The Stacks project

I understand that Krull's intersection theorem is a simple consequence of Artin-Rees, but here what you really need is Krull's intersection theorem, so it might be better to say $f\in J_m$ by Krull's intersection theorem.

In Part (1), I think you are refering to 89.3.5 instead of 89.3.10.

Where did we prove "the set of closed points in $Z$ is dense in $Z$"?

I think the language here is confusing, what if change it into

(1) a topological space $X$, (2) a point $x \in X$, (3) a locally closed subset $E \subset X$, ...

There seems to be an implicit convention (here and in 37.45.5) that $k$ denotes an arbitrary field. I think it should be made explicit somwhere nearby.

Typo in second paragraph of section: "Let $K$ is a field".

This lemma can be proved in the same fashion with first proof of lemma 30.4.2. Moreover, it can be slightly strengthened the statement.

First, reformulate the statement as follows: Let X be a quasi-compact quasi-separated scheme. Let $X = U_1\cup \dots U_t$ be an open covering where $U_i$ is quasi-compact separated. Set $$d = \max_{I\subset \{1,\dots,t\}} (|I| + t(\bigcap_{i\in I} U_i) -1)$$ . Then $H^n(X,F) = 0$ for $n\geq d$ and all quasi-coherent sheaf $F$.

I will prove it by induction on $t$. If $t=1$, the result follows from 30.4.2. If $t>1$, write $X = U\cup V$ with $V=U_t$ quasi-compact separated and $U = U_1\cup \dots \cup U_{t-1}$. Notice that $U\cap V$ can be written as $(U_1\cap V)\cup \dots \cup (U_{t^1}\cup V)$, where $U_i\cap V$ is quasi-compact separated.

Let $a = \max_{I\subset \{1,\dots,t-1\}} (|I|+t(\bigcap_{i\in I} U_i) - 1)$, $b=t(V)$, $c = \max_{I\subset \{1,\dots,t-1\}} (|I|+t(V\cap \bigcap_{i\in I} U_i)-1)$, $d = \max(a,b,c+1) = \max_{I\subset \{1,\dots,t\}} (|I|+(t(\bigcap_{i\in I} U_i)) - 1)$. So, By induction we have $H^i(U,F) =0$ for $i\geq a$, $H^i(V,F)=0$ for $i\geq b$, and $H^i(U\cap V,F) = 0$ for $i\geq c$. By Mayer-Vietoris sequence, $H^i(X,F)=0$ for $i\geq d$.

The proof of (1) is much the same as Lemma 30.4.1. I think it is better to use it.

Namely, we can assume $S$ is affine. I'm going to say $P(U)$ for a quasi-compact open $U\subset X$ when, for any quasi-coherent $O_U$-module $F$, and for any $p\geq 0$, $R^pg_*F$ is quasi-coherent, where $g$ is the composite of inclusion $U\subset X$ and $f$.

I will verify the conditions (1) and (2) of Lemma 30.4.1. First, let $U$ be affine open. $R^0g_*F = g_*F$ is quasi-coherent, because $g_*$ is a quasi-compact quasi-separated morphism. For $p>0$, $R^pg_*F = 0$ by Lemma 30.2.3. So (1) is verified.

Next, let $U$ be a quasi-compact open, $V$ an affine open, and assume $P(U)$, $P(V)$, $P(U\cap V)$ hold. Let $g\colon U\cup V\to S$, $a\colon U\to S$, $b\colon V\to S$, and $c\colon U\cap V\to S$ be the composite of an inclusion to $X$ and $f$. Then, for any quasi-coherent $O_{U\cup V}$-module $F$, we have the relative Mayer-Vietoris sequence $$0\to g_*(F)\to a_*(F|_U)\oplus b_*(F|_V)\to c_*(F|_{U\cap V})\to Rg_*^1(F) \to \dots$$ . By hypothesis $Ra_*^p(F|_U)$, $Rb_*^p(F|_V)$, $Rc_*^p(F|_{U\cap V})$ are quasi-coherent for all $p\geq 0$. Hence $Rg_*^p(F)$ are quasi-coherent for all $p\geq 0$.

Aha! I overlooked the fact that X need not be geometrically irreducible (currently I'm only working with geometrically integral k-schemes). Thanks for taking the time to answer my question.

In Example 6.4.1, should it be initial in the category of presheaves over X? There hasn't been discussion in Ch 6 yet of "the category of presheaves" without specifying the base space.

For the two instances of (insert future reference here), perhaps you intend Lemma 35.37.1 for $G \rightarrow S$ affine and Lemma 35.7.6 for $G = \mathrm{GL}_n$.

Typo in (2): "$f$ is flat and $x$".

If you compute through an example, you will find that your assertion about the base change does not hold --- that is the point of this lemma.

I am confused by (2). Let $f \colon \mathrm{Spec}\, A \to \mathrm{Spec}\, \mathbb Z$ where $A$ is an order in a number field. The fiber of $f$ over $Y' = \mathrm{Spec}\, \mathbb F_p$ is Gorenstein, for its localisation at any maximal ideal $x$ is, after a finite basechange along $k/\mathbb F_p$, of the form $\mathrm{Spec}\, k[x]/(x)^e$ which has the simple socle $kx^{e-1}$, and is therefore Gorenstein by, e.g., Prop. 21.5 of Eisenbud. Assertion (2) implies that $f$ is Gorenstein at $x$ but $x$ was arbitrary, and number fields of degree $>2$ contain non-Gorenstein orders. For the same reason, I am also confused by Lemma 0C05.

There's a typo in (4) with $f\circ g$.

What exactly means by a ring $A$ in which $p=0$ ? Does it mean $A$ is a ring or field of characteristic $0$ ?

I guess preview is not working propery. At first I wrote as this:

\xymatrix{ S \ar[r] & S\ar[r] & S' \\\\ R \ar[r] \ar[u] & R'' \ar[r] \ar[u] & R' }

This works propery in preview, but doesn't work in comment. Then I wrote as this:

\xymatrix{ S \ar[r] & S\ar[r] & S' \\ R \ar[r] \ar[u] & R'' \ar[r] \ar[u] & R' }

This doesn't work propery in preview, but does work in comment.

Hmm... The diagram is not shown propery. I try again.

$$\xymatrix{ S \ar[r] & S \ar[r] & S' \\ R \ar[r] \ar[u] & R'' \ar[r] \ar[u] & R' \ar[u] }$$

In diagram 10.129.1, let $R''$ be a pullback $S\times_{S'} R'$. (not $\otimes$) Then we have following diagram: $$\xymatrix{ S \ar[r] & S \ar[r] & S' \\\\ R \ar[r] \ar[u] & R'' \ar[r] \ar[u] & R' \ar[u] }$$

For any $R''$-derivation $d\colon S\to M$ which is $0$ as $R$-derivation is $0$ also as $R'$-derivation. Hence $\Omega_{S/R}\to \Omega_{S/R''}$ is surjective. Moreover, the kernel of this map is generated as $S$-module by the image of $R''$, by definition of derivation. So, it suffice to show that $\Omega_{S/R''} = \Omega_{S'/R'}$.

Whenever one has a commutative square, there is a morphism from one corner to the fibre product of the rest of the diagram. In the first line of the proof we see that $\nu$ is the normalization of $X$. But in general $X_{k'}$ is not a variety (eg it could have more than $1$ irreducible component and/or could be nonreduced) so what you said will not be true.

As an example consider $X$ given by $T_0^2 + T_1^2 = 0$ in $\mathbf{P}^2$ over the field $k = \mathbf{R}$.