Comments 1 to 20 out of 4327 in reverse chronological order.


On left comment #4616 on Section 107.22 in Exercises

OK, I understand your confusion. When we have a ring $A$ and an element $f \in A$, then the notation $A_f$ and $M_f$ always mean: $S^{-1}A$ and $S^{-1}M$ where $S$ is the multiplicative subset $S = \{1, f, f^2, f^3, \ldots\}$ of $A$ even when $f$ is nilpotent (in which case $A_f$ is the zero ring and $M_f$ is the zero module). Definition 107.22.1 really means for every $\mathfrak p$ there is an $f \in A$, $f \not \in \mathfrak p$ such that $M_f$ is a finite free $A_f$-module. But note that any module over the zero ring is both zero and free and finite and invertible (either by convention or because this follows from the definition of a free or finite or invertible module).

On Cecilia left comment #4614 on Section 107.22 in Exercises

Or is it that in this definition, localization at a point $f \in$ A means that f is a non-nilpotent element and the set S = {$f^n$ : n = 0,1,...}? Thanks in advance!

On Noah Olander left comment #4613 on Theorem 10.28.9 in Commutative Algebra

I think this is a regular induction, not a descending induction.

On Cecilia left comment #4612 on Section 107.22 in Exercises

In Definition 107.22.1 for a finite locally free A-module, it should be "for every p $\in$ Spec(A), $\exists$ an $f \in$ Spec(A)" instead of "f $\in$ A" because otherwise it does not make sense to localize A at f.

On left comment #4610 on Section 13.17 in Derived Categories

Oops, this is very confusing, because there is a mistake, namely, we should replace the condition that $d^i(E^i) = (D^{i + 1} + E^{i + 1}) \cap \text{Ker}(d^{i + 1})$ by the condition that $d^i(E^i) = (D^{i + 1} + E^{i + 1}) \cap \text{Im}(d^i)$. Thanks for pointing this out. I will fix this very soon.

Let me explain a bit more: for any choice of $E^i \subset X^i$ with $d^i(E^i) \subset D^{i + 1} + E^{i + 1}$ we see that setting $Y^i = D^i + E^i$ gives a subcomplex whose terms are in $\mathcal{B}$ and whose cohomology surjects onto the cohomology of $X^\bullet$. Clearly, if $d^i(E^i) = (D^{i + 1} + E^{i + 1}) \cap \text{Im}(d^i)$ then we see that the map on cohomology is also injective. So now start with some really large index $n$. Then we take $E^n$ equal to $0$. By descending induction we can choose $E^i$ for all $i$ with the desired property.

For example, for $i$ the largest integer such that $X^i$ is nonzero we can choose $E^i = 0$ because this will surject onto $(D^{i + 1} + E^{i + 1}) \cap \text{Im}(d^i) = 0$.

On Aolong left comment #4609 on Section 13.17 in Derived Categories

I fail to follow the proof of the claim in Lemma 0FCL. I don't think the injectivity of the induced map on cohomology is obvious.

The construction of E^i is also confusing to me. For, take i=n, which is the largest number that X^i is nonzero. What is E^n?

So I was wondering it might be a good idea to add more details...

On Rex left comment #4608 on Section 58.1 in Pro-étale Cohomology

The citation to WeilII here actually points to Weil I.

On James Waldron left comment #4607 on Lemma 27.20.2 in Properties of Schemes

Unless this appears elsewhere, it might be useful to add that if $X$ is reduced then one can weaken statement (5) to '$\mathcal{F}$ is finite type and $\rho_{\mathcal{F}}$ is locally constant'. i.e. the requirement that $\mathcal{F}_x$ is free can be dropped.

(This appears as Exercise 13.7.K in the 17/11/2017 version of Vakil's notes The Rising Sea or Exercise 5.8 in Hartshorne.)

The affine case Lemma 00NX could also include the corresponding statement for finite type modules over reduced rings.

On Rex left comment #4606 on Lemma 10.142.14 in Commutative Algebra

I believe part (1) is actually proven in 03PA. It may be helpful to include a link to that here.

On Rex left comment #4605 on Section 56.62 in Étale Cohomology

Some typos:

"Thus if P holds for F1 and F2, then it hods for F"

"To finish the proof have to show that Hd(X,i∗I)=Hd(Z,I) ..."

"As G|U=KL⊗n|π−1(U) is invertible ..." (extra mathcal K here)

On left comment #4604 on Section 56.96 in Étale Cohomology

Singular cohomology is outside the ambit of the Stacks project... at least for now.

On Grétar Amazeen left comment #4603 on Remark 26.2.3 in Constructions of Schemes

Of course! Thank you.

On Rex left comment #4602 on Section 56.96 in Étale Cohomology

Minor typo: "On the same underlying category with have a second topology, namely the étale topology"

Is there any section on comparison between etale cohomology and singular cohomology? If not, are there any plans to add such a section?

On left comment #4601 on Remark 26.2.3 in Constructions of Schemes

Yes, it is because $h$ restricted to the open $X_U$ is given to us and $X$ is the union of the opens $X_U$.

On Grétar Amazeen left comment #4600 on Remark 26.2.3 in Constructions of Schemes

Are the morphisms $h:X\to Y$ unique?

On David Tweedle left comment #4599 on Lemma 9.19.1 in Fields

No problem!

On left comment #4598 on Lemma 9.19.1 in Fields

Sorry, my bad. You are of course completely correct and you said it correctly the first time too. I will fix this when I next go through all the comments. Thanks!

On David Tweedle left comment #4597 on Lemma 9.19.1 in Fields

I'm sorry, either I am missing something else, or I did not express myself clearly. Let us assume $F$ is infinite (as in the proof of the lemma, I agree with you that most people would apply this to the parenthetical comment). This is false: a finite union of proper subvector spaces is never a subvector space. Counterexample: If $U,W$ are proper subvector spaces of $V$, then $U\cup W$ is a subvector space as long as $U\subseteq W$ or $W\subseteq U$. This is true (and I think this is what was intended): a vector space cannot be written as a finite union of proper subvector spaces.

On left comment #4596 on Lemma 9.19.1 in Fields

@#4595 Yes, and just two sentences earlier we reduced to the case where $F$ is infinite. I think most people would read the parenthetical remark as applying to that case.

On David Tweedle left comment #4595 on Lemma 9.19.1 in Fields

In the proof of the lemma, the statement (in the second paragraph of the proof) "...(a finite union of proper subvector spaces is never a subvector space; details omitted)" is false. A counterexample is any proper subvector space of a vector space. It is sufficient for the lemma to know that a vector space can never be equal to a finite union of proper subspaces if $F$ is infinite.