The Stacks project

Can I suggest adding to the lemma: (4) If \hat{A} is normal, then so is A (with a reference to 033G as the proof)?

typo "reduced" for "normal" in the proof.

In the definition of $S_k$, the convention that the dimension of the zero module can be made useless by requiring that the prime ideal $\mathfrak p$ belongs to the support of the module.

I am so sorry. I didn't think through enough and didn't realize that $X^n$ is just the product of $n$ copies of $X$, with the obvious action of $S_n$. I am sorry for any inconvenience thus caused. Also feel free to delete the comments to make the site cleaner if that is desired. Thanks in any case for this great site and effort.

I do not understand what is the object $X^n$ in the proof of lemma 0BN2. It seems to come out of nowhere, endowed with a mysterious action of $S_n$. Could someone explain what is it? Thanks.

I do not understand what is the object $X^n$ in the proof of lemma 0BN2. It seems to come out of nowhere, endowed with a mysterious action of $S_n$. Could someone explain what is it? Thanks.

Another case for (3): $G\to S$ is radicial.

The proof also shows that $X \to X/G$ is a $G$-torsor.

the local equation for $E$ has the same name as the morphism $f:X\to S$. Maybe call it $g$ instead?

One could mention in the statement that $E$ is the exeptional divisor of the blowup

couple typos: $H^1(L_{S/R})$ should be $H_1(L_{S/R})$

$\Omega_{R/\lambda}$ should be $\Omega_{R/\Lambda}$

"Hence (3) $\Rightarrow$ (2)" should be "Hence (4) $\Rightarrow$ (2)"

"Write $g$" should be "Write $h$".

7th sentense in the proof.

Why isn't there a mention of the (optional) locally small condition of a (left, right) multiplicative system as in Weibel's homological algebra book? Without this condition $S^{-1}\mathcal{C}$ may not be locally small even though $\mathcal{C}$ is.

In the above comment, I mean $\cup$ whenever I wrote $\cap$.

In the proof of Lemma 01XF, instead of working with a closed point $P \in U$, we can work with an open set $V$ where $U \cap V = X$. Then following the exact same proof we can find $f \in \Gamma(X, \mathcal{O}_ X)$ such that $X_f$ is a principal open subscheme of $U$ and $X_f \cap V = X$.

Since $X$ is quasi-compact, it has a finite affine covering $U_1, \dots , U_n$. Since $U_1 \cap (U_2 \cap \dots \cap U_n) = X$, there exists $f_1$ such that $X_{f_1}$ is affine open and $X_{f_1} \cap (U_2 \cap \dots \cap U_n) = X$. Similarly, since $U_2 \cap (X_{f_1} \cap U_3 \cap \dots \cap U_n) = X$, there exists $f_2$ such that $X_{f_2}$ is affine open and $X_{f_2} \cap (X_{f_1} \cap U_3 \cap \dots \cap U_n) = X$. By iterating such steps we can get $X_{f_1} \cap \dots X_{f_n} = X$ and that each $X_{f_i}$ is affine open.

This approach has the advantage that it avoids Zorn's lemma.

In the statement (7) and (8) of Nakayama lemma, it would be better to point out $x_1,\cdots,x_n$ generate $M/IM$ as an $R$-module, since $M/IM$ can be naturally $R/I$-module as well.

In 30.26.4, the argument will be more transparent if we pick $U$ to be the domain of definition of $f$. A more detailed proof is following:

There exists a largest nonempty open subscheme $U\subset X$ such that $f$ corresponds to a section of $\Gamma(U,\mathcal{O} _ X^ * )$ . Hence any prime divisor $Z$ with generic point $\xi$ s.t. $f\notin\mathcal{O} _ {X,\xi}$ must have $\xi\in X\backslash U$. Since $U$ is irreducible, we have $U\cap Z=\emptyset$. Since $Z$ is of codimension 1, it is easy to see it is an irreducible component of $X\backslash U$. Hence Lemma 30.26.1 gives the desired result.

If $\mathop{\mathrm{ord}} _ Z (f)\neq 0$, equivalently $\mathop{\mathrm{ord}}Z(f)<0$ or $\mathop{\mathrm{ord}} _ Z(f^{-1})<0$. Hence we have either $f\notin \mathcal{O} _ {X,\xi}$ or $f^{-1}\notin \mathcal{O} _ {X,\xi}$. Apply previous argument then we are done.

"Then we see that $V_z=U_{z,z′1}\cap …\cap U_{z,z′r}\cap U_i$ " should be "Then we see that $V_z=V_{z,z′1}\cap …\cap V_{z,z′r}\cap U_i$".

Thanks for pointing this out! I've recorded it on https://github.com/gerby-project/gerby-website/issues/140 and it will be dealt with soon I hope.