The Stacks project

It would be nice if it was stated somewhere that the map $r_{\mathcal{L}, \psi}$ coincides with the map $r_{\psi}$ in https://stacks.math.columbia.edu/tag/07ZG . Namely, https://stacks.math.columbia.edu/tag/01O8 gives us a canonical morphism

$$r_{\mathcal{L}, \psi}\colon U(\psi) \to \underline{\text{Proj}}_S(\mathcal{A})$$

and https://stacks.math.columbia.edu/tag/07ZG gives us a canonical morphism

$$r_{\psi}\colon U(\psi)' \to \underline{\operatorname{Proj}}_X(f^* \mathcal{A}).$$

I think $U(\psi) = U(\psi)'$ and it would be nice if there was a lemma saying that $r_{\mathcal{L}, \psi}$ coincides with the composition of $r_{\psi}$ and the projection

$$\underline{\operatorname{Proj}}_X(f^* \mathcal{A}) \to \underline{\operatorname{Proj}}_S(\mathcal{A}).$$

It seems that this might be implicitly used in the proof of https://stacks.math.columbia.edu/tag/01VR part (4).

Strictly speaking, the first sentence of the section is incorrect: namely, G-unramified maps are called "unramified" in EGA.

More generally: If $X$ is étale over $S$ and $Y$ is unramified over $S$, then $X\to Y$ is étale. See Görtz, Wedhorn, Algebraic Geometry II, Springer, Remark 18.30.(2).

In the first paragraph of the proof of Lemma 37.43.3, should “finite quasi-coherent O_X-algebras” be “finite quasi-coherent O_S-algebras”?

I think there is a shorter proof of a more general statement.

Theorem: Let $R$ be a ring. Suppose an $R$-module $M$ admits a minimal generating set (with respect to inclusion). If $IM=M$ for every annihilator $I$ of a simple $R$-module, then $M=0$.

Proof: By contradiction. If $M\neq 0$, the minimal generating set contains at least one element $m\neq 0$. By Zorn there exists a maximal submodule $N\subset M$ not containing $m$ but containing all other generators from our minimal generating set. The quotient $M/N$ has to be simple. If we let $I$ be its annihilator, then $IM\subset N\neq M$. QED

I think there is a shorter proof of a more general statement.

Theorem: Let $R$ be a ring. Suppose an $R$-module $M$ admits a minimal generating set (with respect to inclusion). If $IM=M$ for every annihilator $I$ of a simple $R$-module, then $M=0$.

Proof: If $M\neq 0$, the minimal generating contains at least one element $m\neq 0$. By Zorn there exists a maximal submodule $N\subset M$ not containing $m$. The quotient $M/N$ has to be simple. If we let $I$ be its annihilator, then $IM\subset N\neq M$. Contradiction. QED

I think there is a shorter proof of a more general statement.

Theorem: Suppose a module $M$ admits a minimal generating set (with respect to inclusion). If $IM=M$ for every annihilator $I$ of a simple module, then $M=0$.

Proof: If $M\neq 0$, the minimal generating contains at least one element $m\neq 0$. By Zorn there exists a maximal submodule $N\subset M$ not containing $m$. The quotient $M/N$ has to be simple. If we let $I$ be its annihilator, then $IM\subset N\neq M$. Contradiction. QED

The passage from $I/I^2 \otimes \kappa(\mathfrak{q})$ to $I_j \otimes \kappa(\mathfrak{q}_j)$ is somewhat terse. Maybe one can cite the item (1) of Tag 08JZ here.

In the second paragraph of the proof, $(J, x_i - f_i)$ should be $(J, x_i - g_i)$.

The converse holds: If $a:F\to G$ is representable, flat, locally of finite presentation and surjective as a map of sheaves, then it is surjective. See Corollary 3.

Typos: "Hence $U\times_Y X$ is a quasi-compact algebraic space" should have $U\times_{\mathcal{Y}}\mathcal{X}$ instead. In the final paragraph, the penultimate sentence should also read "Then $p^{-1}(U) = U\times_{\mathcal{Y}} Z$ and the algebraic space $U\times_{\mathcal{Y}} Z$ is quasi-compact...."

@#6834: your claim is recorded as Corollary 3.

@#6567: I was about to say what #6511 said and then I realized about their comment.

I think it would be better to let the Grassmann $\operatorname{G}(k,n)$ be defined for $0\leq k\leq n$ not only strict inequalities, and the Grassmann should parametrise locally free, rank $k$, quotients. Not $n-k$ quotients. Three reasons, listed below, I know you are fully aware of, so I would like to know why you did not do the natural thing.

1. It is technically the same as with $n-k$ quotients.
2. It is aligned with EGA.
3. And it will parametrise linear spaces (as in geometry, not linear algebra) of dimension k in affine n space.

Regarding Behrend, Conrad, Edidin, Fantechi, Fulton, Göttsche, and Kresch notes: Sebastian Casalaina-Martin held a stacks reading seminar (archived) whose webpage holds the book chapters (archived).

@#11559: Yes, this is nice. Thanks! It turns out that the argument (which is probably the same as your argument) was in Section 110.10. So I have edited that section to add a lemma with the statement and proof. If we ever need the lemma earlier in the Stacks Project, then we'll move it. See this commit.

OK, thanks for these comments! I agree that the original exposition (which will soon be updated) was a bit thin. I made some edits proving the equivalence of the two notions of units and then given some arguments proving enough so that MacLane's paper provides the coherence. I ran through the cases $(C, \phi)$, $(C, \phi, \mathbf{1})$, and $(C, \phi, \psi, \mathbf{1})$ in order which I think clarifies things. At the moment an exposition of MacLane's arguments is omitted. See this, this, and this.

The following lemma fits with this material in this section and I don't know any reference: if $M$ is $I$-adically complete, $I$ finitely generated, and $N$ a closed submodule of $M$, then $N$ is also $I$-adically complete. In a noetherian setting this would of course follow from Artin-Rees... (I can send a short proof if this is of interest.)

I think the construction in the proof gives a homotopy from $\mathrm{id}_{U\times \Delta[1]}$ to $e_1 \circ \pi$. And the homotopy from $e_0\circ \pi$ to $\mathrm{id}_{U\times \Delta[1]}$ should use minimum : $\beta(i)=\min \{ \beta_1(i),\beta_2(i)\}$.

You mention on this page that Cartan-Eilenberg resolutions need not use injectives: what kind of generalization can we make? I'm having trouble writing down a better statement because Lemma 013T seems to rely heavily on the injective assumption in two different ways. Do you have something specific in mind or a reference?