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Comments 1 to 20 out of 8784 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 #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.”


On left comment #9443 on Lemma 5.8.4 in Topology

In the second paragraph, it seems that “take one of the and expand it to an irreducible component ” needs Lemma 5.8.3, point (3), which requires Zorn's lemma. It can be done without the axiom of choice: the second paragraph may be replaced by

Suppose , where is some irreducible subset of . By the argument in the first paragraph, for some . Since the original union does not have redundant members, . Therefore is maximal in the set of irreducible subsets of , that is, is an irreducible component of .


On Branislav Sobot left comment #9442 on Lemma 29.39.3 in Morphisms of Schemes

I belive that in the first part of the proof we should take in the exponent of number instead of , so that total degree is N.


On Ryo Suzuki left comment #9441 on Lemma 10.32.5 in Commutative Algebra

Oops! I rewrite it.

Let be a ring. Let be an ideal of and be a finitely generated ideal of . Suppose . Then for some .


On Ryo Suzuki left comment #9440 on Lemma 10.32.5 in Commutative Algebra

It can be slightly generalized as follows: Let be a ring. Let be an ideal of and be a finitely generated ideal of . Suppose . Then for some .

Then Lemma 00L6 also can be generalized in the same way.


On left comment #9438 on Lemma 12.4.2 in Homological Algebra

Going to leave as is.


On left comment #9437 on Section 14.34 in Simplicial Methods

Thanks and fixed here.


On left comment #9436 on Definition 78.19.3 in Groupoids in Algebraic Spaces

Thanks and fixed here.