Comments 1 to 20 out of 5542 in reverse chronological order.


On Ian Prado left comment #5953 on Section 109.2 in Exercises

:)

On Owen left comment #5952 on Lemma 60.11.6 in Pro-étale Cohomology

'Moreover, every point of $Y$ specializes to a unique point of $Y_0$ as (a) this is true for $X$ and (b) the map $X\to Y$ is separated.' It should be $Y\to X$, not $X\to Y$, but I still don't follow. It's possible for a specialization in $X$ to lift to more than one specialization along a separated map $Y\ra X$.

On Fabian Korthauer left comment #5951 on Lemma 5.8.13 in Topology

The assumption that $f$ is continuous seems to be superfluous. Furthermore one could maybe add a comment (resp. add an analogous lemma somewhere else) that the statement remains true with essentially the same prove if one exchanges the word "irreducible" everywhere by the word "connected".

On Owen left comment #5950 on Lemma 60.11.5 in Pro-étale Cohomology

The reference to tag 0978 suggests that $\operatorname{Spec} A$ is presumed profinite; perhaps that hypothesis should be added to the statement of the lemma? (Definition 08YI of extremally disconnected space doesn't presume compact Hausdorff and I believe there are extremally disconnected spectral spaces that aren't profinite.)

On Dylan left comment #5949 on Lemma 13.16.1 in Derived Categories

I believe it should be "for i<a" in the last sentence of the first paragraph.

On Bjorn Poonen left comment #5948 on Section 29.6 in Morphisms of Schemes

Regarding (1) and (2) after Definition 01R7, there is a counterexample simpler than Example 01QW: the disjoint union $X = \coprod_{n \ge 1} \operatorname{Spec} k[t]/(t^n)$ mapping to $Y=\operatorname{Spec} k[t]$.

(1) Here $\overline{f(X)}$ is a point, but the scheme-theoretic image $Z$ is all of $Y$.

(2) The formation of the scheme-theoretic image does not commute with restriction to $U = \operatorname{Spec} k[t,1/t] \subset Y$ (the preimage of $U$ in $X$ is empty).

On Bjorn Poonen left comment #5947 on Section 29.6 in Morphisms of Schemes

In response to Comment #2119 by David Hansen on July 16, 2016 at 18:45 Comment #2140 by Johan on July 21, 2016 at 19:47

I wonder if the search behavior has changed since then. For me a search for "scheme-theoretic image" does find this section.

In any case, I agree with David Hansen that "scheme theoretic" should be "scheme-theoretic". Likewise, "set theoretic" should be "set-theoretic", not just in this section, but throughout.

On Zhouhang MAO left comment #5946 on Lemma 10.134.6 in Commutative Algebra

"In particular, $H_1(L_{B/A})=I/I^2$. Here it should be $H_1(NL_{B/A})=I/I^2$ (although the statement for the cotangent complex is correct, this is not what you want to say here).

On left comment #5944 on Section 17.15 in Sheaves of Modules

Yes, bilinear maps $f : \mathcal{F} \times \mathcal{G} \to \mathcal{H}$ of sheaves of modules are those maps of sheaves of sets which give $\mathcal{O}_X(U)$-bilinear maps on sets of sections over $U$ for all opens $U \subset X$.

Equivalently you can ask certain diagrams of maps of sheaves of sets commute, immitating the usual axioms for bilinear maps of modules such as the axiom $f(x + y, z) = f(x, z) + f(y, z)$, etc.

If $f$ is a map of sheaves of sets and it induces a bilinar map of modules on stalks for all points of $X$, then $f$ is a bilinear map of sheaves of modules as you can test whether local sections are equal by checking on stalks.

Let $\text{Mor}( - , - )$ denote morphisms in the category of sheaves of sets on $X$. Another way you could define the notion of a bilinear map is this: a map of sheaves of sets $f : \mathcal{F} \times \mathcal{G} \to \mathcal{H}$ is bilinear if given any sheaf of sets $\mathcal{S}$ the rule $\text{Mor}(\mathcal{S}, \mathcal{F}) \times \text{Mor}(\mathcal{S}, \mathcal{G}) \to \text{Mor}(\mathcal{S}, \mathcal{H})$, $(a, b) \mapsto f \circ (a \times b)$ is a bilinear map of modules over the ring $\text{Mor}(\mathcal{S}, \mathcal{O}_X)$. We don't usually take this point of view as it is easier to think about sets of local sections and it is clearly equivalent.

Another way to say the definition: $\mathcal{O}_X$ is a ring object in the category of sheaves of sets and $\mathcal{F}$, $\mathcal{G}$, $\mathcal{H}$ are module objects over this ring. Then a bilinear map can be defined for module objects over a ring object in any category. (To formulate what is a ring object and what is a module object over a ring object, and what is a bilinear map of such in a category it is pleasant --but not strictly necessary-- to assume the category has finite products; and this is true for the category of sheaves of sets.)

On Tim Campion left comment #5943 on Section 17.15 in Sheaves of Modules

Is it fair to omit the definition of a bilinear map of $\mathcal O_X$-modules? After all, a bilinear map of modules $f: M \times N \to P$ is a function such that for every point $m \in M$ the map $f(m,-): N \to P$ is linear, etc. The most naive tranlsation into sheaves would therefore say that $f: \mathcal F \times \mathcal G \to \mathcal H$ is bilinear if for every global section $s$ of $\mathcal F$, the map $f(s,-): N \to P$ is $\mathcal O_X$-linear, etc. This definition is clearly wrong. I think one wants to say that $f|_U: \mathcal F(U) \to \mathcal G(U) \to \mathcal H(U)$ is $\mathcal O_X(U)$-bilinear for every $U$, but I'm not quite sure. I'm also not sure whether bilinearity can be checked on stalks, etc.

On Laurent Moret-Bailly left comment #5942 on Lemma 70.12.3 in Algebraic Spaces over Fields

First line of proof: delete "be a morphism".

On Dario Weißmann left comment #5941 on Lemma 70.12.3 in Algebraic Spaces over Fields

typos: Y_K should be X_K when showing that the fibres are connected. Also in the next sentence "Y is geometrically connected" should be "X is geometrically connected".

On Friedrich Knop left comment #5940 on Section 35.3 in Descent

Typo: "because it does not holds"

On Dario Weißmann left comment #5939 on Lemma 92.10.2 in Algebraic Stacks

@#26: seconded

On Dario Weißmann left comment #5938 on Section 92.8 in Algebraic Stacks

typo: give a morphism -> given a morphism

On Gabriel Ribeiro left comment #5937 on Lemma 12.7.1 in Homological Algebra

Actually we don't need to compose with the isomorphisms to check whether the triangles commute or not. They commute by the very definition of the morphisms in (2) and (3). Also, I would suggest that it is perhaps clearer to use the remark 0103 instead of the lemma 0105 to prove that (1) implies (2) and (3).

On Dario Weißmann left comment #5936 on Section 101.3 in Cohomology of Algebraic Stacks

second paragraph: I suggest to move the second " in "sheaf of O_X" modules to the end and write O_X - modules instead of O_X modules

On Morten left comment #5935 on Section 55.14 in Semistable Reduction

By symmetry the case of the algebra A[𝔪y] is the same as the case of A[𝔪y].

Should one y be replaced by x?

On Bach left comment #5934 on Section 12.6 in Homological Algebra

I am not sure this is important, but in the construction of Baer sum, when I read: "we pull back...and we push out..."makes me think that we start by pulling back the diagonal map, but the following diagram is in the different order.

On Zhouhang MAO left comment #5933 on Section 59.6 in Crystalline Cohomology

In the definition of universal divided power $A$-derivation, there is a typo: "for some $B$-linear map $\Omega_{B/A,\delta}\to M$" - you should add the name $\xi$ before the map. Furthermore, it seems better to stress that such map is unique.

In addition, you call the module "the divided power module of differentials", which I find somehow imprecise. I suggest to say "the module of (divided power) differentials", since it is about differentials preserving the divided power structure.