The Stacks project

I dont get the second to last paragraph, how do you know that adding an element to L won't increase it?

For the implication (1)$\Rightarrow$(2), shouldn't one perhaps mention 10.37.5?

I think the reference to the map 27.10.1.1 that gives rise to the global map $$S\to\bigoplus_{n\geq 0}\Gamma(X,\mathcal O_X(n))$$ is wrong, and should instead be 27.9.0.2.

Sorry for the consecutive posts, there are other two possible typos.

In the fourth paragraph, "$\mathit{QCoh}(\mathcal{O}_ S) \rightarrow \mathit{QCoh}(S_\tau , \mathcal{O})$" should be "$\mathit{QCoh}(\mathcal{O}_ S) \rightarrow \textit{Mod}(S_\tau , \mathcal{O})$".

In the last paragraph, "$\mathcal{F}_1(U) \otimes _ A B \to \mathcal{E}(U) \otimes _ A B \to \mathcal{E}(U)$" should be "$\mathcal{F}_1(U) \otimes _ A B \to \mathcal{E}(U) \otimes _ A B \to \mathcal{E}(V)$"

There is a typo: the last sentence of the second paragraph should be "Thus $(3)$ and its special case $(2)$ hold." I am also a little confused about the statement of $(2)$ and $(3)$. What does "compatible with colimits" mean? Does it mean that $(-)^a$ commutes with colimits?

How does lemma 00J0 imply $R_{\mathfrak{p}_i}$ has finite length over R? $R_{\mathfrak{p}_i}$ may not be a finite $R$-module.

Suggested alternative proof: since $Y_i\to Y$ is quasi-compact quasi-separated, so is the composite $f_i:Y_i\to Y\to X$. Hence, we can take the normalization $X_i'\to X$ of $X$ in $Y_i$. To check that $X'_1\amalg X'_2\to X$ is the normalization of $X$ in $Y$, it suffices to see that the factorization $Y_1\amalg Y_2\to X'_1\amalg X'_2\to X$ of $f$ satisfies parts (1) and (2) of Lemma 29.53.4. Part (2) can be verified using the universal property of the factorization $Y_i\to X'_i\to X$ of $f_i$. We see part (1), i.e., that $\nu:X'_1\amalg X'_2\to X$ is integral: On the one hand, $\nu$ is affine, by Schemes, Lemma 26.6.8. On the other hand, let $U=\operatorname{Spec} A\subset X$ be open affine. Denote $U_i'=\operatorname{Spec} A_i'$ to the inverse image in $X_i'$. The map $A\to A_i'$ is integral. The induced map on global sections by $\nu^{-1}(U)=U_1'\amalg U_2'\to U$ is $A\to A_1'\times A_2'=A'$, which is integral (it suffices to show that the ideals $I_1=A_1'\times\{0\}$ and $I_2=\{0\}\times A_2'$ of $A'$ are integral over $A$, for $I_1+I_2=A'$).

First paragraph proof of Prop. 10.9.10: A general element of $\overline{S}^{-1}(S'^{-1}A)$ is of the form $(x/t')/(s/1)$, right? Not $(x/t')/(s/s')$ for arbitrary $s'$.

Approximately in the middle of the proof, the following is stated: ''Let $U \subset X$ be one of the open strata in the partition." Why is one of the $X_i$ necessarily open? (We did not even assume that $X$ is noetherian.) Maybe the argument should be: Let $X_k \overset{i}{\hookrightarrow} U_k \overset{j}{\hookrightarrow} X$ be a composition, where $i$ is a closed immersion and $j$ an open immersion. It suffices to show the desired isomorphism on the $U_k$'s. Hence, we can assume $X_k$ is closed. Then define $Z:= X_k$ and continue with the proof as described above (using for $U$ the complement of $Z$ instead of ''one of the open strata").

Typo in Lemma 0GML: Missing closing parenthesis at the end of the statement.

why does the zariski freeness of the module follow from Lemma 10.127.6? I feel like a more delicate argument is needed here (albeit a standard one)

:I have searched for the definition of "injective hull". It does not appear until 08Y2 (currently 47.5.1), and then only as an ad hoc version for modules.

In case it is worth of being included, here's the proof of the uniqueness of $h$: Suppose $\tilde{h}:X'\to Z$ makes also the diagram commute. In particular, $\tilde{h}$ is a morphism over $X$, so it is uniquely determined by $\pi_*(\mathcal{O}_X\to\tilde{h}_*\mathcal{O}_{X'})=\pi_*(\tilde{h}^\sharp):\mathcal{B}\to\mathcal{O}'$ (see Constructions, Section 27.4). Since $h\circ f'=g$, the composite $\mathcal{O}_Z\xrightarrow{\tilde{h}^\sharp}\tilde{h}_*\mathcal{O}_{X'}\to g_*\mathcal{O}_X$ equals $g^\sharp$. Mapping this composite through $\pi_*$ gives $\mathcal{B}\xrightarrow{\pi_*(\tilde{h}^\sharp)}\mathcal{O}'\to f_*\mathcal{O}_Y$, i.e., $\pi_*(g^\sharp)=\chi$. Since $\mathcal{O}'\to f_*\mathcal{O}_Y$ is injective, it follows that $\pi_*(\tilde{h}^\sharp)=\pi_*(h^\sharp)$, whence $\tilde{h}=h$.

Another issue is that, should the zero module be called Cohen-Macaulay?

Strictly speaking, part (2) should be rephrased as "for at most one $i$" to include the trivial case $M=0$.

I think one could open with the following sentence (or something similar): "the result follows from Lemmas 27.4.6 and 29.11.3, but we can give a direct proof."

Small typo: In Lemma 03T5, I believe "Similar statements hold for $D^−$ and $D^+$" should say "Similar statements hold for $D^−$ and $D^b$."

Isn't this Artin's theorem on representability of flat groupoids? Perhaps a slogan to that end will be helpful here.

Just to point out the redundancy: part (2) is Schemes, 26.21.14.

In the proof of Lemma 00VE, the notation $\mathcal I_Y$ and $(X,\varphi)$ is used, as opposed to $\mathcal I_V$ and $(U,\varphi)$ used in the rest of the section. I think this should be changed for consistency. Unless $Y$ is the final object of $\mathcal D$ but it is not I don't think?

Also small typo at the beginning: it says "sometimes we omit the subscript $u$" but it should be "superscript".