The Stacks project

Should assume $A\ne 0$ (so $T_{A/k,e} \ne 0$) or else "only if" is not true (fails when $A=0$). Maybe also replace "$d$" with "$m$", to avoid awkward-looking notation such as ${\rm{d}}<script type="math/tex; mode=display">d$ .

A typo in the hypothesis of the second statement: the second sequence should be $f_1,\dots,f_{r-1},g$, not $f_1,\dots,f_{r-1},f$.

"Dense morphism"? (cf. e.g. section 11 of my paper http://www.tac.mta.ca/tac/volumes/27/7/27-07abs.html, which may not be exactly the same but is very closely related)

What is the definition of height 1 prime ideal? Please help me I don't understand.

In addition to Ben's comment, the converse (constructing $\mathcal{G}$ from $\mathcal{F}$) is proved but not stated. Also, in the proof there are 3 instances of an exponent $p$ standing for $n$.

What is the definition of height 1 prime ideal? Please help me I don't understand.

It might be better to either mention that the definition of $D_{I^\infty\mathrm{-torsion}}$ is a special case of the one in that section about Serre subcategories, or incorporate a specialized definition here.

It might be better to mention the name (Matlis)-Greenless-May duality in the text or in the slogan, which facilitates searching and locating.

In the proof of (1), shouldn't $x_i$ be $y_i$?

Hi Johan, In the statement of Lemma 0DV7 there is no relation between the sheaves F and G, which is not per se wrong but looks funny. Maybe better to make the statement more explicit, matching its proof: if G has cohomology then so does F = g^{-1}G; if F has cohomology then so does G = g_*F. Best wishes, Ben

tag 047I. Sorry for spam, the input of asterisk keeps bugging. This is my last try. The assumption of flatness is not needed for the result. See Prop (3.15) of "abelian varieties" by Ben Moonen. In short, define $\tau :G\times_S G\to G\times_S G, (g,h)\mapsto (gh,h)$ which is an automorphism. Denote $p_1$ as the left projection and $p_2$ by the right one. So we have an isomorphism: $$\begin{equation}\tau^{*} p_1^{*} \Omega^1_{G/S}\cong p_1^{*} \Omega^1_{G/S}\end{equation}$$ Let $\psi :G\to G\times_S G,g\mapsto (e,g)$, so we have $$\begin{equation}\psi^{*}\tau^{*} p_1^{*} \Omega^1_{G/S}\cong \psi^{*}p_1^{*} \Omega^1_{G/S}\end{equation}$$ $$\begin{equation}(p_1\circ \tau\circ \psi)^{*} \Omega^1_{G/S}\cong (p_1\circ \psi)^{*} \Omega^1_{G/S}\end{equation}$$ $$\begin{equation}\text{id}\_G^{*} \Omega^1_{G/S}\cong (e\circ f)^{*} \Omega^1_{G/S}\end{equation}$$ $$\begin{equation}\Omega^1_{G/S}\cong f^{*} e^{*}\Omega^1_{G/S}\end{equation}$$ Q.E.D.

tag 047I. The assumption of flatness is not needed for the result. See Prop (3.15) of this link \ref{http://page.mi.fu-berlin.de/elenalavanda/BMoonen.pdf} In short, define $\tau :G\times_S G\to G\times_S G, (g,h)\mapsto (gh,h)$ which is an automorphism. Denote $p_1$ as the left projection and $p_2$ by the right one. So we have an isomorphism: $$\begin{equation}\tau^{\*} p_1^{\*} \Omega^1_{G/S}\cong p_1^{\*} \Omega^1_{G/S}\end{equation}$$ Let $\psi :G\to G\times_S G,g\mapsto (e,g)$, so we have $$\begin{equation}\psi^{\*}\tau^{\*} p_1^{\*} \Omega^1_{G/S}\cong \psi^{\*}p_1^{\*} \Omega^1_{G/S}\end{equation}$$ $$\begin{equation}(p_1\circ \tau\circ \psi)^{\*} \Omega^1_{G/S}\cong (p_1\circ \psi)^{\*} \Omega^1_{G/S}\end{equation}$$ $$\begin{equation}\text{id}\_G^{\*} \Omega^1_{G/S}\cong (e\circ f)^{\*} \Omega^1_{G/S}\end{equation}$$ $$\begin{equation}\Omega^1_{G/S}\cong f^{\*} e^{\*}\Omega^1_{G/S}\end{equation}$$ Q.E.D.

tag 047I. The assumption of flatness is not needed for the result. See Prop (3.15) of this link \ref{http://page.mi.fu-berlin.de/elenalavanda/BMoonen.pdf}Define. In short, define $\tau :G\times_S G\to G\times_S G, (g,h)\mapsto (gh,h)$ which is an automorphism. Denote $p_1$ as the left projection and $p_2$ by the right one. So we have an isomorphism: $$\tau^{\*} p_1^{\*} \Omega^1_{G/S}\cong p_1^{\*} \Omega^1_{G/S}$$ Let $\psi :G\to G\times_S G,g\mapsto (e,g)$, so we have $$\psi^{\*}\tau^{\*} p_1^{\*} \Omega^1_{G/S}\cong \psi^{\*}p_1^{\*} \Omega^1_{G/S}$$ $$(p_1\circ \tau\circ \psi)^{\*} \Omega^1_{G/S}\cong (p_1\circ \psi)^{\*} \Omega^1_{G/S}$$ $$\text{id}\_G^{\*} \Omega^1_{G/S}\cong (e\circ f)^{\*} \Omega^1_{G/S}$$ $$\Omega^1_{G/S}\cong f^{\*} e^{\*}\Omega^1_{G/S}$$ Q.E.D.

Shouldn't a right $A$-module be the same as a map $\mathrm{Tot} (M^\bullet \otimes _R A^\bullet) \to M^\bullet$, not $A^\bullet \otimes _R M^\bullet$ as you have written? It's possible that I'm misremembering a sign convention but this seems not right.

Shouldn't the extension $t$ of $s$ be a morphism over $X$?

In the fourth paragraph of the proof of Lemma 030F you should say that B * is algebraically independent over F, not on E.

After "We spell out what this means", I think the $T_i'$'s should be $T_{i'}$

It's better to mention what $X^{\nu}$ is in Lemma 07TD.

Typo: in 03NY (5), I think either both or non of the inclusions i_x and i_y should be written with a hooked arrow

Again thanks. I have now fixed this here in exactly the manner you suggested.