diff -urN -X stacks-0.2/src/documentation/dontdiff stacks-0.2.orig/src/flat.tex stacks-0.2/src/flat.tex --- stacks-0.2.orig/src/flat.tex 2006-03-05 15:12:12.000000000 +0000 +++ stacks-0.2/src/flat.tex 2006-03-07 19:44:10.000000000 +0000 @@ -100,19 +100,18 @@ \label{subsection-equivalence} \noindent -Let me state our goal now so we have a clear idea of what we are trying to -achieve. It is to show that the fibered category $(QCoh/S)$ of quasi-coherent -sheaves over a scheme $S$ over $(Sch/S)$ the category of schemes over $S$ is -a stack with respect to the flat topology. The idea is that we will be able -to exploit the fact that there is a standard equivalance of categories between -$QCoh (U)$ and $Mod_A$ where $U = \text{Spec}(A)$, and use the following lemma -which I will state without proof. I refer the reader to Vistoli's notes -\cite{Vis2}[Lemma 4.25, page 94] for the proof. - -\begin{lemma} -\label{lemma-zariski-flat} -Let $S$ be a scheme, $\mathcal F$ be a fibered category ovver the category -$(Sch/S)$. Suppose the following conditions are satisfied. +Let me state our goal now so we have a clear idea of what we are +trying to achieve. It is to show that the fibered category $(\text{QCoh}/S)$ +(of quasi-coherent sheaves over a scheme $S$) over $\text{Sch}/S$ (the +category of schemes over $S$) is a stack with respect to the flat +topology. The idea is that we will be able to exploit the fact that +there is a standard equivalence of categories between $\text{QCoh}(U)$ and +$\text{Mod}_A$ where $U = \text{Spec}(A)$, and then use Lemma \ref{lemma-zariski-flat}. + +\begin{lemma}\label{lemma-zariski-flat} +(\cite{Vis2},Lemma 4.25) +Let $S$ be a scheme, $\mathcal F$ be a fibered category over the +category $(Sch/S)$. Suppose the following conditions are satisfied. \begin{enumerate} \item $\mathcal F$ is a stack with respect to the Zariski topology. \item Whenever $V \rightarrow U$ is a flat surjective morphism of affine @@ -122,24 +121,25 @@ \end{enumerate} Then $\mathcal F$ is a stack with respect to the flat (fpqc) topology. \end{lemma} +\begin{proof} + FIXME: Add proof. +\end{proof} \noindent Note that the fpqc topology stands for "fidelment plat et quasi-compact" which means faithfully flat and quasi-compact. This is a finer topology than the fppf topology which is a finer topology than the etale topology. -\smallskip\noindent -In the case of $(QCoh/S)$ over $(Sch/S)$ we can easily see that the first -condition is satisfied (i.e in the Zariski topology our definition for -quasi-coherent sheaves is exactly the conditions needed for the descent datum -to be effective). - -\smallskip\noindent -For the second condition, it will be necessary for me to define certain -notions and to prove certain algebraic results. -\smallskip\noindent -First we begin with some definitions. +\smallskip\noindent In the case of $(\text{QCoh}/S)$ over +$(\text{Sch}/S)$ we can easily see that the first condition is +satisfied (i.e in the Zariski topology our definition for +quasi-coherent sheaves is exactly the conditions needed for the +descent datum to be effective). + +\smallskip\noindent For the second condition, it will be necessary to +define certain notions and to prove certain algebraic results. We +begin with some definitions. \begin{definition} \label{definition-faithfully-flat} @@ -192,7 +192,7 @@ \smallskip\noindent There is an obvious way of composing morphisms, which makes the objects with -descent data the objects of a cateogry which we will denote +descent data the objects of a category which we will denote $\mathcal F(\{U_i \rightarrow U\})$ \smallskip\noindent @@ -207,11 +207,11 @@ \noindent To continue we need to set some conventions. Let $A$ be a commutative ring, -and denote $Mod_A$ as the category of modules over $A$. We also have a ring -homomorphism $f: A \rightarrow B$. We define a cateogry -$Mod_{A\rightarrow B}$ as follows. Let the objects be pairs $(N,\phi)$ +and denote $\text{Mod}_A$ as the category of modules over $A$. We also have a ring +homomorphism $f: A \rightarrow B$. We define a category +$\text{Mod}_{A\rightarrow B}$ as follows. Let the objects be pairs $(N,\phi)$ where $N$ is a $B$-module and $\phi: N \otimes_A B \simeq B \otimes_A N$ -is an isomorphism of $B^{\otimes 2}$-modules such that the following cocycle +is an isomorphism of $B^{\otimes p2}$-modules such that the following cocycle condition is satisfied: \begin{eqnarray} \phi_1: B \otimes_A N \otimes_A B & \rightarrow & B \otimes_A B \otimes_A N, @@ -221,7 +221,7 @@ \phi_3: N \otimes_A B \otimes_A B & \rightarrow & B \otimes_A N \otimes_A B, \nonumber \end{eqnarray} -where $\phi_1 = id_B \otimes \phi$, $\phi_3 = \phi \otimes id_B$, and +where $\phi_1 = \text{id}_B \otimes \phi$, $\phi_3 = \phi \otimes \text{id}_B$, and $\phi_2 = \phi_1\phi_3$. \smallskip\noindent @@ -231,14 +231,14 @@ \( \begin{array}{ccc} N \otimes_A B & \stackrel{\phi}{\longrightarrow} & B \otimes_A N \\ -\beta \otimes id_B \downarrow & & \downarrow id_B \otimes \beta \\ +\beta \otimes \text{id}_B \downarrow & & \downarrow \text{id}_B \otimes \beta \\ N' \otimes_A B & \stackrel{\phi'}{\longrightarrow} & B \otimes_A N' \end{array} \) \end{center} \smallskip\noindent -Given a functor $F: Mod_A \rightarrow Mod_{A\rightarrow B}$ which takes an +Given a functor $F: \text{Mod}_A \rightarrow \text{Mod}_{A\rightarrow B}$ which takes an $A$-module $M$ to the pair $(B\otimes_A M, \phi_M)$ where $\phi_M: (B\otimes_A M)\otimes_A B \rightarrow B \otimes_A (B \otimes_A M)$ maps $b \otimes m \otimes b'$ to $b \otimes b' \otimes m$ (and satisfies the @@ -247,7 +247,7 @@ \begin{theorem} \label{flat-descent-rings} If $B$ is faithfully flat over $A$, the functor -$F: Mod_A \rightarrow Mod_{A\rightarrow B}$ as defined above is an +$F: \text{Mod}_A \rightarrow \text{Mod}_{A\rightarrow B}$ as defined above is an equivalence of categories. \end{theorem} @@ -259,9 +259,9 @@ Let $M$ be an $A$-module. Then the following sequence $$ 0 \rightarrow M \stackrel{\alpha_M}{\longrightarrow} -B \otimes_A M \stackrel{(e_1 - e_2)\otimes id_M}{\longrightarrow} +B \otimes_A M \stackrel{(e_1 - e_2)\otimes \text{id}_M}{\longrightarrow} B^{\otimes 2}\otimes_A M -\stackrel{(e_1-e_2+e_3)\otimes id_M}{\longrightarrow} +\stackrel{(e_1-e_2+e_3)\otimes \text{id}_M}{\longrightarrow} B^{\otimes 3} \otimes_A M \rightarrow \cdots $$ is exact. Where $e_i: B^{\otimes n} \rightarrow B^{\otimes n+1}$ is the @@ -281,86 +281,86 @@ of $A$-modules is exact if and only if the sequence tensored with $B$ is exact. This is fortunate because once we tensor our sequence with $B$ we get the following sequence (with the same maps as in the statement of the lemma -just with tensored with a $id_B$ on the left): +just with tensored with a $\text{id}_B$ on the left): $$ -0 \rightarrow B \otimes_A M \rightarrow B^{\otimes_2} \otimes_A M +0 \rightarrow B \otimes_A M \rightarrow B^{\otimes 2} \otimes_A M \rightarrow B^{\otimes 3}\otimes_A M \rightarrow B^{\otimes 4} \otimes_A M \rightarrow \cdots. $$ \smallskip\noindent -And there is a natural map $B^{\otimes_2} \otimes_A M +And there is a natural map $B^{\otimes 2} \otimes_A M \stackrel{mult}{\rightarrow} B \otimes_A M$ which just takes $b\otimes b' \otimes m$ to the element $bb' \otimes m$. In other words it -is just the multipication map $B\otimes_A B \rightarrow B$ composed with the +is just the multiplication map $B\otimes_A B \rightarrow B$ composed with the identity on $M$. -Now, to prove that the sequence is exact at $B^{\otimes_2} \otimes_A M$, we +Now, to prove that the sequence is exact at $B^{\otimes 2} \otimes_A M$, we pick an element $\Sigma b_i \otimes b'_i \otimes m_i$ in the kernel of the -map $id_B \circ (e_1-e_2) \circ id_M$. This means that in +map $\text{id}_B \circ (e_1-e_2) \circ \text{id}_M$. This means that in $B^{\otimes_3} \otimes_A M$ we have the following relation: $\Sigma b_i\otimes b'_i \otimes 1 \otimes m_i = \Sigma b_i \otimes 1 \otimes b'_i \otimes m_i$. We then apply the map -$mult \circ id_B \circ id_M$ to the equality and get the +$mult \circ \text{id}_B \circ \text{id}_M$ to the equality and get the $\Sigma b_ib'_i \otimes 1 \otimes m_i = b_i \otimes b'_i \otimes m_i$ -in $B^{\otimes_2} \otimes_A M$. So we are getting -$id_B \circ \alpha_M (\Sigma b_ib'_i \otimes m_i) = +in $B^{\otimes 2} \otimes_A M$. So we are getting +$\text{id}_B \circ \alpha_M (\Sigma b_ib'_i \otimes m_i) = \Sigma b_i\otimes b'_i \otimes m_i$ which was the element from the kernel that we chose. Thus every element in the kernel is also in the image of the appropriate map hence the sequence is exact there. The same argument can be made at each step of the sequence to show that the appropriate kernels are contained in the appropriate images. A choice for sections -that work is simply multipication composed with whatever number of identity +that work is simply multiplication composed with whatever number of identity maps are necessary. \end{proof} \begin{proof}[Proof of the theorem] So to prove the theorem we need to first consider the functor -$F: Mod_A \rightarrow Mod_{A\rightarrow B}$ which takes an $A$-module $M$ +$F: \text{Mod}_A \rightarrow \text{Mod}_{A\rightarrow B}$ which takes an $A$-module $M$ to the pair $(B\otimes_A M, \phi_M)$ where $\phi_M: (B\otimes_A M)\otimes_A B \rightarrow B \otimes_A (B \otimes_A M)$ maps $b \otimes m \otimes b'$ to $b \otimes b' \otimes m$. \smallskip\noindent To show that $F$ is an equivalence of categories, we need to show that there -is a functor $G: Mod_{A\rightarrow B} \rightarrow Mod_A$ such that $GF$ and +is a functor $G: \text{Mod}_{A\rightarrow B} \rightarrow \text{Mod}_A$ such that $GF$ and $FG$ are isomorphic to the identity. \smallskip\noindent So let us define a functor $G$ to take pairs $(N, \phi)$ to elements $GN = \{n \in N | 1\otimes n = \phi(n\otimes 1)\}$ and given a morphism -$\beta: (N, \phi) \rightarrow (N', \phi')$ in $Mod_{A\rightarrow B}$ we +$\beta: (N, \phi) \rightarrow (N', \phi')$ in $\text{Mod}_{A\rightarrow B}$ we get a morphism $\beta_G: GN \rightarrow GN'$. \smallskip\noindent So first let us check that $GF$ is isomorphic to the identity. Notice that $$ -((e_1-e_2)\otimes id_M)(b \otimes m) = +((e_1-e_2)\otimes \text{id}_M)(b \otimes m) = b \otimes 1 \otimes m - 1 \otimes b \otimes m = \phi_M(b\otimes m \otimes 1) - 1 \otimes b \otimes m $$ for all $m$ and $b$. For simplicity we can rewrite this as -$((e_1 - e_2)\otimes id_M)(x) = \phi_M(x) - 1\otimes x$ for all +$((e_1 - e_2)\otimes \text{id}_M)(x) = \phi_M(x) - 1\otimes x$ for all $x \in B \otimes_A M$. So then by our definition of the functor $G$ we get -that $G(B \otimes_A M, \phi_M) = ker ((e_1-e_2)\otimes id_M)$. However, due +that $G(B \otimes_A M, \phi_M) = \text{ker} ((e_1-e_2)\otimes \text{id}_M)$. However, due to our lemma, we know that the sequence $$ 0 \rightarrow M \rightarrow^{\alpha_M} B \otimes_A M \rightarrow B^{\otimes 2}\otimes_A M \rightarrow B^{\otimes 3} \otimes_A M \rightarrow \cdots $$ -is exact, thus $ker ((e_1-e_2)\otimes id_M) = im(\alpha_M) \simeq M$. +is exact, thus $\text{ker} ((e_1-e_2)\otimes \text{id}_M) = \text{im}(\alpha_M) \simeq M$. So $M \simeq G(B \otimes_A M) = GF(M)$ as needed. \smallskip\noindent Now we will show that $FG$ is isomorphic to the identity. So we take -$(N, \phi)$ in $Mod_{A\rightarrow B}$ and we set +$(N, \phi)$ in $\text{Mod}_{A\rightarrow B}$ and we set $M = G(N,\phi) = \{n \in N | 1 \otimes n = \phi(n \otimes 1)\}$. Since $M$ is an $A$-submodule of the $B$-submodule $N$ we get a homomorphism of $B$-modules $\theta: B \otimes_A M \rightarrow N$ which takes $b\otimes m$ to $bm$. It is easy to check that this is a morphism in -$Mod_{A\rightarrow B}$. So notice that we can also think of $\theta$ as a +$\text{Mod}_{A\rightarrow B}$. So notice that we can also think of $\theta$ as a map $F(M) \rightarrow N$, thus we can see that $\theta$ defines a natural transformation $id \rightarrow FG$. So to complete the proof we need to show that $\theta$ is an isomorphism. @@ -371,20 +371,20 @@ $m \otimes b$ to $b \otimes m$, and $\alpha, \beta: N \rightarrow B \otimes_A M$ are defined by $\alpha (n) = 1 \otimes n$ and $\beta (n) = \phi(n \otimes 1)$. So by -definition, $M = ker(\alpha - \beta)$. We have the following diagram, +definition, $M = \text{ker}(\alpha - \beta)$. We have the following diagram, where the rows are exact: \begin{center} \( \begin{array}{cccccccc} 0 & \longrightarrow & M \otimes_A B & -\stackrel{i \otimes id_B}{\longrightarrow} & N \otimes_A B & -\stackrel{(\alpha-\beta)\otimes id_B}{\longrightarrow }& +\stackrel{i \otimes \text{id}_B}{\longrightarrow} & N \otimes_A B & +\stackrel{(\alpha-\beta)\otimes \text{id}_B}{\longrightarrow }& B \otimes_A N \otimes_A B \\ & & \downarrow \theta \circ \iota_M & & \downarrow \phi & & \downarrow \phi_1\\ 0 & \longrightarrow & N & \stackrel{\alpha_N}{\longrightarrow} & -B \otimes_A N & \stackrel{(e_1-e_2)\otimes id_N}{\longrightarrow} & +B \otimes_A N & \stackrel{(e_1-e_2)\otimes \text{id}_N}{\longrightarrow} & B \otimes_A B \otimes_A N \end{array} \) @@ -395,12 +395,12 @@ and $\phi_1$ are isomorphisms we are able to get that $\theta$ is an isomorphism. To show the diagram commutes let us focus on one square at a time. For the first square we want -$\phi(i \otimes id_B)(m\otimes b) = \alpha_M \theta \iota_M (m\otimes b)$. +$\phi(i \otimes \text{id}_B)(m\otimes b) = \alpha_M \theta \iota_M (m\otimes b)$. We know that $\alpha_M\theta \iota_M(m\otimes b) = 1 \otimes bm$. So we just need to show that we get the same thing for -$(\phi(i \otimes id_B))(m\otimes b)$. We have +$(\phi(i \otimes \text{id}_B))(m\otimes b)$. We have \begin{eqnarray*} -(\phi (i \otimes id_B))(b \otimes m) & = & \phi(m \otimes b) \\ +(\phi (i \otimes \text{id}_B))(b \otimes m) & = & \phi(m \otimes b) \\ & = & \phi((1\otimes b)(m \otimes 1))\\ & = & (1 \otimes b) \phi (m \otimes 1)\\ & = & (1 \otimes b)(1 \otimes m)\\ @@ -408,15 +408,15 @@ \end{eqnarray*} as needed. Now we just need to show the second square commutes. For the second square it should be clear that -$\phi_1(\alpha \otimes id_B) = (e_2 \otimes id_N) \circ \phi$. So we just +$\phi_1(\alpha \otimes \text{id}_B) = (e_2 \otimes \text{id}_N) \circ \phi$. So we just need to check that -$\phi_1(\beta \otimes id_B) = (e_1 \otimes id_N)\circ \phi$. We have: +$\phi_1(\beta \otimes \text{id}_B) = (e_1 \otimes \text{id}_N)\circ \phi$. We have: \begin{eqnarray*} -\phi_1(\beta \otimes id_B)(n\otimes b) & = & +\phi_1(\beta \otimes \text{id}_B)(n\otimes b) & = & \phi_1(\phi(n\otimes 1) \otimes b) \\ & = & \phi_1 \phi_3(n \otimes 1 \otimes b) \\ & = & \phi_2 (n \otimes 1 \otimes b)\\ -& = & (e_1 \otimes id_N) \phi( n \otimes b) +& = & (e_1 \otimes \text{id}_N) \phi( n \otimes b) \end{eqnarray*} \smallskip\noindent @@ -429,7 +429,7 @@ \begin{theorem} \label{theorem-quasi-coherent-stack} -Let $S$ be a scheme. The fibered category $(QCoh/S)$ over $(Sch/S)$ is +Let $S$ be a scheme. The fibered category $(\text{QCoh}/S)$ over $(Sch/S)$ is stack with respect to the flat (fpqc) topology. \end{theorem} @@ -439,27 +439,27 @@ Let me remind you that we just need to check the second condition of the lemma. So for a flat and surjective morphism $V \rightarrow U$ (corresponding to a faithfully flat ring homomorphism $f: A \rightarrow B$). -We need to show that there is an equivalence of categories between $QCoh(U)$ -and $QCoh(V\rightarrow U)$. We will do this using the previous theorem which -states that there is an equivalence of categories between $Mod_A$ and -$Mod_{A\rightarrow B}$. +We need to show that there is an equivalence of categories between $\text{QCoh}(U)$ +and $\text{QCoh}(V\rightarrow U)$. We will do this using the previous theorem which +states that there is an equivalence of categories between $\text{Mod}_A$ and +$\text{Mod}_{A\rightarrow B}$. \smallskip\noindent -There is a standard equivalence of categories between $QCoh(U)$ and $Mod_A$. +There is a standard equivalence of categories between $\text{QCoh}(U)$ and $\text{Mod}_A$. So we just need to show that there is an equivalence of categories between -$QCoh(V\rightarrow U)$ and $Mod_{A\rightarrow B}$. To do this let us look at -$\mathcal N$ an object in $QCoh(V)$ which corresponds to an $B$-module $N$. +$\text{QCoh}(V\rightarrow U)$ and $\text{Mod}_{A\rightarrow B}$. To do this let us look at +$\mathcal N$ an object in $\text{QCoh}(V)$ which corresponds to an $B$-module $N$. Looking at $p_1^* \mathcal N$ and $p_2^* \mathcal N$ in $V \times_U V = \text{Spec} (B \otimes_A B)$ we get $N \otimes_A B$ and $B \otimes_A N$ respectively. So the descent datum $\psi: p_1^* \mathcal N \simeq p_2^* \mathcal N$ will correspond to the descent data $\phi: N \otimes_A B \simeq B \otimes_A N$ in -$Mod_{A \rightarrow B}$. So $(\mathcal N, \psi)$ is an object of -$QCoh(V\rightarrow U)$ if and only if $\phi$ satisfies the cocycle condition, -thus giving us an equivalence of categories between $QCoh(V\rightarrow U)$ -and $Mod_{A\rightarrow B}$. Thus the functor -$QCoh(U) \rightarrow QCoh(V\rightarrow U)$ corresponds to the functor -$Mod_A \rightarrow Mod_{A\rightarrow B}$! So since, the later is an +$\text{Mod}_{A \rightarrow B}$. So $(\mathcal N, \psi)$ is an object of +$\text{QCoh}(V\rightarrow U)$ if and only if $\phi$ satisfies the cocycle condition, +thus giving us an equivalence of categories between $\text{QCoh}(V\rightarrow U)$ +and $\text{Mod}_{A\rightarrow B}$. Thus the functor +$\text{QCoh}(U) \rightarrow \text{QCoh}(V\rightarrow U)$ corresponds to the functor +$\text{Mod}_A \rightarrow \text{Mod}_{A\rightarrow B}$! So since, the later is an equivalence we get the equivalence of categories that we need. Thus finishing the proof. \end{proof} diff -urN -X stacks-0.2/src/documentation/dontdiff stacks-0.2.orig/src/stacks.tex stacks-0.2/src/stacks.tex --- stacks-0.2.orig/src/stacks.tex 2006-02-11 16:45:26.000000000 +0000 +++ stacks-0.2/src/stacks.tex 2006-03-07 19:44:10.000000000 +0000 @@ -208,8 +208,18 @@ g'^\ast \phi \circ t'^{-1}_{x}) \circ t''^{-1}_{x} \\ & = & [(\text{Isom}(x,y))(g'')]([(\text{Isom}(x,y))(g')](\phi)). \end{eqnarray*} + +Alternatively we can argue with +\hyperref[categories-lemma-fibred-strict]{Lemma~\ref*{categories-lemma-fibred-strict}} +which says that $\mathcal{S}\to\mathcal{C}$ is equivalent to the +category assoicated to a contravariant function $F\colon +\mathcal{C}\to \text{Groupoids}$. Then $t_y$ is the identity +transformation for all $y$ so the restriction map is $g\mapsto g^*\phi += F(g)\phi$ which is clearly functorial making $\text{Isom}(x,y)$ a +presheaf. \end{proof} + \noindent This lemma says that there is no harm in thinking of $g^\ast \phi$ as a morphism ${f'}^\ast x \to {f'}^\ast y$.