Comments 1 to 20 out of 6164 in reverse chronological order.


On Yuto Masamura left comment #6615 on Lemma 27.12.1 in Constructions of Schemes

Thanks for your comment, but I wanted to ask about the part of uniqueness of $\varphi$. More precisely, the part 'we have $\alpha^{\otimes n}(\psi_\varphi^d(x))=(\alpha')^{\otimes n}(\psi_{\varphi'}^d(x))$ and $\psi_\varphi^d(x)=\varphi^\sharp(x/f^n)\psi_\varphi^d(f^n)$ and similarly for $\psi_{\varphi'}^d$, and hence we have $\varphi^\sharp(x/f^n)=(\varphi')^\sharp(x/f^n)$.'

But it's OK since I solved it by myself. At the time I misunderstood that you deduce $\psi_\varphi^d(x)=\psi_{\varphi'}^d(x)$ from the first equation. The correct way is as follows: first we deduce that $\alpha(\varphi^\sharp(x/f^n))\cdot\psi(f)^n=\alpha'((\varphi')^\sharp(x/f^n))\cdot\psi(f)^n$ over $Y_{\psi(f)}$ and hence $\alpha(\varphi^\sharp(x/f^n))=\alpha'((\varphi')^\sharp(x/f^n))$. Since $\alpha$ and $\alpha'$ are the identity in degree $0$ (I overlooked this fact), we get the conclusion.

On Liu S-H left comment #6614 on Section 58.13 in Étale Cohomology

"These two are equal since by assumption $s \in \check H(\mathcal{U}, \mathcal{F})$ and" should be "$s \in \check H^0(\mathcal{U}, \mathcal{F})$"?

On Laurent Moret-Bailly left comment #6613 on Section 15.47 in More on Algebra

About the comment following Definition 07P7: J-1 implies J-0 for nonzero reduced rings. In fact, in the rest of the section at least, the J-0 condition is considered mostly for domains. This suggests that a better definition for J-0 might be that $\mathrm{Reg}(X)$ contains a dense open.

On left comment #6612 on Section 15.47 in More on Algebra

Why is it that rings with empty Reg(X) are ruled out from being J-0? I've tried to glean from the theory here why it might be, but I think I am underinformed.

Also, doesn't the wording of Lemma 07P8 imply that J-1 rings never have an empty regular locus? (The wording: The ring R is J-1 if and only if V(𝔭)∩Reg(X) contains a nonempty open subset[...] ) I thought part of the point was that J-1 rings are allowed to have empty regular loci.

On WhatJiaranEatsTonight left comment #6611 on Lemma 42.18.3 in Chow Homology and Chern Classes

I see. The two schemes are not of the same dimension. Sorry for my ignorance.

On WhatJiaranEatsTonight left comment #6610 on Lemma 42.18.3 in Chow Homology and Chern Classes

Do I miss some details? I think that by lemma 18.1, we know that $c_*div(f)=div(g)$. And obviously g is a unit on $Spec(k)$. Thus $div(g)=0$.

But the proof seems much longer than I expected. I don't know if I miss some crucial details or misunderstand something.

On Yijin Wang left comment #6609 on Section 15.108 in More on Algebra

There are two typos in Lemma 0BRJ. In the first paragraph, 'Since R⊂R^{G} is an integral ring extension' should be 'Since R^{G}⊂R is an integral ring extension.' 'We may replace R by B' should be 'We may replace R^{G} by A'

On Jonas Ehrhard left comment #6608 on Section 103.5 in Introducing Algebraic Stacks

The cocycle condition for the $\alpha_{ij}$ misses a ${}^*$ I think.

On left comment #6607 on Lemma 10.136.13 in Commutative Algebra

The point is that $R$ will be the filtered union of Noetherian rings for which the result is true. Then you use that if $R = colim_{i \in I} R_i$ is a filtered colimit and if $f_1, \ldots, f_r \in R_0$ for some $0 \in I$ form a regular sequence in each $R_i$ for $i \geq 0$, then $f_1, \ldots, f_r$ form a regular sequence in $R$.

On WhatJiaranEatsTonight left comment #6606 on Lemma 10.136.13 in Commutative Algebra

(2) is equivalent to that $R[x_1,\ldots,x_n]/(f_1,\ldots,f_c)$ is flat over $R$. And since flatness is preserved under base change, we can reduce (2) to Noetherian case.

But I don't know how to reduce the case to Noetherian for (1).

On Yuto Masamura left comment #6605 on Lemma 30.14.3 in Cohomology of Schemes

@6604 Sorry, my typo: "How the map ... is obtained" should be "How is the map ... obtained".

On Yuto Masamura left comment #6604 on Lemma 30.14.3 in Cohomology of Schemes

I have a problem reading "we have a canonical isomorphism $\widetilde N\to\widetilde M$ such that $M_n\to N_n=\Gamma(X,\widetilde M(n))$ is the canonical map." How the map $M_n\to N_n$ is obtained from the isomorphism $\widetilde N\to\widetilde M$? (I think we want to say that the induced map $\widetilde M\to\widetilde N$ by the map $M\to N$ is equal to the inverse of the isomorphism $\widetilde N\to \widetilde M$, but...)

On Jonas Ehrhard left comment #6603 on Lemma 15.21.3 in More on Algebra

In the second sentence of the proof it should be $P_i(x_i) = 0$ (the index $i$ is missing).

On Jonas Ehrhard left comment #6602 on Lemma 10.36.2 in Commutative Algebra

Is that not just an application of Lemma 05BT to the morphism $\varphi: M \to M, x \mapsto y \cdot x$? From there we get a polynomial $P$ with $P(\varphi) = 0$, and then $P(y) = P(y) \cdot 1 = P(\varphi)(1) = 0$.

On WhatJiaranEatsTonight left comment #6601 on Lemma 10.103.11 in Commutative Algebra

The last inequality should be $\text{depth}(M_{\mathfrak p})\geq \dim (\text{Supp}(M_{\mathfrak p}))$. The right bracket is omitted.

On WhatJiaranEatsTonight left comment #6600 on Lemma 10.103.9 in Commutative Algebra

"Then any maximal chain of ideals". It should be "Then any maximal chain of primes".

On left comment #6599 on Definition 10.102.5 in Commutative Algebra

On WhatJiaranEatsTonight left comment #6598 on Definition 10.102.5 in Commutative Algebra

Does $I(\phi)$ means the ideal generated by the determinants of $r\times r$ minors?

On suggestion_bot left comment #6597 on Lemma 10.115.4 in Commutative Algebra

Suggested tag: Noether normalization

On Hunter left comment #6596 on Section 5.26 in Topology

curious why "quasi-compact" is used instead of "compact" here everywhere since all spaces on the page are Hausdorff