The Stacks project

now i just learned a new things, tysm sensei!

That was a very delightful table of content, tysm!

Typo? "The functor F:C→Finite-G-Sets (58.3.5.1) an equivalence. " Is there a verb missing in this sentence?

I agree with #9981. Specifically, one has: a morphism of schemes $f$ is smooth if and only if it is locally of finite presentation, flat and all fibers of $f$ are geometrically regular [GWII, Corollary 18.57]. Equivalently, $f$ is smooth if and only if it is locally of finite presentation and regular (More on Morphisms, Lemma 37.21.2).

Typos in the proof of 06IW (Lemma 90.13.3): any instances of $A/J$, $A/K$, $A/(J\cap K)$ should quotients of $B$.

There are some mistakes in the commutative diagram. The right square is not well defined since we do not have a morphism $X\times_Y X \to V \times_Y V$. After dropping the right square, what we need is to show that $U\times_V U\to X\times_Y X$ is étale, which follows from a composition of étale morphisms $U\times_V U\to (X\times_Y V)\times_V (X\times_Y V)\to X\times_Y X$.

There seems to be a small mistake in the proof in the assertion $\operatorname{Spec}(K')=\operatorname{Spec}(A')\times_{\operatorname{Spec}(A)}\operatorname{Spec}(K).$ For example, if $A=K=k$ and $A'=k[t]_{(t)}$, then the right hand side is $\operatorname{Spec}(A')$, whereas $\operatorname{Spec}(K')=\operatorname{Spec}(k(t))$.

However, the argument only seems to require that the morphism $\operatorname{Spec}(K')\longrightarrow X_K$ factors through the section $\operatorname{Spec}(K)\longrightarrow X_K.$ Since a section of $X_K\to\operatorname{Spec}(K)$ is a closed immersion, it follows that the scheme theoretic image over $K$ is $\operatorname{Spec}(K)$. Hence $Z_K=\operatorname{Spec}(K),$ which is what is needed.

'restriction' is misspelled in the last line of Lemma 6.21.8 proof.

Typo: in the statement's second diagram, the object in the lower left corner has bad parentheses.

In the proof of Lemma 101.28.6., in order to show it's essentially surjective we should use 2(b) instead of 2(a).

In the proof of Lemma 30.20.4 (tag 02OB), should “Lemma 30.20.3 part (4)” be “Lemma 30.20.3 part (3)”?

Indeed, by part (3) and the fact that $n+m-c\geq n$, we have $\ker(\beta)\subseteq\ker(\alpha)$, while the reverse inclusion follows immediately from the factorization of $\beta$ through $\alpha$. Hence $\ker(\beta)=\ker(\alpha)$.

On the other hand, I am not sure how part (4) directly gives this under the assumption $m\geq c$, since the relevant statement in part (4) seems to require an index at least $2c$.

Maybe one could mention that the empty subscheme is also an effective Cartier divisor. The remark after the definition of effective Cartier is slightly misleading since the empty subscheme is usually not considered to be of codimension 1. Of course, it is not wrong, just non-sensical, since the notion of codimension is defined only for irreducible closed subschemes.

I think there may be a typo in the proof of Lemma 29.40.8 (Tag 0FVC).

In the sentence

“Since $j$ is an immersion, so is $(j,j_{d-d_0})$,”

should “$j$” be “$j_{d-d_0}$” ?

I think there may be a typo in the sentence

“assume we have affine opens $U,V⊂S_{0}$ such that $U∩V$ is affine too.”

Should this be $U_{0},V_{0} \subset S_{0}?$ In the next sentence, $U,V \subset S$ are defined as the inverse images, and the subsequent notation also uses $U_{0},V_{0}$.

there is a typo: In condition (3), $f^* y$ should be $h^* y$.

Thank you for providing this detailed explanation. The presentation is clear and helpful, especially for readers working through the definitions and examples. I found this section useful as a reference for studying the topic.site\ref{https://textrepeater.tech/}

Thank you for providing this detailed explanation. The presentation is clear and helpful, especially for readers working through the definitions and examples. I found this section useful as a reference for studying the topic.site\ref{https://textrepeater.tech/}

Thanks for the education, i will bookmarks this note!

that was long explanation with perfect execution, regards !

Just to set it out with the verb-based terminology (and better have it fixed or discarded for good). To describe the relation $x'\rightsquigarrow x$, are the following expressions correct?

$x'$ generalizes (from) $x$

$x'$ specializes to $x$