The Stacks project

Or don't change the definition of cěch cohomology，but X is Hausdorff and paracompact?

Sorry can I ask a question about lemma 20.16.2? That's: We only know X is a Hausdorff topological space(like a manifold)，but we also only take locally finite open covers as a direct system when define cěch cohomology，is the conclusion still right?

@7414: Dear Vincent, isn't lemma 10.107.9 what you are looking for?

Is it possible to remove the noetherian hypothesis on $\spec R$ by a limit argument of schemes when we replace the absolute ampleness condition by a relative one? It is probably true that the higher direct image cohomology commutes with limits of schemes.

Can we add the following lemma? For each monomorphism $f:\text{Spec } A\to \text{Spec } B$, in other words each ring epimorphism $g:B\to A$, the inverse function $g^{-1}$ is an injection from the prime ideals of $A$ to the prime ideals of $B$.

Lemma 08Z5 is not quite right without assuming that $\mathfrak{m}_R S=\mathfrak{m}_S$.

Lemma 08Z5 is not quite right without assuming that $\mathfrak{m}_R S=\\mathfrak{m}_S$.

Typo in the proof of Lemma 28.21.2: In the last but two line, M×{A}B should be M\otimes{A}B.

Sorry for the last one, typo. I think the definition of $u_p$ between Lemma 7.5.2 and Lemma 7.5.3 should be $$u_ p\mathcal{F}(V) = \mathop{\mathrm{colim}}\nolimits _{(U,\phi)\in \mathcal{I}_ V^{opp}} \mathcal{F}{(U)}$$ I don't know if this is true.

I think the definition of $u_p$ between Lemma 7.5.2 and Lemma 7.5.3 should be $$u_ p\mathcal{F}(V) = \mathop{\mathrm{colim}}\nolimits _{(U,\phi)\in \mathcal{I}_ V^{opp}} \mathcal{F}_ {U}$$ I don't know if this is true.

It is not true that if $a : K \to L$ is zero on cohomology, then $a = 0$. A simple counterexample is to consider any nonzero map $\mathbf{Z}/2\mathbf{Z} \to \mathbf{Z}[1]$ in $D(\mathbf{Z})$. Such nonzero maps exist as the correspond $1$-to-$1$ to nonzero extensions of $\mathbf{Z}/2\mathbf{Z}$ by $\mathbf{Z}$.

Is the condition $H^i(K^\bullet_0)=0$ for $i>0$ necessary?

I think if $a:K^\bullet\to L^\bullet$ induces zero maps on cohomologies. Then $a=0$ in $D(\mathcal A)$ by considering the distinguished triangle $K^\bullet \to L^\bullet \to C(a)^\bullet$. Hence by the distinguished triangle $\tau_{\leq -1} K^\bullet_1 \to K_1^\bullet \to\tau_{\geq 0} K^\bullet_1$ and the fact that $Hom_{D(A)}(K_0^\bullet,-)$ is a homological functor, we know $K_0^\bullet\to K_1^\bullet$ factorizes through $\tau_{\leq -1} K^\bullet_1 \to K_1^\bullet$.

Typo: where it says $f = x^{n+m} + a_1x^{n-1}+\cdots+a_{n+m}$ it should say $f = x^{n+m} + a_1x^{n+m-1}+\cdots+a_{n+m}$

"Given schemes X, Y, Y" should be "[..] X, Y, Z"

In addition to #7400, we can simplifiy the exposition of (1)$\Leftrightarrow$(2) as follows. First, (1)$\Rightarrow$(2) is clear. Next, $E\cap Z$ is a finite union of locally closed subsets $Z_i$ of $Z$, so we can assume $Z=X$ and forget about $X$. Then since $Z$ is irreducible, one of the $Z_i$ must be dense in $Z$, hence open because it is open in its closure.

I think the end of the proof is more complicated then need be. If $\xi \in E$, then $\xi\in E\cap Z$, and we can immediately conclude that $E\cap Z$ is dense in $Z$, no need to introduce any $U_i$.

It is not a mathematical statement to say that "result A is an improvement of result B", so I think it is fine to leave the slightly imprecise formulation of that English sentence for now. And I really do think this result is an improvement since sheaves on a space are the same as sheaves on the corresponding sobrification and the sobrification of a Noetherian topological space is Noetherian. Cheers!

Well, already in order to say what it means that $f$ is affine using Bootstrap, Definition 80.4.1 we need to know that $f$ is representable by algebraic spaces. The factorization of $X \times_Y T \to X$ through some $X_\lambda$ follows from Lemma 87.9.4. So I think the proof is fine.

If $R \subset S$ is an inclusion of domains, then the notation $\text{trdeg}_R(S)$ means the transcendence degree of the induced extension of fraction fields. I am going to leave this alone for now.

@#7365. OK, I hope that people enjoy a trip into the past! Of course, one cannot cite things that once existed in a scholarly work, so maybe we should still remove this discussion from the list of books?