The Stacks project

in lemma 02UT，upper * should be upper -1

I suggest we add a remark or lemma about the completion of preadditive categories into additive categories by adding zero objects and finite biproducts. So people can understand the relation between them. See https://math.stackexchange.com/questions/4552586/examples-of-preadditive-categories-that-is-not-additive#4552592.

I suggest we add a remark or lemma about the completion of preadditive categories into additive categories by adding zero objects and finite biproducts. So people can understand the relation between them. See \ref{https://math.stackexchange.com/questions/4552586/examples-of-preadditive-categories-that-is-not-additive#4552592}.

The category of even dimension vector space over a field $K$ is an easier example that is additive but not abelian, since the kernel of the linear map $K^2\to K^2,e_1\mapsto e_1,e_2\mapsto 0$ is one dimensional. The word 'filtered vector spaces' may scare people.

There's a missing parenthesis at "$X_{s'} = (\varphi ')^{-1}(D_+(s')$"

There are $2\times 4\times 2=16$ equivalent definitions of abelian varieties. Let {projective, proper}, {geometrically irreducible, irreducible, geometrically connected, connected}, {smooth, geometrically reduced} be three sets of properties, pick one from each of them, and let $A/K$ be a group scheme with the chosen properties. Then it agrees with the definition of abelian variety. I suggest we add this somewhere just for the sake of new students who read too many books and get confused with the different definitions of abelian varieties. Actually I have the following proof in my thesis.

Proof: Clearly projectivity implies properness; and geometrically irreducibility implies the other three which in turn implies connectivity. By tag 056T we know that smoothness of K-schemes implies geometrically reducedness. Thus it suffices to show that "abelian varieties are projective, geometrically irreducible and smooth", and "proper connected and geometrically reduced group schemes over $K$ is an abelian variety".

By tag 0BF9, abelian varieties are projective and smooth. Geometrically irreducibility comes from the definition. Next we show that proper connected and geometrically reduced group schemes over $K$ are abelian varieties, i.e. they are proper and geometrically integral.

It suffices to show geometrically irreducibility. A group scheme over the field $K$ must contain a $K$-rational point, the unit section. In this case, $A$ is connected implies it is geometrically connected by tag 04KV. Hence after base change to any field extension $K′$ of $K$, $A_{K′}$ is connected. Plus, since every connected group scheme over a field is irreducible by tag 0B7Q, $A_{K′}$ is irreducible. So $A$ is geometrically irreducible. The result follows.

If $\underline{A}(U) = A$ for all non-empty $U$, then doesn't that mean $\underline{A}$ is flabby and hence acyclic? That makes proof much simpler. Or am I missing something?

I find Property (o) confusing. Is anything lost by saying, '' . . . there exists a largest open subscheme $S' \subset S$ such that $f|_{f^{-1}(S')}: f^{-1}(S') \to S'$ has $P$. Moreover, formation of $S'$ commutes with arbitrary base change.'' ?

In the second-to-last sentence, I think it should be $\prod I_n^\bullet[-n]$.

Unless I have misunderstood something, it also seems to me that the natural map $K_0(\mathcal{B}) \rightarrow K_0(D^b_{\mathcal{B}}(\mathcal{A}))$ given by $[B] \mapsto [B[0]]$ is a group isomorphism, with inverse as described in this lemma.

To see that the composite $K_0(D^b_{\mathcal{B}}(\mathcal{A})) \rightarrow K_0(\mathcal{B}) \rightarrow K_0(D^b_{\mathcal{B}}(\mathcal{A}))$ is the identity map, we can use the distinguished triangle of Remark 13.12.4 and repeatedly take canonical truncations for any bounded complex $X$ in $D^b_{\mathcal{B}}(\mathcal{A})$.

This would generalize Lemma 13.28.2.

Should have checked before posting #7784 above: the preservation of the Noetherian property (as well as being a DVR) is in [EGA I 2nd ed., ($\mathbf{0}_{I}$.6.8.3)]. The proof is a quick lemma ((6.8.3.1), loc.cit.).

Could add the following remark after this lemma 03C3: if A is Noetherian then can assume B to be both Noetherian and complete, and if A is DVR then B can be assumed to be a (complete) DVR [EGA I 2nd ed., ($\mathbf{0}_{I}$.6.8.3)]. The proof is a quick lemma ((6.8.3.1), loc.cit.). (This applies in the proof of Lemma 0328.)

Lemma 33.38.3. : Do you mean $x_i \in U_i$ instead of $x_i \in X$?

Same thing as Comment 7826, but also for Section 20.34 and Section 20.21.

I second Ryo Suzuki's comment, but here is a slightly more genral property that immediately implies (3) and (4): "If $\mathscr{T}$ is a nonempty set of connected subsets with nonempty intersection, then $E=\bigcup_{T\in\mathscr{T}}T$ is connected." Proof (like Ryo's proof, this does not use Zorn's lemma): Fix $x\in\bigcap_{T\in\mathscr{T}}T$. Then $x\in E$ since $\mathscr{T}\neq\emptyset$, hence $E\neq\emptyset$. Let $E_0\subset E$ be open, closed and nonempty. Then $E_0$ must meet some $T_0\in\mathscr{T}$, hence contain it since $E_0\cap T_0$ is nonempty, open and closed in $T_0$. Therefore $E_0$ contains $x$, hence meets (and contains) every $T$. QED

I also suggest to state explicitly that the first paragraph of the proof proves (1) and (2).

In this remark, maybe a reference to Proposition 30.19.1 is better than the currently referenced Lemma 30.16.3? (Proper vs locally projective.)

I think the title of this section is missing a word. Perhaps it should be "closed subset" ?

Is there a reason why only the first order infinitesimal neighborhood is defined? What about the $n^{\mathrm{th}}$ order infinitesimal neighborhood? Can this be defined and does it satisfy the same universal property or would something go wrong? I apologise in advance if this comment is not appropriate.

On the proof of (4), uniqueness is not proven. Also, (3) is not proven. I write proposed amendment:

Proof of (3). Let $E$ be connected. Consider the set $A$ of connected subsets containing $E$. Let $T = \bigcup A$. We claim $T$ is connected. Namely, supposed that $T = T_1 \coprod T_2$ is a disjoint union of two open and closed subsets of $T$. Because $E$ is connected and $E\subset T$, $E\subset T_1$ or $E\subset T_2$. Without loss of generality, we can suppose $E\subset T_1$. So $E'\cap T_1 \neq \varnothing$ for all $E'\in A$. By connectedness of $E'$, $E'\subset T_1$. Hence $T = T_1$. Immediately, $T$ is a unique connected components containing $E$.

At (62), I and ideal should be changed to I an ideal?