The Stacks project

Second paragraph of Lemma 09H9 may be needs a line saying that now we prove that $K/F$ is separable.

I think there's a small typo in the last paragraph of the proof, in the third to last sentence: I think you want to say IS_f \neq S_f to conclude that (S/I)_f is not zero.

Also, perhaps I missed something, but I could not find anywhere in the proof refereeing to parts (8) or (9) of the lemma.

very very small issue, on the first line you had 768g, but on the second line you were saying 728g.

Trivial remark: in (45), should the article "a" be used instead of "an"?

Usually, in a situation like this when we are definining a blah which is unique up to isomorphism in the definiition we say "a blah" and then in a comment after the definition we say that because the thing is unique up to unique isomorphism we will from now on use the terminology "the blah". See the text following this definition in Section 12.3.

Two minor typos: $$\mathcal O_{S_i}(V_{i,j_1})\otimes_{\mathcal O_S(S)}\mathcal O_{S_i}(V_{i,j_2})\longrightarrow\mathcal O_{S_i}(V_{i,j_1}\cap V_{i,j_2})$$

should be

$$\mathcal O_{S_i}(U_{i,j_1})\otimes_{\mathcal O_S(S)}\mathcal O_{S_i}(U_{i,j_2})\longrightarrow\mathcal O_{S_i}(U_{i,j_1}\cap U_{i,j_2}).$$

And $$\mathcal O(U_)\otimes\mathcal O(V)$$ should be $$\mathcal O(U)\otimes\mathcal O(V).$$

A few trivial typos: Let F be a abelian sheaf Extra parenthesis: The family of functors $H_i((X,−)$ forms [...] from $Ab(X)\rightarrow Ab$ Extra parenthesis*: The family of functors $H_i((X,−)$ forms [...] from $Mod(\mathcal{O}_X)\rightarrow Mod_{\mathcal{O}_X(X)}$

This is a very minor point, but is there a reason for using the terminology "a cokernel", etc. instead of "the cokernel", etc. when these objects are unique upto unique iso?

Yes very good. The key is that here $u$ is surjective whereas in Lemma 10.98.1 it doesn't need to be.

To show $u$ is injective, may be we can also argue as follows. Let $K$ denote the kernel of $u$, it is a finite $S$ module. Since $M$ is flat over $R$, applying $-\otimes_RR/\mathfrak{m}$ we get $K\otimes_RR/\mathfrak{m}=0$. It follows that $K\otimes_SS/\mathfrak{m}S=0$, which shows, by Nakayama's lemma, that $K=0$.

The sections of the sheaf $\mathcal{F}$ over an object $U$ of the site are the elements $s \in \mathcal{I}(U)$ whose image in $\mathcal{Q}(U)$ are equal to the restriction $q|_U$. Now you have to define a simply transitive action of $\mathcal{H}(U)$ on this set provided it is nonempty and the construction has to be compatible with restriction mappings. Finally, you have to show that if $\mathcal{F}(U)$ is empty, then you can find a covering $\{U_i \to U\}$ such that each $\mathcal{F}(U_i)$ is nonempty. Does this help?

In the last paragraph, could someone explain why it is easy to verify that $\mathcal{F}$ is a $\mathcal{H}$-torsor. I don't see it, thanks!

Just a typo: in the statement of the Lemma, after item (2), the "Then second" should be "The second".

Yes, that was confusing. Thanks. Fixed here.

In the proof, $L_{B \otimes _ A B/B} = 0$ should be replaced by $L_{B/B \otimes _ A B} = 0$.

Also in the case $n>1,$ in the third sentence where you cite Lemma 00PN, it seems like you mean to cite Lemma 0C6H.

An alternative way to see that $I\cap R\ne(0)$ is as follows: Take a non-zero $x\in I$. Since $L/K$ is finite, we can take the minimal polynomial $p$ of $x$ over $K$. Write it as $a_nx^n+\cdots+a_0=0$ with $a_i\in K$. By multiplying by some elements in $R$ if necessary, we may assume $a_i\in R,\,\forall i$. Also $a_0\ne0$ as $p$ is minimal. Now $a_nx^n+\cdots+a_1x\in I$, so $a_0\in I\cap R$.

The proof of the case n=1 is confusing. Once you assume $R$ is integrally closed in $S$ you have $S'= R$, yet at the end you write "$\phi (C) \subset S'$..." In fact since $R \subset R[x]/I \subset S$ and $R$ is integrally closed in $S$ so $R = C$ so it seems like you shouldn't even introduce $C$.

Thanks and fixed here.

Looks like in the second to last sentence in the proof you should subtract by $u\phi (a_k x^k)$ not $u \phi (a_k)$.