The Stacks project

This lemma can be strengthened to say that $R \to S$ is an epimorphism if and only if $\mathrm{Mod}_S \to \mathrm{Mod}_R$ is fully faithful. The point is that the full faithfulness implies that the two different $S$-module structures on $S \otimes_R S$ coincide, and now tag 04VN implies that $R \to S$ is epic.

Sorry, I don't see why this definition does not depend on the choice of the K-flat resolution. For two K-flat resolutions $K_1^{\bullet}$ and $K_2^{\bullet}$ of $F^{\bullet}$, is there always a quasi-isomorphism $K_1^{\bullet}\rightarrow K_2^{\bullet}$ (in order to use the Lemma 06YG above)?

The following result follows easily from the results that have been already included, but I think it's useful to have for reference.

Lemma. Let $X$ be a scheme. The following are equivalent: 1. $X$ is separated. 2. Every morphism of schemes $X\to Y$ is separated. 3. There is a separated morphism of schemes $X\to Y$ with $Y$ affine. 4. There is a separated morphism of schemes $X\to Y$ with $Y$ separated.

Moreover, one also obtains equivalent conditions if one replaces every instance of “separated” by “quasi-separated.”

Proof. 1$\Rightarrow$2. Apply Lemma 26.21.13 to the composition $X\to Y\to\operatorname{Spec}\mathbb{Z}$.

2$\Rightarrow$3. Trivial.

3$\Rightarrow$4. It follows from Lemma 26.21.15.

4$\Rightarrow$1. Apply Lemma 26.21.12 to the composition $X\to Y\to\operatorname{Spec}\mathbb{Z}$. $\square$

Curiously, the section does not discuss when the limit of an algebraic stack is also an algebraic stack / limits of morphisms of algebraic stacks and which properties behave well with respect to limits. It would be nice to have such a section in this chapter?

Since $\Lambda$ is not assumed to be Notherian, should we revise the argument as follows: $H_{A/R}$ is finitely generated and $\mathfrak q \Lambda_{\mathfrak q}$ is locally nilpotent?

Also, in the calculation of $B_z$, should it be $t_{ij}$ in place of $y_1,\ldots,y_r$?

I think in order to check the commutativity of the last diagram, it would be better to draw a dotted line from B to C_lambda, as a bridge to show the commutativity

Corrections: In #10968, Point 2 of Lemma might not be true in general, and #10969 might not be true in general either. Both are nonetheless true if $R,M$ are Noetherian; about the non-Noetherian case I don't know, but I asked about it here.

In the paragraph where you assume $i$ is $H_1$-regular or Koszul-regular, you put "quasi-regular" instead of Koszul-regular in a couple of places, right?

• what about fpqc morphism? Is that universally open?

Not saying this should be included in the Stacks Project, but just for the reader's information:

More generally, the completion functor preserves exact sequences of strict morphisms of first-countable topological abelian groups (a morphism $f$ of topological abelian groups is said to be strict if $\operatorname{Coim}f\to\operatorname{Im}f$ is an isomorphism), see [BouCA, Ch. III, §2, no. 12, Lemma 2]. Note that strict morphisms are not a subcategory (i.e., strictness is not stable under composition) [BouTop, Ch. III, §2, no. 8, Remark 2].

References

[BouCA]. N. Bourbaki, Commutative Algebra

[BouTop]. N. Bourbaki, General Topology

I don’t see why this last implication holds. In general, a uniform bound on the degrees of finite subextensions does not imply that the algebraic closure of $k$ in $K$ is finite over $k$; see e.g.\ \url{https://math.stackexchange.com/questions/2835406/in-a-field-extension-if-each-elements-degree-is-bounded-uniformly-by-n-is-t}. In the finitely generated situation of Lemma~037J, what is the additional argument that shows the algebraic closure of $k$ in $K$ is finite?

At the end of the proof of Lemma 03JA, it seems to me that by using a different surjective $A$-module map, i.e.: $$A^{\oplus n}\longrightarrow B,\ \ (a_1,\ldots,a_n)\mapsto \sum_{i=1}^n a_ix_i$$ and then continuing the proof the same way as above, one can see that the integer $n$ (number of generators of the $A$-module $B$) also works in the definition of a morphism with universally bounded fibres. (Assuming that I didn't overlook anything) is there a reason for preferring the integer $d_1\ldots d_n$ to $n$?

In the definition there is a typo, I believe it should say $Z \subset Y$ rather than $Z \subset X$.

Sorry, I just can't see how to use lemma 061R. It seems to me that you need to know that $Y\cap U$ is locally around $y$ cut by $f_1,...,f_n$ inside $U$, which doesn't have to hold if $Y->X$ is not assumed to be locally of finite presentation.

Reference: EGA, Chapter 0, Sect. 7.7.

From the second paragraph, regarding “we say the topology on $M$ is linear if there exists a fundamental system of neighbourhoods of $0$ consisting of subgroups. If so then these subgroups are also open.” They are actually clopen. Quoting Atiyah, Macdonald, Introduction to Commutative Algebra, pp. 102-103:

[Let $G$ be a topological abelian group that has a fundamental system of open neighborhoods of $0$ given by subgroups $G_i$]. The subgroups $G_i$ of $G$ are both open and closed. In fact if $g\in G_i$ then $g + G_i$ is a neighborhood of $g$; since $g + G_i \subset G_i$ this shows $G_i$ is open. Hence for any $h$ the coset $h + G_i$ is open and therefore $\bigcup_{h\not\in G_i}(h + G_i)$ is open; since this is the complement of $G_i$ in $G$ it follows that $G_i$ is closed.

Regarding "we leave it to the reader to check that this condition is independent of the choice of fundamental system of open subgroups $\{U_i\}_{i\in I}$ chosen above" and "in fact the topological abelian group $M^\wedge=\lim_{i\in I}M/U_i$ is independent of this choice."

One can show $\lim_{i\in I} M/U_i\cong\lim_{j\in J} M/V_j$ for another fundamental system of open subgroups $\{V_j\}_{j\in J}$ in the following way: replacing $\{V_j\}_{j\in J}$ by $\{U_i\}_{i\in I}\cup\{V_j\}_{j\in J}$, we may assume $\{U_i\}_{i\in I}\subset\{V_j\}_{j\in J}$. Then $\{U_i\}_{i\in I}\to\{V_j\}_{j\in J}$ is an initial functor of cofiltered posetal categories (by the dual of 4.19.3), so the result follows from 4.17.4.

The image of the sequence $f_1,...,f_r$ in the ring $\mathca{O}_X$ might not be regular since these elements could generated the unit ideal, but it is "weakly regular" (just forget this condition) and these are Koszul regular.

I think the sequence of (9-d) should be $x = x_0 \rightsquigarrow x_1 \rightsquigarrow \cdots \rightsquigarrow x_n = x'$.

I suggest making the first paragraph more transparent by the following adjustment.

Suppose that for every $\xi \in \pi_0(W)$ we have found affine open $U_\xi \subseteq W$ such that $f(F_\xi) \subseteq U_\xi$. As $\psi$ is a continuous map of profinite topological spaces, $\psi(T \setminus f^{-1}(U_\xi))$ is a closed subset of $\pi_0(W)$ that does not contain $\xi$. Hence its complement $V_\xi$ is an open neighborhood of $\xi$. Thus the $V_\xi$'s form an open covering of $\pi_0(W)$. Moreover, by construction $\psi^{-1}(V_\xi) \subseteq f^{-1}(U_\xi)$. As $\pi_0(W)$ is profinite, this open covering admits a refinement by a finite covering $\sqcup_{i=1}^n V_i$ of disjoint sets $V_i$ that are simultaneously closed and open. For every $i=1,\dots,n$ pick $\xi_i = \xi$ such that $V_i \subseteq V_{\xi_i}$. As $V_i$ is closed and $U_{\xi_i}$ affine, $U_{\xi_i} \cap \pi^{-1}(V_i)$ is affine as well. Hence replacing $U_{\xi_i}$ by $U_{\xi_i} \cap \pi^{-1}(V_i)$ we have $\psi^{-1}(V_i) = f^{-1}(U_{\xi_i})$. As the $V_i$'s cover $\pi_0(W)$, the $f^{-1}(U_{\xi_i})$'s cover $T$ and so $f(T) \subseteq \sqcup_{i=1}^n U_{\xi_i}=:U$. Being the disjoint union of finitely many affine opens, $U$ is an affine open.