The Stacks project

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.

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

@#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.

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

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.

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 $(i,k,m)$.

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

@#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.

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.

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 $$c:\Hom(\Spec R,X)\to \\{ (x,\varphi),x\in X,\varphi\in \Hom_{local}(\mathcal{O}\_{X, x},R)\\},f\mapsto (f(\mathfrak{m}),f^\sharp)$$ 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 $$\\{ (x,\varphi),x\in U,\varphi\in \Hom_{local}(\mathcal{O}\_{U, x},R)\\}\to \\{ (x,\varphi),x\in X,\varphi\in \Hom_{local}(\mathcal{O}\_{X, x},R)\\}$$ 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 $g=h$.

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).

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

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

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...

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 $m$.

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 $m$.

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?

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).

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.

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