The Stacks project

Concerning the definition of the maps $\nabla:M\otimes \Omega^i \to M\otimes \Omega^{i+1}$:

Several $\otimes$ should be $\wedge$, i.e., $$\begin{align} \nabla(bm\otimes \omega) - \nabla(m\otimes b\omega) & = \nabla(bm)\wedge \omega + bm\otimes d\omega - \nabla(m)\wedge b\omega - m\otimes d(b\omega)\\ & = b\nabla(m)\wedge \omega + m\otimes db \wedge \omega + bm\otimes d\omega - b\nabla(m)\wedge \omega - bm\otimes d\omega - m\otimes db\wedge \omega = 0 \end{align}$$

Concerning the second paragraph of the proof. I suggest replacing "We will show this by induction ...." (till the end of the paragraph) by the shorter (and easier) direct calculation: The claim is clear for $n=1$. Assume $n>1$. Note that $\delta_i(z-y)$ lies in $(K\cap J(1))^{[2]}$ for $i>1$. Calculating modulo $K^2 + (K\cap J(1))^{[2]}$ we have $$\delta_n(z)=\delta_n(z-y+y)=\sum_{i=0}^n \delta_i(z-y)\delta_{n-i}(y)=\delta_{n-1}(y)\delta_1(z-y)+\delta_n(y).$$ The claim follows.

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Typo in the definition of the multiplication: $f \omega_2'+f'\omega_2'$ has too many $'$s

There is also an issue how the multiplication formula is displayed. It overlaps with the sidebar. Works fine in the pdf version though.

(And I gave a short proof of the fact that I don't know how to count...)

4th sentence of the proof: sch -> such.

It seems better to clarify the second property of graded tensor product: the $p,q$ in $x\in M^p,y\in N^q$ are different from those in the direct sum $\bigoplus_{p+q=n}$. In other words, here $p+q$ is only assumed to be no greater than $n$. It seems better to choose different names for them.

I came here to make the same comment as Wessel, but then I saw Wessel's comment, so I won't (maybe I just did...). So maybe count this as another vote for changing it??

For the last sentence of the proof, $V$ has been used to denote an open of $Y$, so it's not a subspace of $Z$ indeed, maybe replace it by $V\prime$?

Small point: after replacing S by f(X), you still use the assumption that S is affine. I think this can be fixed by instead base changing to an arbitrary affine open contained in f(X), after which the rest of the proof goes through.

I think it should be: $\text{Spec}(S)\to\text{Spec}(R)$ hits all the minimal primes.

0B3P follows immediately from 01R3 (1). Maybe we can use this as proof instead of "omitted"?

Kunz theorem is pretty good, will other properties about Frobenius action (and singularity) be added in the future? For example, Remark 13.6. in Ofer Gabber. Notes on some t-structures. In Geometric aspects of Dwork theory. Vol. I, II, pages 711–734. Walter de Gruyter GmbH & Co. KG, Berlin, 2004. shows that F-finiteness on a noetherian ring $A$ with $p=0$ will imply $A$ is quotient by a regular ring hence the existence of dualizing complex. The proof is very short and elegant.

The second last paragraph: "we obtain a specialization \eta' to t_j with \eta' \in X^0". Here \eta' is in T^0 (not X^0), right?

（2）Is it be f|_{U_i}: U_i \rightarrow Y ?

Suggested slogan: "Henselization of a ring inherits good properties of formal fibers"

In lemma 00RW, at some point the conclusion is that $\Omega _{S \otimes _ R S/S} = \Omega _{S/R} \otimes _ S (S \otimes _ R S)$ which is said to follow from lemma 00RV, but in lemma 00RV the tensor product is over the base change $R\to S$, and in the 00RW the tensor product is not over the base change but over the induced map $S\to S\otimes_R S$. Is that a problem or am I missing something?

I have a hard time understanding how Lemma 0959 is used here to prove that $j_!f_* f^{-1}\mathcal{G} = \overline{f}_* j'_!f^{-1}\mathcal{G}$. Instead, wouldn't it be better to combine Propositions 03QP and 03S5 to show that both sides agree on the stalks? Then one also doesn't need to argue that the diagram is cartesian. (Also, at least in the preview, the LaTeX doesn't render properly in this comment.)

In Lemma 81.4.5 in Section 81.4 the topology on the submodule $K$ is the topology inherited from $M$ (it is given by the submodules $K \cap I^nM$). But in the current Section 10.95 there is no mention whatsoever of topologies or completion with respect to any topology. We are just considering $I$-adic completion straight up. Maybe we should be a little bi more careful in the statement of Lemma 81.4.5. Thanks for the comment.

@#4200 Well, it may be that we can shorten the proof at the end there, but flat surjections of rings aren't necessarily isomorphisms.