The Stacks project

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.

Ah, never mind. Since it asks about its irreducible component, there will be nothing to do if $X \cap Y$ is empty. I should notice the statement more carefully... Have a nice day!

Adding on to Comment #11001, a few remarks on the second proof given of Lemma 0F5S:

1. Is it worth inserting a sentence or so clarifying that/how the application of $F$ to the map $\bigoplus _ { j \in J } R \overset { \varphi } { \to } \bigoplus _ { i \in I } R$ (of which $M$ is presented as the cokernel) is well-defined in terms of the canonical coproduct inclusions of the domain and codomain by virtue of $F$'s commuting with coproducts? This is a separate question from the functoriality/presentation-independence of $F$, and as far as I can tell not quite trivial—any argument I can think of implicitly uses that $R$ is a compact object of $R\text{-mod}$, or at least its consequence that for any map $R \to \bigoplus _ { i \in I } R$ there exists a finite subset $I' \subseteq I$ such that the former factors through the evident inclusion $\bigoplus _ { i \in I' } R \to \bigoplus _ { i \in I } R$. (This subtlety is common to various instances of the Eilenberg-Watts argument.)

2. Relatedly, is it worth specifying how $F$ acts on morphisms of $R\text{-mod}$? Imho this is the actual meat of the argument (specifically in that the equivalence of two presentations of the morphism is witnessed by a "chain homotopy", which $F$ then carries into $\mathcal{A}$).

3. In the case as hand, we don't actually need to quantify/choose over generic presentations $\bigoplus _ { j \in J } R \overset { \varphi } { \to } \bigoplus _ { i \in I } R$ and then verify well-definition: Instead we can for each $M$ construct the specific presentation where (I) $I$ is the set of elements of $M$, (II) $J$ is the set of elements of the kernel of the evident map $\bigoplus _ { i \in I } R \to M$, (III) $\varphi$ is the evident insertion of kernel elements, and (IV) it's evident how each morphism should be presented. It's easy to see that this construction yields a functor $R\text{-mod} \to \text{Ar} (\mathcal{A})$ (here $\text{Ar}$ denotes the arrow category), which can in turn be composed with the cokernel functor $\text{Ar} (\mathcal{A}) \to \mathcal{A}$ to give a functor $R\text{-mod} \to \mathcal{A}$, no further verification of well-definition required.

If any of these ideas are worth implementing at this time I'm happy to try to help...

Hi, what if $X \cap Y$ is empty?

In (3), should it not have been that " $X$ is a closed subscheme of $X'$ " ?

Hi Johan, there is a typo in the subscript of the colim in part (4). It should be something like j geq i. (Cf. with the statement of part (4) in the lemma preceding.)

I believe we also need property (B) here to make the conclusion in the second to last sentence?

I guess in the formulation of part (1) there is a completion missing for $R'_{\mathfrak{p}'}$.

Typos: It should be $(R_n)$ instead of $(S_n)$ in a couple of places, and you switched the notation for $R$ into $A$ at one point.

In the second proof of Lemma 19.14.1, does the independence of the definition of $F$ of the choice of free presentation of $M$, as well as the functoriality of $F$, follow from any result in this website?

The proof of the "claim" inside the proof of the Lemma will be more transparent if it is mentioned that the open immersion $W \hookrightarrow G$ is a quasicompact morphism, as $W$ is quasicompact and $G$ is separated (Lemma 047L).

Typo: In the second to last paragraph it should be $\mathfrak{r}\subseteq\mathfrak{q}\subseteq R[x]$

Small question: doesn't Artins' Axioms typically do more and show that $$\mathcal{X}$$ is an algebraic stack locally of finite type?

The reference to Lemma 00DV (Nakayama's lemma) in the proof here strikes me as a red herring. I think(!) one can excise it from the proof at zero cost:

Lemma 00J8. Let $R$ be Artinian. The Jacobson radical of $R$ is a nilpotent ideal.

Proof. Let $I \subset R$ be the Jacobson radical. Note that $I \supset I^2 \supset I^3 \supset \ldots$ is a descending sequence. Thus $I^n = I^{n + 1}$ for some $n$. Set $J = \\{ x\in R \mid xI^n = 0\\}$. We have to show $J = R$. Suppose otherwise; by the Artinian property $J$ then has a minimal proper superideal $J'$. In particular, $J'/J$ is simple, so by Lemma 00J2 isomorphic to $R/\mathfrak{m}$ for some maximal $\mathfrak{m} \subset R$. We conclude that $$J \subset J' \subset \\{ x\in R \mid x\mathfrak{m}I^n = 0\\} \subset \\{ x\in R \mid xI^{n+1} = 0\\} = J$$ whence $J=J'$, contradiction. $\Box$

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P.s. (feel free to ignore what follows unless something is wrong!):

I was initially a little baffled what the motivation could be for considering $I^n$'s annihilator. Perhaps this is banal/well-known, but it strikes me that the annihilators of products of maximal ideals of $R$ are precisely what arise in succession as one attempts to iteratively "build-up" $R$ from $(0) \subset R$ via those elements currently minimal above the current ideal with some particular (maximal) denominator $\mathfrak{m}$ (just as one built $J'$ upon $J$ in the above proof).

I.e., having by Lemma 00J7 that $R$ has finitely many maximal ideals $\mathfrak{m}_ 0 , \dots , \mathfrak{m}_ {k-1}$, this view leads us naturally to the observation that $\text{Ann}(\prod_{i=0}^{k-1}\mathfrak{m}_ i^{d_ i}) = R$ (as the former cannot have a minimal proper superideal) where $d_ i$ is the minimal exponent for which $\mathfrak{m}_ i^{d_ i+1}=\mathfrak{m}_ i^{d_ i}$; this is essentially the statement/proof of Lemma 00J8 above.

Moreover the quotients of the obvious (size-$\sum_{i=0}^{k-1} d_ i$) filtration of $\text{Ann}(\prod_{i=0}^{k-1}\mathfrak{m}_ i^{d_ i})$ are each $R/\mathfrak{m}_ i$-modules for some suitable (variable) choice of $i$, and as each must have finite vector space dimension in order to maintain the Artinian property of $R$ we see that $R$ must have finite length over itself; this is essentially (half) the statement/proof of Lemma 00JB.