The Stacks project

I found myself being pretty confused after reading the proof, mainly because it seems that in the (2) => (1) part we don't really make use of condition (a) on the ranks, so I had to spend time thinking where the proof falls apart without it.

The answer to my confusion (I think) is that we implicitly use it to prove the dimension 0 case and get the induction rolling. I think it would be beneficial to have a sentence on why the dimension 0 case follows trivially from (a).

Lemma 0BNK works when you replace f by a finitely generated ideal I and take M to be an I-power torsion R-module, right? Can it be stated in this generality?

The consequence $M \cong M \otimes_R \widehat{R}^I$ also continues to hold because of Tag 05GG.

For the final assertion, would we not also want $Tor_{2}^{B}(\Omega_{B/A},C)=0$? Let P be a presentation of B over A, with kernel I. The condition $Tor_{1}^{B}(\Omega_{B/A},C)=0$ ensures we have an injection $\text{Im}\frac{I}{I^{2}}\bigotimes_{B}C\rightarrow\Omega_{P/A}\bigotimes_{A}C$, so that to compute $H^{1}(NL_{B/A}\bigotimes_{B}C)=H^{1}(L_{B/A})\bigotimes_{B}C$ It's enough to show the sequence $0\rightarrow H^{1}(L_{B/A})\bigotimes_{B}C\rightarrow\frac{I}{I^{2}}\bigotimes_{B}C\rightarrow\text{Im}\frac{I}{I^{2}}\bigotimes_{B}C\rightarrow0$ is exact. The obstruction here is $Tor_{1}^{B}(\text{Im}\frac{I}{I^{2}},C)$. The tor sequence associateded to the exact sequence $0\rightarrow\text{Im}\frac{I}{I^{2}}\rightarrow\Omega_{P/A}\bigotimes_{A}B\rightarrow\Omega_{B/A}\rightarrow0$ shows that we have a surjection $Tor_{2}^{B}(\Omega_{B/A},C)\rightarrow Tor_{1}^{B}(\text{Im}\frac{I}{I^{2}},C)\rightarrow0$ and hence $Tor_{2}^{B}(\Omega_{B/A},C)=0$ is the necessary condition to get what we need.

Typo in the TeX code: after equation 17.23.1.1, there is a missing '\ref' in 'equation-inclusion-of-annihilator-ideals'.

I just wanted to remark that by Noether normalization, it is enough to see that the affine line over k has infinitely many closed points.

Thanks for pointing out this insufficient argument. Of course this is easily fixed by saying that an infinite Jacobson space has infinitely many closed points; I will do this when I next go through all the comments again. Does anybody have another suggestion for fixing this?

Why does the ring S have infinitely many maximal ideals? Lemma 0ALW only implies that S has infinitely many prime ideals, but not maximal ideals.

I think the first paragraph has some typos. In the third sentence, instead of a_i it should be a_j. The end of the paragraph should also display a relation on s, not on st.

Maybe in (7) we should have $f^!(-) = Lf^\ast \otimes_{\mathcal{O}_X} f^\ast \mathcal{O}_Y(X)[-1]$? (There is a restriction to $Y$ missing, and I think we need the normal sheaf, cf. 48.14.2 and the beginning of 48.14.)

There is a "compactififcatins" in the paragraph that starts with "Consider two compactifications $j_ i : X \to \overline{X}_ i$, $i=1,2$ of $X$ over $Y$."

Could more details be added to the proof? I've spent some time on the second paragraph and I cannot understand most of the justifications there. Particularly, I don't see why the two asserted tensor products are equal, why the map between vector spaces is surjective, and why the expressions you find for the x_i imply the ideal (x_1,...x_n) is idempotent.

I think the three cases on the lemma should be for objects $K \in D_{\mathit{QCoh}}(\mathcal{O}_Y)$ instead of $\mathcal{O}_X$.

In the statement of the lemma, I think we should write $Rf_\ast\colon D_\mathit{QCoh}(\mathcal{O}_X) \to D_\mathit{QCoh}(\mathcal{O}_Y)$ to be consistent with the rest of the section (and with the conventions in 36.2).

LHS of the formula below Lemma 12.31.7 should be lim Q ⊗_{A} M_n

Can this statement be changed to 'essentially' of finite type ring map $R \to R'$ instead of finite type? Obviously, any localization of a regular ring map remains regular, but this would make for an easy citation (old Matsumura book also states this for finite type instead of essentially of finite type base change.

Why not just say $B \to C$ is faithfully flat?

Other than overkilling with Chevalley's theorem, here is an more elementary argument. $K$ is of finite type over $\operatorname{Frac}(R)$, thus finite and thus integral. Choose generators of $K$ over $R$ and consider their minimal polynomials with coefficients in $\operatorname{Frac}(R)$. By inverting the product $f$ of the (finite number of) denominators, we conclude the generators of $K$ are integral elements over $R_f$. Thus $K$ is integral over $R_f$, and $R_f$ is a field, as desired.

In Lemma 2 I think you missed the indexes of the sheafs. So it should read "$\mathcal{O}_Y (V) \rightarrow \mathcal{O}_X (U)$ is regular", instead of "$\mathcal{O} (V) \rightarrow \mathcal{O} (U)$ is regular". Atleast I have not seen this particular abuse of notation anywhere.

In Lemma 2 I think you missed the indexes of the sheafs. So it should read "$\O_Y (V) \rightarrow \O_X (U)$ is regular", instead of "$\O (V) \rightarrow \O (U)$ is regular". Atleast I have not seen this particular abuse of notation anywhere.

In STEP 2 what is $\pi_X*$? May be $\pi_X^*$?