Comments 1 to 20 out of 5480 in reverse chronological order.


On left comment #5890 on Lemma 13.16.7 in Derived Categories

Let $A^\bullet$ be a bounded below complex of acyclic objects. I think we can replace the assumption that $RF$ is everywhere defined by the assumption that $RF$ is defined at $A^\bullet$. I will make this change the next time I go through all the comments.

I do not see is how to prove that $RF$ is defined at $A^\bullet$. The problem, I think, is the following. Let $s : A^\bullet \to M^\bullet$ be a quasi-isomorphism. Then I think the arguments in the proof and Lemma 13.5.8 allow us to conclude that given an integer $n$ there is a quasi-isomorphism $s' : M^\bullet \to N^\bullet$ such that the composition $F(A^\bullet) \to F(M^\bullet) \to F(N^\bullet)$ is an isomorphism on cohomology in degrees $\leq n$. But in order to show that $RF$is defined at $A^\bullet$ with value $F(A^\bullet)$ we would need to show that we can do this in all degrees simultaneously!

Thanks for the interesting question.

On left comment #5889 on Lemma 13.16.7 in Derived Categories

Can we remove the hypothesis that $RF$ is everywhere defined? At least in the bounded case it's obvious that we can.

On Zhenhua Wu left comment #5888 on Lemma 26.13.1 in Schemes

@#5882, denote the functor as $c$, for me the problem of confusion is that the relation between $c$ restricted to open affine $U$ and $c$ on $X$ is not clear. For example, before we prove that $U\hookrightarrow X$ is a monomorphism in the category of schemes in tag 01L7 (so for this tag not proved yet), we cannot use the fact $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,U)$ injects into $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,X)$. So in this case I prefer we emphasis the difference between the set $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,U)$ and the set $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,X)$. Actually I would drow a square for it if I know how to draw it in this editor.

And we don't really need the injection in this proof. For the shortest suggestion on the proof, I would add something like:

We proved every element of $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,X)$ can be lifted to an element in $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,U)$ for some open affine $U$. Providing $c$ is functorial. Then the injectivity of $c$ on $X$ reduces to the injectivity of $c$ on $U$, which is already proved.

On Abhiram Natarajan left comment #5887 on Section 10.16 in Commutative Algebra

Slight abuse of notation in point 11 of lemma 00E0. It should be V((f)).

On Anna Cadoret left comment #5886 on Section 12.8 in Homological Algebra

This is a stupid comment but the notion of kernel in 10.12.5 is only introduced for an additive functor between Abelian categories while, here, it is used in the more general context of an additive functor between additive categories.

On Laurent Moret-Bailly left comment #5884 on Lemma 28.23.3 in Properties of Schemes

In the statement, delete the second (or the third) occurrence of "quasi-coherent". Also, I had some trouble understanding the definition of $S'_j$. For more clarity, I think the union should run over triples $(i,k,m)$.

On left comment #5883 on Lemma 86.7.1 in Restricted Power Series

The proof of this lemma is wrong unfortunately. Please refrain from using this lemma for now.

On left comment #5882 on Lemma 26.13.1 in Schemes

@#5879: In stead of giving your own proof, could you comment on what sentence(s) of the current proof should be clarified? I'm a bit surprised you are complaining about the deduction of the general case from the affine case, as that seems to me is explained quite well.

@#5880. Yes. I'm sorry about the mismatch between the editor and what readers get to see. As you say, this has been a problem for a while. We don't quite know how to fix this.

On Zhenhua Wu left comment #5880 on Lemma 26.13.1 in Schemes

The typo above is an example of this issue \ref{https://stackoverflow.com/questions/29014573/mathjax-how-to-deal-with-this-strange-behavior/29040570#29040570}. Note that adding underscore after right curly bracket instantly turned into italc form (you can see the italic button on the tool row becomes highlighted). I added backslash sign before underscore sign to make it looks right in preview. But now in real output it screws up. Let me test again. One backslash: $\mathcal{O}\_{X,x}, \mathcal{O}\_{X,x}$. Correct in preview, probably wrong in real output. No backslash: $\mathcal{O}_{X,x}, \mathcal{O}_{X,x}$. Wrong in preview, probably correct in real output.

On Zhenhua Wu left comment #5879 on Lemma 26.13.1 in Schemes

I don't see how bijection is proved in the general scheme case. It's not that the proof is wrong, but morelike it's not enough. The rest of the proof still needs some non-trivial logic. $\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\Spec}{Spec}$ Define as above. We first show it's surjective. Given any $(x,\varphi)$, there exists an open affine $U=\Spec A$ containing $x$, and a map $A \to \mathcal{O}\_{X, x}$, by composing with $\varphi:\mathcal{O}\_{X, x}\to R$, we have $f:\Spec R\to \Spec A\to X$. It's easy to see $c(f)=(x,\varphi)$. Next we show it's injective. Given any two maps $g,h:\Spec R\to X$, s.t. $c(g)=c(h)=(x,\varphi)$. We know $g(\mathfrak{m})=h(\mathfrak{m})=x$. Pick an open affine $U=\Spec A$ containing $x$ we know both $g$ and $h$ factor through $U$ using the orginal argument. So we have $g_0,h_0:\Spec R\to U$. It suffices to show $g_0=h_0$. Clearly $c$ is functorial w.r.t. the open immersion $U\to X$, and the canonical map is an identity map. We have shown $c_U$ is a bijection (defined as $c$ with $U$ replacing $X$). So $c\_U (g\_0)=c\_U(h\_0)$ implies that $g\_0=h\_0$, which in turn implies $g=h$.

On Paolo left comment #5878 on Section 10.41 in Commutative Algebra

I think that in Lemma 04KM it would be better to use two different letters for $d=[K:K_{sep}]$ and $d=\mathrm{deg}(P)$ in order to avoid confusion (they are different things).

On Sasha left comment #5876 on Section 37.20 in More on Morphisms

In the statement of Lemma 054T it is not clear what is $W$.

On Guillermo Barajas Ayuso left comment #5875 on Lemma 13.9.12 in Derived Categories

I think it should read $b'\circ b$ in the last sentence of the proof.

On Rankeya Datta left comment #5874 on Lemma 10.154.8 in Commutative Algebra

The notation $\frak{m}^h$ feels a little strange to use in the statement of the Lemma because the notation $\frak{m}$ does not appear anywhere else in the statement. That being said, I understand that $(\frak{p}R_{\frak{p}})^h$ looks clumsy...

On Dibya left comment #5873 on Section 10.152 in Commutative Algebra

This note is really helpful, I just have a small query: Proving the "$8 \Rightarrow 10$" of lemma 152.3, when we have $R \rightarrow R'$ an etale map and $\tau : R' \rightarrow R$ as a section, should we not get $\tau^{-1} (m) = m'$ instead of $\tau^{-1}(m') = m$. Since we are applying 8 on $R \rightarrow R'$ and $m'$ is the prime ideal of $R'$ lying over $m$.

On Dibya left comment #5872 on Section 10.152 in Commutative Algebra

This note is really helpful, I just have a small query: Proving the "$8 \Rightarrow 10$" of lemma 152.3, when we have $R \rightarrow R'$ an etale map and $\tau : R' \rightarrow R$ as a section, should we not get $\tau^{-1} (m) = m'$ instead of $\tau^{-1}(m') = m$. Since we are applying 8 on $R \rightarrow R'$ and $m'$ is the prime ideal of $R'$ lying over $m$.

On Josh left comment #5871 on Section 22.12 in Differential Graded Algebra

Should we not speak of $A$-balanced maps, rather than $A$-bilinear maps, in a context where $Q$ is not an $A$-module but merely an $R$-module?

On MCO left comment #5870 on Section 96.20 in Artin's Axioms

It may be worth pointing out (unless it is already in here somewhere) that strong formal effectiveness holds for qcqs algebraic spaces (Theorem 1.1 in Bhatt-Algebraize).

On Bjorn Poonen left comment #5869 on Definition 10.104.1 in Commutative Algebra

As worded, the definition sounds as if it is requiring the finite maximal chains to have equal length, while imposing no condition on infinite maximal chains. I suspect that the definition should instead say

A ring $R$ is said to be catenary if for every pair of prime ideals $\mathfrak{p} \subset \mathfrak{q}$, there exists an integer bounding the lengths of all finite chains of primes $\mathfrak{p} = \mathfrak{p}_0 \subset \mathfrak{p}_1 \subset \cdots \subset \mathfrak{p}_e = \mathfrak{q}$, and the maximal such chains all have the same length.

Less importantly, I wonder also whether it would be clearer to define "saturated" chain and to use that word instead of "maximal" since I could imagine that some people could misunderstand maximal to mean maximal-length (even though they would quickly realize that that could be not be what was meant).

Even less important: Note that it is better to use \cdots instead of \ldots in the chain.

On Rein Janssen Groesbeek left comment #5868 on Section 4.21 in Categories

The sentence "Before we do so ... difference between" before tag 0032 seems to be cut off, is this intentional?