. Assume that is exact, is everywhere defined, and that every complex admits a quasi-isomorphism into a complex such that computes . Then computes for every (hence is everywhere defined) and the morphism coming from Lemma 13.14.16 is an isomorphism.

Proof. Just apply #9980. Every complex of computes for is exact.


On left comment #9980 on Section 13.22 in Derived Categories

I propose to add the following result to this section. It is the translation of J. Lipman, Corollary 2.2.7, to the Stacks Project definitions and results.

Lemma. Let be additive functors between abelian categories. Assume that is everywhere defined and that every complex admits a quasi-isomorphism into a complex computing . In particular, is everywhere defined. The following are equivalent: 1. computes for every . 2. computes for every (hence is everywhere defined) and the morphism coming from Lemma 13.14.16 is an isomorphism.

Proof. Assume 2. Then evaluated at becomes . Thus computes . Conversely, suppose computes . To show that computes , we check the hypotheses of Lemma 13.14.15 for this composite and the class . The first hypothesis is clear. To see the second one, given a quasi-isomorphism , , we have that is a quasi-isomorphism since both and compute (Lemma 13.14.4). Hence is a quasi-isomorphism since and compute (Lemma 13.14.4 again). Lastly, is an isomorphism evaluated at each . Hence it is globally an isomorphism since each is isomorphic to in .


On pause706 left comment #9979 on Section 9.8 in Fields

A trick you could use to avoid missing edge cases is to make a statement chain iff iff and checking each equivalence of statements. So in your case, there is some relation AND X is not sent to 0 via the indeterminate substitution iff there is a relation (via the definition of an integral domain). For lemma 0BID, an alternative proof: clarification: equivalent to iff is algebraic the slightly different part: is isomorphic to (a field) for a minimal polynomial P of and this is necessary for to be a field by Lemma 09G1. The advantage is that this argument can be generalized to not dependent on the ambient field K, since we directly construct one. Lemma 0BMD, alternative proof: A field morphism must send roots of to roots of . Note that both sets are finite. In the case where the morphism is an endomorphism, it follows that it induces a permutation of the roots of since is injective and there are finitely many roots; that is a F-linear field automorphism of E. The advantage is that it avoids linear algebra. Misc: Should "the union of algebraic extensions over a specific field is algebraic" be added? This is very simple to prove since algebraicity is an "element-based" property, and it can useful for Section 9.10 on algebraic closures. Lemma 09GG: "In general, it is very false that an algebraic extension is finite." though every element is contained in a finite extension; it's the union of its finite subextensions (gives intuition for the profinite galois group).


On Joe Lamond left comment #9978 on Section 9.8 in Fields

It seems that my "correction" also has a minor error in it. Suppose are not distinct. Then, for two distinct integers . Then, either or is a root of unity. In my initial comment, I forgot that may be zero. But regardless, the proof seems essentially correct; it just fails to account for these edge cases.


On left comment #9977 on Lemma 15.91.3 in More on Algebra

Ignore #9966. The easy way to see short exactness of is because it is the pushout (in the sense defined in 12.6) of the s.e.s. (given in the proof of 15.91.1) along the map that sends to .


On pause706 left comment #9976 on Section 9.8 in Fields

Joe Lamond I think the argument in Lemma 09GG as is only works if the element chosen is a primitive element. The simplest example where that might fail is the extension adjoining to which has dimension 4 but contains which only has dimension 2. So the statement that we can chose such an element in finite separable extensions seems to be equivalent to the primitive element theorem. The overall statement is still true though, just as you have noted. A simple fix is to consider the multi-set S of powers of instead. Then it is true that S is linearly dependent iff is the root of a polynomial of degree at most (else the extension has degree larger than ). Explicitely, in the case where there are repeats: since there is some such that = 1, then is a root of the polynomial and thus algebraic. A multi-set makes sense since even if , the relation is non-trivial.


On Joe Lamond left comment #9975 on Section 9.8 in Fields

I think the proof of Lemma 09GG has a minor error in it (please correct me if I am mistaken, though). If is a field extension of finite degree n, and , then the elements are not necessarily distinct, hence may be linearly independent. But if the elements are not distinct, then is a root of unity, and hence algebraic over .