The Stacks project

Maybe we should add the lemma that any product of local rings is the product of the localizations at its maximal ideals.

Excellent, I saw other comments about the formula, which is excellent! As a beauty machine manufacturer, Litonlaser is very appreciative.

Some possible typos in the proof of lemma10.149.1:

line -4: $\beta'(g)=\beta(g)+\sum_{i}\delta_{i}\cdot \beta(\frac{\partial g}{\partial x_{i}})$

line -3: $d: J'/J^{2} \rightarrow \Omega_{P/R}\otimes_{P} S, g\mapsto \sum_{i} \pi(\frac{\partial g}{\partial x_{i}})\cdot dx_{i}$, where $\pi: P\rightarrow S$ is the natural quotient map.

We need to replace $P^{\bullet}$ by $P_{\bullet}$.

Lindo

For reference, this is Kashiwara, Schapira, Categories and Sheaves, Proposition 3.2.4.

I think "graduation" should be "gradation". I'm not sure that "graduation" is wrong, many people use it too, but since the page also uses "gradation" I think it's better to stick with that one

How to prove the injectivity of Lemma37.4.2? If we assume $H^{0}(X',G_{m})/n\rightarrow H^{0}(X,G_{m})/n$ is surjective, then it suffices to show $H^{1}(X',\mu_{n})\rightarrow H^{1}(X,\mu_{n})$ is injective by Kummer sequence for etale cohomology.

Since $H^{0}(X,\mathcal{I})[n]$, $H^{0}(X,\mathcal{I})/n$ and $H^{1}(X,\mathcal{I})[n]$ are zero, we have $H^{0}(X',G_{m})[n]=H^{0}(X,G_{m})[n]$ and similarly for etale base change. So we have $i_{*}\mu_{n}=\mu_{n}$ as etale sheaves, where $i:X\rightarrow X'$.

Finally we get the desired injection by the spectral sequence $H^{p}(X',R^{q}i_{\star}\mu_{n}) \Rightarrow H^{p+q}(X,\mu_{n})$

perhaps I am missing something, but Lemma 00JA doesn't seem to prove that R is a product of it's localizations at its maximal ideals, just a product of loval artinian rings?

perhaps I am missing something, but Lemma 00JA doesn't seem to prove that R is a product of it's localizations at its maximal ideals, just a product of loval artinian rings?

Isn't the consequence that all local rings of A are fields equivalent to the other assertions? (It seems to me that it implies (3)). In this case, it might be useful to state it as a fourth equivalent condition.

Apologies if this is wrong, but why is this not just the projection formula. Currently what the stacksproject calls the projection formula uses that $K$ is perfect which is a very strong assumption and not very useful in practice (when one specializes to schemes or algebraic stacks). It might be useful to mention that this formula exists in the projection formula section. Also unless I am mistaken, a similar statement holds for qcqs representable morphisms of algebraic stacks. This is Corollary 4.12 (combined with lemma 2.5) of Hall-Rydh 'Perfect complexes on algebraic stacks'.

Ctrl+F "topoological"

The second sentence of the proof is missing a bracket at the end.

Just a couple of comments on the choice of $M'$, since to me it is not obvious to me why it exists. We consider the family $\mathcal{A}$ of all finitely generated submodules $M'\le M$ such that $z\in M'$. We claim that it has a minimal element, so we may use Zorn's lemma. Take any chain $\{M_i:i\in I\}$ of elements of $\mathcal{A}$ and we need to show that it has a lower bound. We notice that $$z\in\bigcap_{i\in I}(M_i\otimes_kK)=\left(\bigcap_{i\in I}M_i\right)\otimes_kK$$ where the equality holds because $K$ is (obviously) a free $k$-module (easy to check equality). Now we can write $$z=\sum m_j\otimes t_j$$ with $m_j$ living in the intersection $\bigcap_{i\in I}M_i$, so we just take the $A$-submodule generated by the $m_j$.

I wonder what the application of the skyscraper sheaf is in physics. MacLane uses the notation Sky_x(A)(U). I am imagining x is a particle, U is a potential well, and A is an associated potential.

In 0GPY, and the final paragraph of the proof of Lemma 08S6 there are several instances of NL_{A'/A} which should be replaced with NL_{A/A'}.

In the statement of Lem. 09VR, should the morphism of sites be in the other direction, or am I just getting confused?

it should be "b= b \circle a \circle b"

it should be "b= b \cdot a \cdot b"