The Stacks project

In the proof: Is the fact that $B \subset B'$ is a finite free extension used anywhere?

The freeness of the map $R \rightarrow S$ of example 00SR, is explicitly used in the proof of Lemma 03GD. Thus I suggest replacing "locally free" with "free" in the statement of the current Lemma.

End of proof: the claim is "obvious" to the same extent as the axiom of choice; in fact it is exactly the "dependent choice axiom".

The hypothesis that $M$ be a Mittag-Leffler $S$-module is redundant: by 10.88.8, this follows from from $M$ being a flat $S$-module.

I hope it is a suitable place to ask questions. In the definition of proper morphism, you require finite type but not finite presentation. Does it mean that there are proper morphisms which are not of finite presentation?

I hope it is a suitable place to ask questions. In the definition of proper morphism, you require finite type but not finite presentation. Does it mean that there are proper morphisms which are not of finite presentation?

The first displayed equation in the proof, $X_K$ should be $X_k$?

(Original code here.) For reference, this is the actual result:

Lemma. Let $(F,\xi),(G,\zeta):\mathcal{D}\rightrightarrows\mathcal{D}'$ be triangulated functors between pre-triangulated categories. For each object $A\in\mathcal{D}$, suppose given a morphism $\eta_A:F(A)\to G(A)$ and denote $\eta=\{\eta_A\}_{A\in\mathcal{D}}$. The following are equivalent:

1. $\eta:F\Rightarrow G$ is a natural transformation and we have a commutative triangle of natural transformations

$$\require{AMScd} \begin{CD} F\circ [1]@>{\xi}>>[1]\circ F\\ @V\eta_T VV@VVT'\eta V\\ G\circ [1]@>>\zeta> [1]\circ G \end{CD}$$

1. For every distinguished triangle $A\to B\to C\xrightarrow{h} A[1]$ in $\mathcal{D}$, the diagram $$\require{AMScd} \begin{CD} F(A)@>>> F(B) @>>> F(C)@>>> F(A)[1]\\ @VVV@VVV@VVV@VVV\\ G(A)@>>> G(B) @>>> G(C)@>>> G(A)[1] \end{CD}$$ commutes, where the vertical maps are given by $\eta_A,\eta_B,\eta_C$, and the last top and bottom arrows are $\xi_A\circ F(h)$ and $\zeta_A\circ G(h)$, respectively.

Succintly, $\eta$ is a transnatural transformation (a 2-morphism in the category of categories with translation) if and only if it is a trinatural transformation (a 2-morphism in the category of pre-triangulated categories). I wrote the proof in this question (see the Lemma at the end).

I am wondering where the hypothesis $\mathcal{F}$ is an étale sheaf is used. It seems to me that this Lemma holds true even for presheaves on $\mathcal{X}$. ((1)(3)(4) are true for presheaves since the comparison morphisms are constructed for presheaves and (2) can be verified by a direct computation.) Maybe I missed some point?

I think in the proof of lemma 27.10.4 in second paragraph third last line it should be $g\in (S_h)_{d\text{deg(g)}}$ instead of $(S_h)_{-d\text{deg(g)}}$

Sorry the formatting of my previous comment was messed up. $A$ is supposed to be the union of the $Y_i$ that don't contain $x$.

Lemma 5.9.6 (04MF): At the end of the proof, one can exhibit the open connected neighborhood of $x$ directly without having to mention $E''$. Let $A=\bigcup\\{Y_i:x\notin Y_i\\}$, which is closed in $E\cap E'$. So $V=(E\cap E')\setminus A$ is an open neighborhood of $x$. Claim: $V$ is connected. Reason: $V$ is the union of the $Y_j\setminus A$ for all the $Y_j$ containing $x$. Every open subset of an irreducible set is irreducible. So for $Y_j$ containing $x$, the set $Y_j\setminus A$ is also irreducible, hence connected. And so is their union since they all contain the common point $x$.

(math free comment) Since they are consecutives, it would be nice to put lemmas 10.50.4 and 10.50.5 together as two equivalent conditions. Also, with 10.50.15 and 10.50.16 one could produce a list of equivalent definitions like many others on the site.

In (2), it is better to replace (05Y9) by a link to this tag.

There is a minor typo in the proof, since $L$ is bounded below and $K$ is bounded above...

This implication is in fact an equivalence, and it may be worth stating explicitly, especially since these conditions are often given as the definition of smoothness.

What is the definition of "locally equidimensional" in the proof?

I am confused about the proof for Lemma 10.57.7. Which states the following:\

Let $S$ be a graded ring and let $p\subset S$ be a prime ideal. Let $q$ be the homogeneous ideal generated by homogeneous elements of $p$. Then $q$ is a prime. \

To check if a homogeneous ideal $q$ is prime, it is enough to check if $fg\in q$ then $f\in q$ or $g\in q$ for homogeneous elements $f,g$. Let $f,g$ be homogeneous and $fg\in q$. By definition of $q$, it means $fg=\Sigma a_if_i$ for $f_i\in p$ homogeneous. But that imply $fg\in p$ and thus $f\in p$ or $g\in p$. Then $f,g$ are homogeneous elements in $p$ and hence generator of $q$. Thus $f\in q$ or $g\in q$. Then we are done, right? Not sure we need to assume $f,g$ non-homogeneous.

In the statement of the lemma, you want to call the modules M^a,...,M^b rather than M_a,...,M_b

@9797 I just realized the converse is also true. That is, if the diagram $$\require{AMScd} \begin{CD} F(A)@>>> F(B) @>>> F(C)@>>> F(A)[1]\\ @VVV@VVV@VVV@VVV\\ G(A)@>>> G(B) @>>> G(C)@>>> G(A)[1] \end{CD}$$ commutes for all distinguished triangles $A\to B\to C\to A[1]$ of $\mathcal{D}$, then $\eta:F\Rightarrow G$ is trinatural. Indeed, just apply the hypothesis to the distinguished triangle $A\to 0\to A[1]\xrightarrow{\operatorname{id}}A[1]$.