The Stacks project

Typo: I believe the the second $\mathbb{P}^1_T$ in the morphism $g':\mathbb{P}^1_T \to \mathbb{P}^1_T$ in Theorem 095T should be $\mathbb{P}^1_S$.

In the proof of Lemma 02MJ it says "In this case the left hand side of the formula"; I think this should be replaced by the "right hand side of the formula".

By using direction (2) to (1) of Lemma 28.11.3, one can give a more schematic proof as follows. By hypothesis we have locally $\mathcal{L}\cong\mathcal{O}_X$, therefore the open immersion $\iota: X_s \hookrightarrow X$ is locally given by a principal open set (in an affine neighborhood). By the lemma cited, $\iota$ is affine, therfore $U \cap X_ s = \iota^{-1}(U)$ is affine whenever $U$ is affine.

By the claimed exactness of the diagram we have $0 = \mathrm{Ker}(M \otimes K \rightarrow M^{(I)})$, and

• $M'' \otimes N = \mathrm{Coker}(M''\otimes K \rightarrow (M'')^{(I)})$
• $M' \otimes N = \mathrm{Coker}(M'\otimes K \rightarrow (M')^{(I)})$
• $M \otimes N = \mathrm{Coker}(M \otimes K \rightarrow ((M)^{(I)})$

Then the snake connects $0 = \mathrm{Ker}(M \otimes K \rightarrow M^{(I)}) \rightarrow \mathrm{Coker}(M''\otimes K \rightarrow (M'')^{(I)}) \rightarrow \mathrm{Coker}(M'\otimes K \rightarrow (M')^{(I)}) \rightarrow \mathrm{Coker}(M \otimes K \rightarrow ((M)^{(I)})\rightarrow 0$.

Hi, I think it's better to say "diagram chasing" here than "snake lemma". Seriously though, can somebody explain where the snake is?

Lemma 00JD (10.54.1) needs $S$ to be Artinian. I don't see why this should be the case, as the ideals

$$J_k = \bigoplus_{d \ge k} I^d / I^{d+1}$$

seem to give a possibly infinite descending sequence. $S = J_0 \supset J_1 \supset J_2 \supset \cdots$. This sequence stabilises iff $I^d = I^{d+1}$ for $d \gg 0$, which must not be true. For example consider the localisation $k[x]_{(x)}$ of the polynomial ring at the maximal ideal $(x)$ and $I = (x)$.

Dear Ravi, fair question. It is the 5th condition in Lemma 15.70.2. I have edited the proof to clarify. See here.

OK, yes thanks. Fixed here.

Thanks very much! Fixed here.

Thanks and fixed here.

Yes, this is a mistake. Thanks very much for pointing this out. I have fixed this here.

OK, I have checked this proof and it seems OK to me.

But I think there should be another proof of this lemma as well. Namely, we should be able to show directly that given a $D$-module $M$ an $A$-derivation $\vartheta : B \to M$ is the same thing as a divided power $A$-derivation $\theta : D \to M$ using the universal property of $D$. To do this consider the square zero thickening $D \oplus M$ of $D$. There is a divided power structure on $\overline{J} \oplus M$ if we set the higher divided power operations zero on $M$. Consider the $A$-algebra map $B \to D \oplus M$ whose first component is the given map $B \to M$ and second component is $\vartheta$. By the universal property we get a corresponding map $D \to D \oplus M$ whose second component should be the map $\theta$ corresponding to $\vartheta$.

I didn't check this completely, but this should work and we should replace the proof given in the Stacks project by this argument.

Thanks for this. See change here.

Thanks very much. Added your version here.

Thanks for the fix. See changes here.

Lemma 45.2.3 is stated as used later, so I think it is fine as is.

This is good. Thanks. Change is here.

OK, I have made this change in this case. However, when writing these sections I tried to have some consistency in the numbering of the conditions listed in the lemmas. So I want to be careful in changing this in other lemmas, etc. So unless somebody is going to review the chapter as a whole and look more carefully to see if things can be simplified and/or strengthened (as may very well be the case), I would prefer to leave things as is for now.

Thanks and fixed here.

Yes, of course. It is impossible to keep lists up to date; the only solution seems to be to not have lists of things. For this kind of thing, I strongly encourse people to just edit the latex file and email me. In this case I minimally added this here. Thanks.

OK, I added the conclusion from Nakyama's lemma. But I kept the other statement as well because it is how I think about it. See change here. Thanks very much.