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Comments 261 to 280 out of 9050 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 left comment #9468 on Lemma 13.31.7 in Derived Categories

Suggested alternative proof: Use Lemmas 13.31.6 and 13.14.14.


On left comment #9467 on Lemma 13.31.2 in Derived Categories

One could add to the statement:

(4) Any quasi-isomorphism is a homotopy equivalence.

Proof of the equivalence. (2)(4). Precomposition by induces a bijection . This implies that has a unique left inverse in . In a preadditive category, this implies invertibility.

(4)(3). A morphism in is of the form , where and is a quasi-isomorphism. Hence is onto. It is also injective: if is a map in that becomes zero in , then by Homology, Comment #9465 there is a quasi-isomorphism with in . Thus in .


On left comment #9466 on Lemma 13.18.8 in Derived Categories

Suggested slogan: A bounded below complex of injectives is K-injective.


On left comment #9465 on Section 12.8 in Homological Algebra

For future reference, and in case anyone wants to chime in. To this section, I would add the following result:

Lemma. Let be a preadditive category. Let be a LMS in . Let be a morphism in . Then if and only if there exists in with .

Proof. This is a particular case of Categories, Lemma 4.27.6.

This was originally proposed in #8372.


On left comment #9464 on Lemma 13.31.6 in Derived Categories

For anyone that might care: from Lemma 13.31.2 we get that is bijective. In particular, has a left inverse. This alone already implies that is a final object of .

Even though one does not need it in the proof, actually is invertible in . Bijectivity of \eqref{1} means that has a unique left inverse, and this implies existence of an inverse in a preadditive category. The dual result is proven here.


On nkym left comment #9463 on Example 22.26.6 in Differential Graded Algebra

Sorry, forget about


On nkym left comment #9462 on Example 22.26.6 in Differential Graded Algebra

You need to assume has countable direct sums and products.


On left comment #9461 on Lemma 10.120.7 in Commutative Algebra

Suggestion: it would be easier to rememeber the statement if was written with the word "unit" at the same place as in i.e. if it was written:

"(2) is prime if and only if the image of in is a prime element or a unit in ."


On left comment #9460 on Lemma 10.120.7 in Commutative Algebra

The final statement should probably be: "Moreover, then is a UFD if and only if every nonzero nonunit element of has a factorization into irreducibles and is a UFD."


On nkym left comment #9459 on Proposition 29.27.2 in Morphisms of Schemes

In (c) in the proof, should be and is flat over .


On left comment #9458 on Section 43.24 in Intersection Theory

@#9456 Hints: use 43.23.3, use that , and use that the maps on the for restrict to finite maps on the .


On AprilGrimoire left comment #9457 on Section 43.23 in Intersection Theory

Lemma 0B2V: Irreducibility of polynomials corresponds to integralness of closed subschemes. Therefore, I think irreducibility in the proof concerning schemes should be replaced with integralness.


On AprilGrimoire left comment #9456 on Section 43.24 in Intersection Theory

How does the induction for (4) in the proof of [0B0E] work? I couldn't figure it out. In the reference of [Roberts], it is done by an analysis of the dimension in the Grassmannian variety, instead of by induction.

Thanks!


On Guoquan Gao left comment #9455 on Section 43.24 in Intersection Theory

  1. In the proof of Lemma[0B0E], the sentence "For any irreducible component of Ti∩Z contained in Ei we have the desired dimension bound" should be changed to "For any irreducible component of Ti∩Zj contained in Ei we have the desired dimension bound.
  2. On the second to last line of the proof of Lemma[0B1U], the word "transversally" should be changed to "properly".

On Branislav Sobot left comment #9454 on Lemma 37.49.1 in More on Morphisms

If one wishes, it is faster to conclude part (4) from part (5) via lemma 29.37.7


On AprilGrimoire left comment #9452 on Section 43.23 in Intersection Theory

Lemma 0B2V: I think here maybe its better to explain this element is geometrically irreducible, and reduce to an algebraically closed field so image could be considered at closed points, which are decent linear morphisms.


On AprilGrimoire left comment #9451 on Section 43.23 in Intersection Theory

Lemma 0B2T: did not exist in the conclusion of the statement.


On Ivan left comment #9450 on Lemma 37.67.4 in More on Morphisms

Typo in the proof: says "universall" instead of "universally"


On left comment #9445 on Section 10.63 in Commutative Algebra

I think it could be interesting to add the following result to this section:

Lemma. Let be a finite module over a Noetherian ring . Then

Proof. Since is finite, then , see Algebra, Lemma 10.40.5. Thus, by Algebra, Lemma 10.17.2, point 7, we have . Suppose we have shown that for , every prime contains a minimal prime of . Then it will follow from Lemma 10.63.6. On the one hand, the assertion is true for by Lemma 10.63.5. On the other hand, each point in is contained in some irreducible component of by Topology, Lemma 5.9.2 ( is Noetherian since is). Since is sober (use Topology, Lemma 5.8.7 and that is sober), this means that every point in generalizes (in the sense of Topology, Definition 5.19.1) to a minimal prime of , as we wanted to show.


On left comment #9444 on Lemma 5.9.2 in Topology

In the second paragraph, first sentence, I think we should replace “which do not have finitely many irreducible components” by “which cannot be written as a finite union of irreducible closed subsets” (and in “by construction and are finite unions of their irreducible components,” change components by subsets). Otherwise, unless we appeal to Lemma 5.8.3, point 3 (which requires Zorn's lemma), we cannot write and as a union of irreducible closed subsets.

In the statement, point 2, one could add “and it is the union of these.”

At the end of second paragraph, the last sentence could be replaced by “since is empty, is a finite union of some irreducible subsets ; removing redundant 's, we win by Lemma 5.8.4.”