The Stacks project

I think we can also include a lemma about the behavior of the relative normalization under the composition of quasi-compact quasi-separated morphisms.

I suggest to include https://math.stanford.edu/~conrad/papers/kmpaper.pdf 2.2.5 as well. This result is closely related to the topic of this section.

In the proof of Lemma 02s0: "The arguments above therefore reduce us to the case of a since integral closed subscheme" The word "since" should be replaced by "single".

Typo: in (4)(c), we want $K$ in $D_{Coh}^{-}(\mathcal{O}_X)$.

Proof of Lemma 08XV: replace the union symbols by direct sums (three times).

Perhaps a followup lemma would be worth including: an unramified morphism of algebraic stacks is locally quasi-finite.

This follows from Lemma 101.36.6 (this lemma), Lemma 101.23.7 (about local quasi-finiteness), and Lemma 67.38.7 (the corresponding result for algebraic spaces).

Everyone: please go to the page of the lemma you are commenting on and mention the exact sentence you are asking about. The reason for it being zero is, with notation as in the proof of Lemma 75.6.3, that a local section of $f_*\mathcal{O}_X$ corresponds to a local section of $\mathcal{O}_{S'}$ which of course maps to zero in $\Omega_{X/S'} = \Omega_{X/S}$.

@#7767. Look at what happens when $i = j = k$ for the cocycle condition.

Shouldn't one , in the definition of glueing data, require $\varphi_{ii} = id_{\mathcal{F}_i}$? or is it somehow already implicit in the definition?

Shouldn't one , in the definition of glueing data, require $\varphi_{ii} = \textup{id}_{\mathcal{F}_i}$? or is it somehow already implicit in the definition?

There is a typo after the very first diagram: "we can then consider take".

Throughout Lemma 61.6.7, "section" (applied to ring maps) actually means "retraction".

On lemma 0G8F, I would love to ask why do the map: $f_* O_X \to f_* \Omega_{X/S'}$ is zero ? It seems not easy.

In the definition of subcategory (001D) it is normally also required that for each object in the subcategory the identity morphism in the subcategory matches the one in the containing category. (see wikipedia or many category books)

Here is a different proof where we only need $x$ to be a "decent point" (rather than $X$ being decent): We may assume $X$ quasi-compact. By 68.8.3, $x$ is a point of $T\subset U\subset X$ where $T\subset U$ (resp. $U\subset X$) is a closed (resp. open) immersion, and $T$ is a scheme. By 66.4.9, $f$ factors through $U$, so we may assume $U=X$. Then $f$ factors through $T$ because $Y$ is reduced (consider the ideal sheaf of $T$). So we may assume that $X=T$ is a scheme.

Apologies: when I wrote my previous comment, I had missed this section. The example in my comment is of course a "counterexample" to this lemma in the non-decent case.

Here is another example in the "weird" category; it should probably come later, since it is mostly about "bad points". Let $k$ be a field. Let $G$ be an infinite profinite group. Let $X$ be $G$ viewed as a zero-dimensional affine $k$-group scheme, i.e. $X=\mathrm{Spec} \lbrace\text{locally constant maps } G\to k \rbrace$. Let $\Gamma$ be $G$ viewed as a discrete $k$-group scheme, acting on $X$ by translations. Put $\overline{X}:=X/\Gamma$. This is a one-point algebraic space, with projection $q:X\to\overline{X}$.

Let $e\in G$ be the origin (any element would do), and view it as a $k$-point of $X$. We get a $k$-point $\overline{e}:\mathrm{Spec}(k)\to\overline{X}$ which is a monomorphism since it is a section of $\overline{X}\to\mathrm{Spec}(k)$.

I claim that (although $X$ is affine and reduced and $\vert \overline{X}\vert$ is a point), the morphism $q$ does not factor through any morphism $\mathrm{Spec}(K)\to \overline{X}$, where $K$ is a field. Otherwise it would factor through $\overline{e}$ since this is a monomorphism. Now the pullback of $q$ by $\overline{e}$ is $\Gamma\to\mathrm{Spec}(k)$, with the projection $\Gamma \to X$ being the orbit map $g\mapsto g.e$. The latter has no section, whence the claim.

In connection with Comment #7747, what is really proved here is this: if $\varphi:\mathrm{Spec}(k)\to X$ is a monomorphism, then every $\mathrm{Spec}(K)\to X$ defining the same point factors (uniquely) through $\varphi$. The stated result follows immediately.

After the definition 010K I guess that you mean "by our conventions $\mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ is a set"

In the statement we put $s' = f'(x')$ instead of $g'$.