Comments 1 to 20 out of 4268 in reverse chronological order.


On Ashutosh left comment #4535 on Section 9.12 in Fields

Second paragraph of Lemma 09H9 may be needs a line saying that now we prove that $K/F$ is separable.

On BB left comment #4531 on Lemma 10.56.3 in Commutative Algebra

I think there's a small typo in the last paragraph of the proof, in the third to last sentence: I think you want to say IS_f \neq S_f to conclude that (S/I)_f is not zero.

Also, perhaps I missed something, but I could not find anywhere in the proof refereeing to parts (8) or (9) of the lemma.

On 李时璋 left comment #4530 on Section 54.18 in Semistable Reduction

very very small issue, on the first line you had 768g, but on the second line you were saying 728g.

On Aniruddh Agarwal left comment #4529 on Section 10.3 in Commutative Algebra

Trivial remark: in (45), should the article "a" be used instead of "an"?

On left comment #4528 on Definition 12.3.9 in Homological Algebra

Usually, in a situation like this when we are definining a blah which is unique up to isomorphism in the definiition we say "a blah" and then in a comment after the definition we say that because the thing is unique up to unique isomorphism we will from now on use the terminology "the blah". See the text following this definition in Section 12.3.

On awllower left comment #4527 on Lemma 31.4.14 in Limits of Schemes

Two minor typos:

should be

And should be

On Théo de Oliveira Santos left comment #4526 on Section 20.3 in Cohomology of Sheaves

A few trivial typos: Let F be a abelian sheaf Extra parenthesis: The family of functors $H_i((X,−)$ forms [...] from $Ab(X)\rightarrow Ab$ Extra parenthesis*: The family of functors $H_i((X,−)$ forms [...] from $Mod(\mathcal{O}_X)\rightarrow Mod_{\mathcal{O}_X(X)}$

On Aniruddh Agarwal left comment #4525 on Definition 12.3.9 in Homological Algebra

This is a very minor point, but is there a reason for using the terminology "a cokernel", etc. instead of "the cokernel", etc. when these objects are unique upto unique iso?


On left comment #4524 on Lemma 10.98.4 in Commutative Algebra

Yes very good. The key is that here $u$ is surjective whereas in Lemma 10.98.1 it doesn't need to be.

On Ronnie left comment #4523 on Lemma 10.98.4 in Commutative Algebra

To show $u$ is injective, may be we can also argue as follows. Let $K$ denote the kernel of $u$, it is a finite $S$ module. Since $M$ is flat over $R$, applying $-\otimes_RR/\mathfrak{m}$ we get $K\otimes_RR/\mathfrak{m}=0$. It follows that $K\otimes_SS/\mathfrak{m}S=0$, which shows, by Nakayama's lemma, that $K=0$.

On left comment #4522 on Section 21.5 in Cohomology on Sites

The sections of the sheaf $\mathcal{F}$ over an object $U$ of the site are the elements $s \in \mathcal{I}(U)$ whose image in $\mathcal{Q}(U)$ are equal to the restriction $q|_U$. Now you have to define a simply transitive action of $\mathcal{H}(U)$ on this set provided it is nonempty and the construction has to be compatible with restriction mappings. Finally, you have to show that if $\mathcal{F}(U)$ is empty, then you can find a covering $\{U_i \to U\}$ such that each $\mathcal{F}(U_i)$ is nonempty. Does this help?

On Simon left comment #4521 on Section 21.5 in Cohomology on Sites

In the last paragraph, could someone explain why it is easy to verify that $\mathcal{F}$ is a $\mathcal{H}$-torsor. I don't see it, thanks!

On damiano left comment #4519 on Lemma 9.12.1 in Fields

Just a typo: in the statement of the Lemma, after item (2), the "Then second" should be "The second".

On left comment #4518 on Theorem 10.122.12 in Commutative Algebra

Yes, that was confusing. Thanks. Fixed here.

On Manuel Hoff left comment #4517 on Lemma 88.8.4 in The Cotangent Complex

In the proof, $L_{B \otimes _ A B/B} = 0$ should be replaced by $L_{B/B \otimes _ A B} = 0$.

On Noah Olander left comment #4516 on Theorem 10.122.12 in Commutative Algebra

Also in the case $n>1,$ in the third sentence where you cite Lemma 00PN, it seems like you mean to cite Lemma 0C6H.

On awllower left comment #4515 on Lemma 10.118.12 in Commutative Algebra

An alternative way to see that $I\cap R\ne(0)$ is as follows: Take a non-zero $x\in I$. Since $L/K$ is finite, we can take the minimal polynomial $p$ of $x$ over $K$. Write it as $a_nx^n+\cdots+a_0=0$ with $a_i\in K$. By multiplying by some elements in $R$ if necessary, we may assume $a_i\in R,\,\forall i$. Also $a_0\ne0$ as $p$ is minimal. Now $a_nx^n+\cdots+a_1x\in I$, so $a_0\in I\cap R$.

On Noah Olander left comment #4514 on Theorem 10.122.12 in Commutative Algebra

The proof of the case n=1 is confusing. Once you assume $R$ is integrally closed in $S$ you have $S'= R$, yet at the end you write "$\phi (C) \subset S'$..." In fact since $R \subset R[x]/I \subset S$ and $R$ is integrally closed in $S$ so $R = C$ so it seems like you shouldn't even introduce $C$.

On left comment #4513 on Lemma 10.122.6 in Commutative Algebra

Thanks and fixed here.

On Noah Olander left comment #4512 on Lemma 10.122.6 in Commutative Algebra

Looks like in the second to last sentence in the proof you should subtract by $u\phi (a_k x^k)$ not $u \phi (a_k)$.