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adequate.tex
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\input{preamble}
% OK, start here.
%
\begin{document}
\title{Adequate Modules}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
For any scheme $X$ the category $\textit{QCoh}(\mathcal{O}_X)$
of quasi-coherent modules is abelian and a weak Serre subcategory
of the abelian category of all $\mathcal{O}_X$-modules. The same
thing works for the category of quasi-coherent modules on
an algebraic space $X$ viewed as a subcategory of the category
of all $\mathcal{O}_X$-modules on the small \'etale site of $X$.
Moreover, for a quasi-compact and quasi-separated morphism
$f : X \to Y$ the pushforward $f_*$ and higher direct images
preserve quasi-coherency.
\medskip\noindent
Next, let $X$ be a scheme and let $\mathcal{O}$ be the structure
sheaf on one of the big sites of $X$, say, the big fppf site.
The category of quasi-coherent $\mathcal{O}$-modules is abelian
(in fact it is equivalent to the category of usual quasi-coherent
$\mathcal{O}_X$-modules on the scheme $X$ we mentioned above)
but its imbedding into $\textit{Mod}(\mathcal{O})$ is not exact.
An example is the map of quasi-coherent modules
$$
\mathcal{O}_{\mathbf{A}^1_k}
\longrightarrow
\mathcal{O}_{\mathbf{A}^1_k}
$$
on $\mathbf{A}^1_k = \Spec(k[x])$ given by multiplication by $x$.
In the abelian category of quasi-coherent sheaves this map is injective,
whereas in the abelian category of all $\mathcal{O}$-modules on the
big site of $\mathbf{A}^1_k$ this map has a nontrivial kernel as we
see by evaluating on sections over $\Spec(k[x]/(x)) = \Spec(k)$.
Moreover, for a quasi-compact and quasi-separated morphism
$f : X \to Y$ the functor $f_{big, *}$ does not preserve quasi-coherency.
\medskip\noindent
In this chapter we introduce a larger category of modules, closely related
to quasi-coherent modules, which ``fixes'' the two problems mentioned above.
\section{Conventions}
\label{section-conventions}
\noindent
In this chapter we fix
$\tau \in \{Zar, \etale, smooth, syntomic, fppf\}$
and we fix a big $\tau$-site $\Sch_\tau$ as in
Topologies, Section \ref{topologies-section-procedure}.
All schemes will be objects of $\Sch_\tau$.
In particular, given a scheme $S$ we obtain sites
$(\textit{Aff}/S)_\tau \subset (\Sch/S)_\tau$.
The structure sheaf $\mathcal{O}$ on these sites is defined by
the rule $\mathcal{O}(T) = \Gamma(T, \mathcal{O}_T)$.
\medskip\noindent
All rings $A$ will be such that $\Spec(A)$ is (isomorphic to) an
object of $\Sch_\tau$. Given a ring $A$ we denote
$\textit{Alg}_A$ the category of $A$-algebras whose objects are the
$A$-algebras $B$ of the form $B = \Gamma(U, \mathcal{O}_U)$
where $S$ is an affine object of $\Sch_\tau$. Thus given an
affine scheme $S = \Spec(A)$ the functor
$$
(\textit{Aff}/S)_\tau \longrightarrow \textit{Alg}_A,
\quad
U \longmapsto \mathcal{O}(U)
$$
is an equivalence.
\section{Adequate functors}
\label{section-quasi-coherent}
\noindent
In this section we discuss a topic closely related to
direct images of quasi-coherent sheaves. Most of this material
was taken from the paper \cite{Jaffe}.
\begin{definition}
\label{definition-module-valued-functor}
Let $A$ be a ring. A {\it module-valued functor} is a functor
$F : \textit{Alg}_A \to \textit{Ab}$ such that
\begin{enumerate}
\item for every object $B$ of $\textit{Alg}_A$ the group
$F(B)$ is endowed with the structure of a $B$-module, and
\item for any morphism $B \to B'$ of $\textit{Alg}_A$ the map
$F(B) \to F(B')$ is $B$-linear.
\end{enumerate}
A {\it morphism of module-valued functors} is a transformation of
functors $\varphi : F \to G$ such that $F(B) \to G(B)$ is $B$-linear
for all $B \in \Ob(\textit{Alg}_A)$.
\end{definition}
\noindent
Let $S = \Spec(A)$ be an affine scheme.
The category of module-valued functors on $\textit{Alg}_A$ is
equivalent to the category
$\textit{PMod}((\textit{Aff}/S)_\tau, \mathcal{O})$
of presheaves of $\mathcal{O}$-modules. The equivalence is given
by the rule which assigns to the module-valued functor $F$ the
presheaf $\mathcal{F}$ defined by the rule
$\mathcal{F}(U) = F(\mathcal{O}(U))$.
This is clear from the equivalence
$(\textit{Aff}/S)_\tau \to \textit{Alg}_A$, $U \mapsto \mathcal{O}(U)$
given in Section \ref{section-conventions}.
The quasi-inverse sets $F(B) = \mathcal{F}(\Spec(B))$.
\medskip\noindent
An important special case of a module-valued functor comes about as follows.
Let $M$ be an $A$-module. Then we will denote $\underline{M}$ the
module-valued functor $B \mapsto M \otimes_A B$ (with obvious $B$-module
structure). Note that if $M \to N$ is a map of $A$-modules then there is an
associated morphism $\underline{M} \to \underline{N}$ of module-valued
functors. Conversely, any morphism of module-valued functors
$\underline{M} \to \underline{N}$ comes from an $A$-module map $M \to N$
as the reader can see by evaluating on $B = A$. In other words
$\text{Mod}_A$ is a full
subcategory of the category of module-valued functors on $\textit{Alg}_A$.
\medskip\noindent
Given and $A$-module map $\varphi : M \to N$ then
$\text{Coker}(\underline{M} \to \underline{N}) =
\underline{Q}$ where $Q = \text{Coker}(M \to N)$ because $\otimes$
is right exact. But this isn't the case
for the kernel in general: for example an injective map of
$A$-modules need not be injective after base change. Thus the following
definition makes sense.
\begin{definition}
\label{definition-adequate-functor}
Let $A$ be a ring. A module-valued functor $F$ on $\textit{Alg}_A$ is
called
\begin{enumerate}
\item {\it adequate} if there exists a
map of $A$-modules $M \to N$ such that $F$ is isomorphic to
$\text{Ker}(\underline{M} \to \underline{N})$.
\item {\it linearly adequate} if $F$ is isomorphic to the
kernel of a map $\underline{A^{\oplus n}} \to \underline{A^{\oplus m}}$.
\end{enumerate}
\end{definition}
\noindent
Note that $F$ is adequate if and only if there exists an
exact sequence $0 \to F \to \underline{M} \to \underline{N}$ and
$F$ is linearly adequate if and only if there exists an exact sequence
$0 \to F \to \underline{A^{\oplus n}} \to \underline{A^{\oplus m}}$.
\medskip\noindent
Let $A$ be a ring. In this section we will show the category of adequate
functors on $\textit{Alg}_A$ is abelian
(Lemmas \ref{lemma-cokernel-adequate} and \ref{lemma-kernel-adequate})
and has a set of generators
(Lemma \ref{lemma-adequate-surjection-from-linear}).
We will also see that it is a weak Serre subcategory of the category
of all module-valued functors on $\textit{Alg}_A$
(Lemma \ref{lemma-extension-adequate})
and that it has arbitrary colimits
(Lemma \ref{lemma-colimit-adequate}).
\begin{lemma}
\label{lemma-adequate-finite-presentation}
Let $A$ be a ring.
Let $F$ be an adequate functor on $\textit{Alg}_A$.
If $B = \colim B_i$ is a filtered
colimit of $A$-algebras, then $F(B) = \colim F(B_i)$.
\end{lemma}
\begin{proof}
This holds because for any $A$-module $M$ we have
$M \otimes_A B = \colim M \otimes_A B_i$ (see
Algebra, Lemma \ref{algebra-lemma-tensor-products-commute-with-limits})
and because filtered colimits commute with exact sequences, see
Algebra, Lemma \ref{algebra-lemma-directed-colimit-exact}.
\end{proof}
\begin{remark}
\label{remark-settheoretic}
Consider the category $\textit{Alg}_{fp, A}$ whose objects are $A$-algebras
$B$ of the form $B = A[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$ and whose
morphisms are $A$-algebra maps. Every $A$-algebra $B$ is a filtered colimit
of finitely presented $A$-algebra, i.e., a filtered colimit of objects of
$\textit{Alg}_{fp, A}$. By
Lemma \ref{lemma-adequate-finite-presentation}
we conclude every adequate functor $F$ is determined by its restriction to
$\textit{Alg}_{fp, A}$. For some questions we can therefore restrict to
functors on $\textit{Alg}_{fp, A}$. For example, the category of adequate
functors does not depend on the choice of the big $\tau$-site
chosen in
Section \ref{section-conventions}.
\end{remark}
\begin{lemma}
\label{lemma-adequate-flat}
Let $A$ be a ring.
Let $F$ be an adequate functor on $\textit{Alg}_A$.
If $B \to B'$ is flat, then $F(B) \otimes_B B' \to F(B')$
is an isomorphism.
\end{lemma}
\begin{proof}
Choose an exact sequence $0 \to F \to \underline{M} \to \underline{N}$.
This gives the diagram
$$
\xymatrix{
0 \ar[r] & F(B) \otimes_B B' \ar[r] \ar[d] &
(M \otimes_A B)\otimes_B B' \ar[r] \ar[d] &
(N \otimes_A B)\otimes_B B' \ar[d] \\
0 \ar[r] & F(B') \ar[r] &
M \otimes_A B' \ar[r] &
N \otimes_A B'
}
$$
where the rows are exact (the top one because $B \to B'$ is flat).
Since the right two vertical arrows are isomorphisms, so is the
left one.
\end{proof}
\begin{lemma}
\label{lemma-adequate-surjection-from-linear}
Let $A$ be a ring.
Let $F$ be an adequate functor on $\textit{Alg}_A$. Then there exists a
surjection $L \to F$ with $L$ a direct sum of linearly adequate functors.
\end{lemma}
\begin{proof}
Choose an exact sequence $0 \to F \to \underline{M} \to \underline{N}$
where $\underline{M} \to \underline{N}$ is given by
$\varphi : M \to N$. By
Lemma \ref{lemma-adequate-finite-presentation}
it suffices to construct $L \to F$ such that $L(B) \to F(B)$ is surjective
for every finitely presented $A$-algebra $B$. Hence it suffices to construct,
given a finitely presented $A$-algebra $B$ and an element $\xi \in F(B)$
a map $L \to F$ with $L$ linearly adequate such that $\xi$ is in the image
of $L(B) \to F(B)$.
(Because there is a set worth of such pairs $(B, \xi)$ up to isomorphism.)
\medskip\noindent
To do this write $\sum_{i = 1, \ldots, n} m_i \otimes b_i$ the image of
$\xi$ in $\underline{M}(B) = M \otimes_A B$. We know that
$\sum \varphi(m_i) \otimes b_i = 0$ in $N \otimes_A B$.
As $N$ is a filtered colimit of finitely presented $A$-modules, we can
find a finitely presented $A$-module $N'$, a commutative diagram
of $A$-modules
$$
\xymatrix{
A^{\oplus n} \ar[r] \ar[d]_{m_1, \ldots, m_n} & N' \ar[d] \\
M \ar[r] & N
}
$$
such that $(b_1, \ldots, b_n)$ maps to zero in $N' \otimes_A B$.
Choose a presentation $A^{\oplus l} \to A^{\oplus k} \to N' \to 0$.
Choose a lift $A^{\oplus n} \to A^{\oplus k}$ of the map
$A^{\oplus n} \to N'$ of the diagram. Then we see that there exist
$(c_1, \ldots, c_l) \in B^{\oplus l}$ such that
$(b_1, \ldots, b_n, c_1, \ldots, c_l)$ maps to zero in $B^{\oplus k}$
under the map $B^{\oplus n} \oplus B^{\oplus l} \to B^{\oplus k}$.
Consider the commutative diagram
$$
\xymatrix{
A^{\oplus n} \oplus A^{\oplus l} \ar[r] \ar[d] & A^{\oplus k} \ar[d] \\
M \ar[r] & N
}
$$
where the left vertical arrow is zero on the summand $A^{\oplus l}$.
Then we see that $L$ equal to the kernel of $\underline{A^{\oplus n + l}}
\to \underline{A^{\oplus k}}$ works because the element
$(b_1, \ldots, b_n, c_1, \ldots, c_l) \in L(B)$ maps to $\xi$.
\end{proof}
\noindent
Consider a graded $A$-algebra $B = \bigoplus_{d \geq 0} B_d$. Then there are
two $A$-algebra maps $p, a : B \to B[t, t^{-1}]$, namely $p : b \mapsto b$ and
$a : b \mapsto t^{\deg(b)} b$ where $b$ is homogeneous. If $F$ is a
module-valued functor on $\textit{Alg}_A$, then we define
\begin{equation}
\label{equation-weight-k}
F(B)^{(k)} = \{\xi \in F(B) \mid t^k F(p)(\xi) = F(a)(\xi)\}.
\end{equation}
For functors which behave well with respect to flat ring extensions
this gives a direct sum decomposition. This amounts to the fact that
representations of $\mathbf{G}_m$ are completely reducible.
\begin{lemma}
\label{lemma-flat-functor-split}
Let $A$ be a ring.
Let $F$ be a module-valued functor on $\textit{Alg}_A$.
Assume that for $B \to B'$ flat the map
$F(B) \otimes_B B' \to F(B')$ is an isomorphism.
Let $B$ be a graded $A$-algebra. Then
\begin{enumerate}
\item $F(B) = \bigoplus_{k \in \mathbf{Z}} F(B)^{(k)}$, and
\item the map $B \to B_0 \to B$ induces map $F(B) \to F(B)$
whose image is contained in $F(B)^{(0)}$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $x \in F(B)$. The map $p : B \to B[t, t^{-1}]$ is free
hence we know that
$$
F(B[t, t^{-1}]) =
\bigoplus\nolimits_{k \in \mathbf{Z}} F(p)(F(B)) \cdot t^k =
\bigoplus\nolimits_{k \in \mathbf{Z}} F(B) \cdot t^k
$$
as indicated we drop the $F(p)$ in the rest of the proof.
Write $F(a)(x) = \sum t^k x_k$ for some $x_k \in F(B)$.
Denote $\epsilon : B[t, t^{-1}] \to B$
the $B$-algebra map $t \mapsto 1$. Note that the compositions
$\epsilon \circ p, \epsilon \circ a : B \to B[t, t^{-1}] \to B$ are
the identity. Hence we see that
$$
x = F(\epsilon)(F(a)(x)) = F(\epsilon)(\sum t^k x_k) = \sum x_k.
$$
On the other hand, we claim that $x_k \in F(B)^{(k)}$. Namely, consider
the commutative diagram
$$
\xymatrix{
B \ar[r]_a \ar[d]_{a'} &
B[t, t^{-1}] \ar[d]^f \\
B[s, s^{-1}] \ar[r]^-g &
B[t, s, t^{-1}, s^{-1}]
}
$$
where $a'(b) = s^{\deg(b)}b$, $f(b) = b$, $f(t) = st$ and
$g(b) = t^{\deg(b)}b$ and $g(s) = s$. Then
$$
F(g)(F(a'))(x) = F(g)(\sum s^k x_k) =
\sum s^k F(a)(x_k)
$$
and going the other way we see
$$
F(f)(F(a))(x) = F(f)(\sum t^k x_k) = \sum (st)^k x_k.
$$
Since $B \to B[s, t, s^{-1}, t^{-1}]$ is free we see that
$F(B[t, s, t^{-1}, s^{-1}]) =
\bigoplus_{k, l \in \mathbf{Z}} F(B) \cdot t^ks^l$ and
comparing coefficients in the expressions above we find
$F(a)(x_k) = t^k x_k$ as desired.
\medskip\noindent
Finally, the image of $F(B_0) \to F(B)$ is contained in $F(B)^{(0)}$
because $B_0 \to B \xrightarrow{a} B[t, t^{-1}]$ is equal to
$B_0 \to B \xrightarrow{p} B[t, t^{-1}]$.
\end{proof}
\noindent
As a particular case of
Lemma \ref{lemma-flat-functor-split}
note that
$$
\underline{M}(B)^{(k)} = M \otimes_A B_k
$$
where $B_k$ is the degree $k$ part of the graded $A$-algebra $B$.
\begin{lemma}
\label{lemma-lift-map}
Let $A$ be a ring. Given a solid diagram
$$
\xymatrix{
0 \ar[r] &
L \ar[d]_\varphi \ar[r] &
\underline{A^{\oplus n}} \ar[r] \ar@{..>}[ld] &
\underline{A^{\oplus m}} \\
& \underline{M}
}
$$
of module-valued functors on $\textit{Alg}_A$
with exact row there exists a dotted arrow making the diagram commute.
\end{lemma}
\begin{proof}
Suppose that the map $A^{\oplus n} \to A^{\oplus m}$ is given by the
$m \times n$-matrix $(a_{ij})$. Consider the ring
$B = A[x_1, \ldots, x_n]/(\sum a_{ij}x_j)$. The element
$(x_1, \ldots, x_n) \in \underline{A^{\oplus n}}(B)$ maps to zero in
$\underline{A^{\oplus m}}(B)$ hence is the image of a unique element
$\xi \in L(B)$. Note that $\xi$ has the following universal property:
for any $A$-algebra $C$ and any $\xi' \in L(C)$ there exists an $A$-algebra
map $B \to C$ such that $\xi$ maps to $\xi'$ via the map $L(B) \to L(C)$.
\medskip\noindent
Note that $B$ is a graded $A$-algebra, hence we can use
Lemmas \ref{lemma-flat-functor-split} and \ref{lemma-adequate-flat}
to decompose the values of our functors on $B$ into graded pieces.
Note that $\xi \in L(B)^{(1)}$ as $(x_1, \ldots, x_n)$ is an element
of degree one in $\underline{A^{\oplus n}}(B)$. Hence we see that
$\varphi(\xi) \in \underline{M}(B)^{(1)} = M \otimes_A B_1$.
Since $B_1$ is generated by $x_1, \ldots, x_n$ as an $A$-module we
can write $\varphi(\xi) = \sum m_i \otimes x_i$. Consider the map
$A^{\oplus n} \to M$ which maps the $i$th basis vector to $m_i$.
By construction the associated map
$\underline{A^{\oplus n}} \to \underline{M}$
maps the element $\xi$ to $\varphi(\xi)$. It follows from the
universal property mentioned above that the diagram commutes.
\end{proof}
\begin{lemma}
\label{lemma-cokernel-into-module}
Let $A$ be a ring.
Let $\varphi : F \to \underline{M}$ be a map of module-valued functors
on $\textit{Alg}_A$ with $F$ adequate.
Then $\text{Coker}(\varphi)$ is adequate.
\end{lemma}
\begin{proof}
By
Lemma \ref{lemma-adequate-surjection-from-linear}
we may assume that $F = \bigoplus L_i$ is a direct sum of linearly adequate
functors. Choose exact sequences
$0 \to L_i \to \underline{A^{\oplus n_i}} \to \underline{A^{\oplus m_i}}$.
For each $i$ choose a map $A^{\oplus n_i} \to M$ as in
Lemma \ref{lemma-lift-map}.
Consider the diagram
$$
\xymatrix{
0 \ar[r] &
\bigoplus L_i \ar[r] \ar[d] &
\bigoplus \underline{A^{\oplus n_i}} \ar[r] \ar[ld] &
\bigoplus \underline{A^{\oplus m_i}} \\
& \underline{M}
}
$$
Consider the $A$-modules
$$
Q =
\text{Coker}(\bigoplus A^{\oplus n_i} \to M \oplus \bigoplus A^{\oplus m_i})
\quad\text{and}\quad
P = \text{Coker}(\bigoplus A^{\oplus n_i} \to \bigoplus A^{\oplus m_i}).
$$
Then we see that $\text{Coker}(\varphi)$ is isomorphic to the
kernel of $\underline{Q} \to \underline{P}$.
\end{proof}
\begin{lemma}
\label{lemma-cokernel-adequate}
Let $A$ be a ring.
Let $\varphi : F \to G$ be a map of adequate functors on $\textit{Alg}_A$.
Then $\text{Coker}(\varphi)$ is adequate.
\end{lemma}
\begin{proof}
Choose an injection $G \to \underline{M}$.
Then we have an injection $G/F \to \underline{M}/F$. By
Lemma \ref{lemma-cokernel-into-module}
we see that $\underline{M}/F$ is adequate, hence we can find an injection
$\underline{M}/F \to \underline{N}$.
Composing we obtain an injection $G/F \to \underline{N}$. By
Lemma \ref{lemma-cokernel-into-module}
the cokernel of the induced map $G \to \underline{N}$ is adequate
hence we can find an injection $\underline{N}/G \to \underline{K}$.
Then $0 \to G/F \to \underline{N} \to \underline{K}$ is exact and
we win.
\end{proof}
\begin{lemma}
\label{lemma-kernel-adequate}
Let $A$ be a ring.
Let $\varphi : F \to G$ be a map of adequate functors on $\textit{Alg}_A$.
Then $\text{Ker}(\varphi)$ is adequate.
\end{lemma}
\begin{proof}
Choose an injection $F \to \underline{M}$ and an injection
$G \to \underline{N}$. Denote $F \to \underline{M \oplus N}$
the diagonal map so that
$$
\xymatrix{
F \ar[d] \ar[r] & G \ar[d] \\
\underline{M \oplus N} \ar[r] & \underline{N}
}
$$
commutes. By
Lemma \ref{lemma-cokernel-adequate}
we can find a module map $M \oplus N \to K$ such that
$F$ is the kernel of $\underline{M \oplus N} \to \underline{K}$.
Then $\text{Ker}(\varphi)$ is the kernel of
$\underline{M \oplus N} \to \underline{K \oplus N}$.
\end{proof}
\begin{lemma}
\label{lemma-colimit-adequate}
Let $A$ be a ring.
An arbitrary direct sum of adequate functors on $\textit{Alg}_A$
is adequate. A colimit of adequate functors is adequate.
\end{lemma}
\begin{proof}
The statement on direct sums is immediate.
A general colimit can be written as a kernel of a map between
direct sums, see
Categories, Lemma \ref{categories-lemma-colimits-coproducts-coequalizers}.
Hence this follows from
Lemma \ref{lemma-kernel-adequate}.
\end{proof}
\begin{lemma}
\label{lemma-flat-linear-functor}
Let $A$ be a ring.
Let $F, G$ be module-valued functors on $\textit{Alg}_A$.
Let $\varphi : F \to G$ be a transformation of functors. Assume
\begin{enumerate}
\item $\varphi$ is additive,
\item for every $A$-algebra $B$ and $\xi \in F(B)$ and unit
$u \in B^*$ we have $\varphi(u\xi) = u\varphi(\xi)$ in $G(B)$, and
\item for any flat ring map $B \to B'$ we have
$G(B) \otimes_B B' = G(B')$.
\end{enumerate}
Then $\varphi$ is a morphism of module-valued functors.
\end{lemma}
\begin{proof}
Let $B$ be an $A$-algebra, $\xi \in F(B)$, and $b \in B$. We have to show
that $\varphi(b \xi) = b \varphi(\xi)$. Consider the ring map
$$
B \to B' = B[x, y, x^{-1}, y^{-1}]/(x + y - b).
$$
This ring map is faithfully flat, hence $G(B) \subset G(B')$. On the
other hand
$$
\varphi(b\xi) = \varphi((x + y)\xi) =
\varphi(x\xi) + \varphi(y\xi) = x\varphi(\xi) + y\varphi(\xi)
= (x + y)\varphi(\xi) = b\varphi(\xi)
$$
because $x, y$ are units in $B'$. Hence we win.
\end{proof}
\begin{lemma}
\label{lemma-extension-adequate-key}
Let $A$ be a ring.
Let $0 \to \underline{M} \to G \to L \to 0$ be a short exact sequence
of module-valued functors on $\textit{Alg}_A$ with $L$ linearly adequate.
Then $G$ is adequate.
\end{lemma}
\begin{proof}
We first point out that for any flat $A$-algebra map
$B \to B'$ the map $G(B) \otimes_B B' \to G(B')$ is an isomorphism.
Namely, this holds for $\underline{M}$ and $L$, see
Lemma \ref{lemma-adequate-flat}
and hence follows for $G$ by the five lemma. In particular, by
Lemma \ref{lemma-flat-functor-split}
we see that $G(B) = \bigoplus_{k \in \mathbf{Z}} G(B)^{(k)}$
for any graded $A$-algebra $B$.
\medskip\noindent
Choose an exact sequence
$0 \to L \to \underline{A^{\oplus n}} \to \underline{A^{\oplus m}}$.
Suppose that the map $A^{\oplus n} \to A^{\oplus m}$ is given by the
$m \times n$-matrix $(a_{ij})$. Consider the graded $A$-algebra
$B = A[x_1, \ldots, x_n]/(\sum a_{ij}x_j)$. The element
$(x_1, \ldots, x_n) \in \underline{A^{\oplus n}}(B)$ maps to zero in
$\underline{A^{\oplus m}}(B)$ hence is the image of a unique element
$\xi \in L(B)$. Observe that $\xi \in L(B)^{(1)}$. The map
$$
\Hom_A(B, C) \longrightarrow L(C), \quad
f \longmapsto L(f)(\xi)
$$
defines an isomorphism of functors. The reason is that $f$ is
determined by the images $c_i = f(x_i) \in C$ which have to
satisfy the relations $\sum a_{ij}c_j = 0$. And $L(C)$ is the
set of $n$-tuples $(c_1, \ldots, c_n)$ satisfying the relations
$\sum a_{ij} c_j = 0$.
\medskip\noindent
Since the value of each of the functors $\underline{M}$, $G$, $L$
on $B$ is a direct sum of its weight spaces (by the lemma mentioned
above) exactness of $0 \to \underline{M} \to G \to L \to 0$ implies
the sequence $0 \to \underline{M}(B)^{(1)} \to G(B)^{(1)} \to L(B)^{(1)} \to 0$
is exact. Thus we may choose an element $\theta \in G(B)^{(1)}$ mapping
to $\xi$.
\medskip\noindent
Consider the graded $A$-algebra
$$
C = A[x_1, \ldots, x_n, y_1, \ldots, y_n]/
(\sum a_{ij}x_j, \sum a_{ij}y_j)
$$
There are three graded $A$-algebra homomorphisms $p_1, p_2, m : B \to C$
defined by the rules
$$
p_1(x_i) = x_i, \quad
p_1(x_i) = y_i, \quad
m(x_i) = x_i + y_i.
$$
We will show that the element
$$
\tau = G(m)(\theta) - G(p_1)(\theta) - G(p_2)(\theta) \in G(C)
$$
is zero. First, $\tau$ maps to zero in $L(C)$ by a direct calculation.
Hence $\tau$ is an element of $\underline{M}(C)$.
Moreover, since $m$, $p_1$, $p_2$ are graded algebra maps we see
that $\tau \in G(C)^{(1)}$ and since $\underline{M} \subset G$
we conclude
$$
\tau \in \underline{M}(C)^{(1)} = M \otimes_A C_1.
$$
We may write uniquely
$\tau = \underline{M}(p_1)(\tau_1) + \underline{M}(p_2)(\tau_2)$ with
$\tau_i \in M \otimes_A B_1 = \underline{M}(B)^{(1)}$ because
$C_1 = p_1(B_1) \oplus p_2(B_1)$.
Consider the ring map $q_1 : C \to B$ defined by $x_i \mapsto x_i$ and
$y_i \mapsto 0$. Then
$\underline{M}(q_1)(\tau) =
\underline{M}(q_1)(\underline{M}(p_1)(\tau_1) + \underline{M}(p_2)(\tau_2)) =
\tau_1$.
On the other hand, because
$q_1 \circ m = q_1 \circ p_1$ we see that
$G(q_1)(\tau) = - G(q_1 \circ p_2)(\tau)$. Since $q_1 \circ p_2$ factors as
$B \to A \to B$ we see that $G(q_1 \circ p_2)(\tau)$ is in
$G(B)^{(0)}$, see
Lemma \ref{lemma-flat-functor-split}.
Hence $\tau_1 = 0$ because it is in
$G(B)^{(0)} \cap \underline{M}(B)^{(1)} \subset
G(B)^{(0)} \cap G(B)^{(1)} = 0$.
Similarly $\tau_2 = 0$, whence $\tau = 0$.
\medskip\noindent
Since $\theta \in G(B)$ we obtain a transformation of functors
$$
\psi : L(-) = \Hom_A(B, - ) \longrightarrow G(-)
$$
by mapping $f : B \to C$ to $G(f)(\theta)$. Since $\theta$ is a lift of
$\xi$ the map $\psi$ is a right inverse of $G \to L$. In terms of
$\psi$ the statements proved above have the following meaning:
$\tau = 0$ means that $\psi$ is additive and
$\theta \in G(B)^{(1)}$ implies that for any $A$-algebra $D$ we have
$\psi(ul) = u\psi(l)$ in $G(D)$ for $l \in L(D)$ and $u \in D^*$ a unit.
This implies that $\psi$ is a morphism of module-valued functors, see
Lemma \ref{lemma-flat-linear-functor}.
Clearly this implies that $G \cong \underline{M} \oplus L$ and we win.
\end{proof}
\begin{remark}
\label{remark-linearly-adequate}
Let $A$ be a ring.
The proof of
Lemma \ref{lemma-extension-adequate-key}
shows that any extension $0 \to \underline{M} \to E \to L \to 0$
of module-valued functors on $\textit{Alg}_A$
with $L$ linearly adequate splits. It uses only the following properties
of the module-valued functor $F = \underline{M}$:
\begin{enumerate}
\item $F(B) \otimes_B B' \to F(B')$ is an isomorphism
for a flat ring map $B \to B'$, and
\item
$F(C)^{(1)} = F(p_1)(F(B)^{(1)}) \oplus F(p_2)(F(B)^{(1)})$
where $B = A[x_1, \ldots, x_n]/(\sum a_{ij}x_j)$ and
$C = A[x_1, \ldots, x_n, y_1, \ldots, y_n]/
(\sum a_{ij}x_j, \sum a_{ij}y_j)$.
\end{enumerate}
These two properties hold for any adequate functor $F$; details omitted.
Hence we see that $L$ is a projective object of the abelian category of
adequate functors.
\end{remark}
\begin{lemma}
\label{lemma-extension-adequate}
Let $A$ be a ring.
Let $0 \to F \to G \to H \to 0$ be a short exact sequence of
module-valued functors on $\textit{Alg}_A$.
If $F$ and $H$ are adequate, so is $G$.
\end{lemma}
\begin{proof}
Choose an exact sequence $0 \to F \to \underline{M} \to \underline{N}$.
If we can show that $(\underline{M} \oplus G)/F$ is adequate, then
$G$ is the kernel of the map of adequate functors
$(\underline{M} \oplus G)/F \to \underline{N}$, hence
adequate by
Lemma \ref{lemma-kernel-adequate}.
Thus we may assume $F = \underline{M}$.
\medskip\noindent
We can choose a surjection $L \to H$ where $L$ is a direct sum of
linearly adequate functors, see
Lemma \ref{lemma-adequate-surjection-from-linear}.
If we can show that the pullback $G \times_H L$ is adequate, then
$G$ is the cokernel of the map $\text{Ker}(L \to H) \to G \times_H L$
hence adequate by
Lemma \ref{lemma-cokernel-adequate}.
Thus we may assume that $H = \bigoplus L_i$ is a direct sum of
linearly adequate functors. By
Lemma \ref{lemma-extension-adequate-key}
each of the pullbacks $G \times_H L_i$ is adequate. By
Lemma \ref{lemma-colimit-adequate}
we see that $\bigoplus G \times_H L_i$ is adequate.
Then $G$ is the cokernel of
$$
\bigoplus\nolimits_{i \not = i'} F \longrightarrow
\bigoplus G \times_H L_i
$$
where $\xi$ in the summand $(i, i')$ maps to
$(0, \ldots, 0, \xi, 0, \ldots, 0, -\xi, 0, \ldots, 0)$
with nonzero entries in the summands $i$ and $i'$.
Thus $G$ is adequate by
Lemma \ref{lemma-cokernel-adequate}.
\end{proof}
\begin{lemma}
\label{lemma-base-change-adequate}
Let $A \to A'$ be a ring map. If $F$ is an adequate functor on
$\textit{Alg}_A$, then its restriction $F'$ to
$\textit{Alg}_{A'}$ is adequate too.
\end{lemma}
\begin{proof}
Choose an exact sequence $0 \to F \to \underline{M} \to \underline{N}$.
Then $F'(B') = F(B') = \text{Ker}(M \otimes_A B' \to N \otimes_A B')$.
Since $M \otimes_A B' = M \otimes_A A' \otimes_{A'} B'$ and similarly
for $N$ we see that $F'$ is the kernel of
$\underline{M \otimes_A A'} \to \underline{N \otimes_A A'}$.
\end{proof}
\begin{lemma}
\label{lemma-pushforward-adequate}
Let $A \to A'$ be a ring map. If $F'$ is an adequate functor on
$\textit{Alg}_{A'}$, then the module-valued functor
$F : B \mapsto F'(A' \otimes_A B)$ on $\textit{Alg}_A$ is adequate too.
\end{lemma}
\begin{proof}
Choose an exact sequence $0 \to F' \to \underline{M'} \to \underline{N'}$.
Then
\begin{align*}
F(B) & = F'(A' \otimes_A B) \\
& = \text{Ker}(M' \otimes_{A'} (
A' \otimes_A B) \to N' \otimes_{A'} (A' \otimes_A B)) \\
& = \text{Ker}(M' \otimes_A B \to N' \otimes_A B)
\end{align*}
Thus $F$ is the kernel of
$\underline{M} \to \underline{N}$
where $M = M'$ and $N = N'$ viewed as $A$-modules.
\end{proof}
\begin{lemma}
\label{lemma-adequate-product}
Let $A = A_1 \times \ldots \times A_n$ be a product of rings.
An adequate functor over $A$ is the same thing as a sequence
$F_1, \ldots, F_n$ of adequate functors $F_i$ over $A_i$.
\end{lemma}
\begin{proof}
This is true because an $A$-algebra $B$ is canonically a product
$B_1 \times \ldots \times B_n$ and the same thing holds for $A$-modules.
Setting $F(B) = \coprod F_i(B_i)$ gives the correspondence.
Details omitted.
\end{proof}
\begin{lemma}
\label{lemma-adequate-descent}
Let $A \to A'$ be a ring map and let $F$ be a module-valued functor on
$\textit{Alg}_A$ such that
\begin{enumerate}
\item the restriction $F'$ of $F$ to the category of $A'$-algebras is
adequate, and
\item for any $A$-algebra $B$ the sequence
$$
0 \to F(B) \to F(B \otimes_A A') \to F(B \otimes_A A' \otimes_A A')
$$
is exact.
\end{enumerate}
Then $F$ is adequate.
\end{lemma}
\begin{proof}
The functors $B \to F(B \otimes_A A')$ and
$B \mapsto F(B \otimes_A A' \otimes_A A')$ are adequate, see
Lemmas \ref{lemma-pushforward-adequate} and
\ref{lemma-base-change-adequate}.
Hence $F$ as a kernel of a map of adequate functors is adequate, see
Lemma \ref{lemma-kernel-adequate}.
\end{proof}
\section{Higher exts of adequate functors}
\label{section-higher-ext}
\noindent
Let $A$ be a ring. In
Lemma \ref{lemma-extension-adequate}
we have seen that any extension of adequate functors in the category
of module-valued functors on $\textit{Alg}_A$ is adequate. In this
section we show that the same remains true for higher ext groups.
\begin{lemma}
\label{lemma-adjoint}
Let $A$ be a ring.
For every module-valued functor $F$ on $\textit{Alg}_A$
there exists a morphism $Q(F) \to F$ of module-valued functors on
$\textit{Alg}_A$ such that (1) $Q(F)$ is adequate and (2) for every
adequate functor $G$ the map $\Hom(G, Q(F)) \to \Hom(G, F)$
is a bijection.
\end{lemma}
\begin{proof}
Choose a set $\{L_i\}_{i \in I}$ of linearly adequate functors such that
every linearly adequate functor is isomorphic to one of the $L_i$.
This is possible. Suppose that we can find $Q(F) \to F$ with (1) and
(2)' or every $i \in I$ the map $\Hom(L_i, Q(F)) \to \Hom(L_i, F)$
is a bijection. Then (2) holds. Namely, combining
Lemmas \ref{lemma-adequate-surjection-from-linear} and
\ref{lemma-kernel-adequate}
we see that every adequate functor $G$ sits in an exact sequence
$$
K \to L \to G \to 0
$$
with $K$ and $L$ direct sums of linearly adequate functors. Hence (2)'
implies that
$\Hom(L, Q(F)) \to \Hom(L, F)$
and
$\Hom(K, Q(F)) \to \Hom(K, F)$
are bijections, whence the same thing for $G$.
\medskip\noindent
Consider the category $\mathcal{I}$ whose objects are pairs
$(i, \varphi)$ where $i \in I$ and $\varphi : L_i \to F$ is a morphism.
A morphism $(i, \varphi) \to (i', \varphi')$ is a map
$\psi : L_i \to L_{i'}$ such that $\varphi' \circ \psi = \varphi$.
Set
$$
Q(F) = \colim_{(i, \varphi) \in \Ob(\mathcal{I})} L_i
$$
There is a natural map $Q(F) \to F$, by
Lemma \ref{lemma-colimit-adequate}
it is adequate, and by construction it has property (2)'.
\end{proof}
\begin{lemma}
\label{lemma-enough-injectives}
Let $A$ be a ring. Denote $\mathcal{P}$ the category of module-valued
functors on $\textit{Alg}_A$ and $\mathcal{A}$ the category of adequate
functors on $\textit{Alg}_A$. Denote $i : \mathcal{A} \to \mathcal{P}$
the inclusion functor. Denote $Q : \mathcal{P} \to \mathcal{A}$
the construction of Lemma \ref{lemma-adjoint}.
Then
\begin{enumerate}
\item $i$ is fully faithful, exact, and its image is a weak Serre subcategory,
\item $\mathcal{P}$ has enough injectives,
\item the functor $Q$ is a right adjoint to $i$ hence left exact,
\item $Q$ transforms injectives into injectives,
\item $\mathcal{A}$ has enough injectives.
\end{enumerate}
\end{lemma}
\begin{proof}
This lemma just collects some facts we have already seen so far.
Part (1) is clear from the definitions, the characterization of
weak Serre subcategories (see
Homology, Lemma \ref{homology-lemma-characterize-weak-serre-subcategory}),
and
Lemmas \ref{lemma-cokernel-adequate}, \ref{lemma-kernel-adequate},
and \ref{lemma-extension-adequate}.
Recall that $\mathcal{P}$ is equivalent to the category
$\textit{PMod}((\textit{Aff}/\Spec(A))_\tau, \mathcal{O})$.
Hence (2) by
Injectives, Proposition \ref{injectives-proposition-presheaves-modules}.
Part (3) follows from
Lemma \ref{lemma-adjoint}
and
Categories, Lemma \ref{categories-lemma-adjoint-exact}.
Parts (4) and (5) follow from
Homology, Lemmas \ref{homology-lemma-adjoint-preserve-injectives} and
\ref{homology-lemma-adjoint-enough-injectives}.
\end{proof}
\noindent
Let $A$ be a ring. As in
Formal Deformation Theory, Section
\ref{formal-defos-section-tangent-spaces-functors}
given an $A$-algebra $B$ and an $B$-module $N$ we set $B[N]$ equal to
the $R$-algebra with underlying $B$-module $B \oplus N$ with multiplication
given by $(b, m)(b', m ') = (bb', bm' + b'm)$. Note that this construction
is functorial in the pair $(B, N)$ where morphism $(B, N) \to (B', N')$
is given by an $A$-algebra map $B \to B'$ and an $B$-module map
$N \to N'$. In some sense the functor $TF$ of pairs defined in the following
lemma is the tangent space of $F$.
Below we will only consider pairs $(B, N)$ such that
$B[N]$ is an object of $\textit{Alg}_A$.
\begin{lemma}
\label{lemma-tangent-functor}
Let $A$ be a ring. Let $F$ be a module valued functor.
For every $B \in \Ob(\textit{Alg}_A)$ and $B$-module $N$
there is a canonical decomposition
$$
F(B[N]) = F(B) \oplus TF(B, N)
$$
characterized by the following properties
\begin{enumerate}
\item $TF(B, N) = \text{Ker}(F(B[N]) \to F(B))$,
\item there is a $B$-module structure $TF(B, N)$
compatible with $B[N]$-module structure on $F(B[N])$,
\item $TF$ is a functor from the category of pairs $(B, N)$,
\item
\label{item-mult-map}
there are canonical maps $N \otimes_B F(B) \to TF(B, N)$
inducing a transformation between functors defined on the category
of pairs $(B, N)$,
\item $TF(B, 0) = 0$ and the map $TF(B, N) \to TF(B, N')$ is
zero when $N \to N'$ is the zero map.
\end{enumerate}
\end{lemma}
\begin{proof}
Since $B \to B[N] \to B$ is the identity we see that $F(B) \to F(B[N])$
is a direct summand whose complement is $TF(N, B)$ as defined in (1).
This construction is functorial in the pair $(B, N)$ simply because
given a morphism of pairs $(B, N) \to (B', N')$ we obtain a commutative
diagram
$$
\xymatrix{
B' \ar[r] & B'[N'] \ar[r] & B' \\
B \ar[r] \ar[u] & B[N] \ar[r] \ar[u] & B \ar[u]
}
$$
in $\textit{Alg}_A$. The $B$-module structure comes from the $B[N]$-module
structure and the ring map $B \to B[N]$. The map in (4) is the
composition
$$
N \otimes_B F(B) \longrightarrow
B[N] \otimes_{B[N]} F(B[N]) \longrightarrow F(B[N])
$$
whose image is contained in $TF(B, N)$. (The first arrow uses the inclusions
$N \to B[N]$ and $F(B) \to F(B[N])$ and the second arrow is the multiplication
map.) If $N = 0$, then $B = B[N]$
hence $TF(B, 0) = 0$. If $N \to N'$ is zero then it factors as
$N \to 0 \to N'$ hence the induced map is zero since $TF(B, 0) = 0$.
\end{proof}
\noindent
Let $A$ be a ring. Let $M$ be an $A$-module. Then the module-valued functor
$\underline{M}$ has tangent space $T\underline{M}$ given by the rule
$T\underline{M}(B, N) = N \otimes_A M$. In particular, for $B$ given, the
functor $N \mapsto T\underline{M}(B, N)$ is additive and right exact. It turns
out this also holds for injective module-valued functors.
\begin{lemma}
\label{lemma-tangent-injective}
Let $A$ be a ring. Let $I$ be an injective object of the category
of module-valued functors. Then for any $B \in \Ob(\textit{Alg}_A)$
and short exact sequence
$0 \to N_1 \to N \to N_2 \to 0$
of $B$-modules the sequence
$$
TI(B, N_1) \to TI(B, N) \to TI(B, N_2) \to 0
$$
is exact.
\end{lemma}
\begin{proof}
We will use the results of
Lemma \ref{lemma-tangent-functor}
without further mention.
Denote $h : \textit{Alg}_A \to \textit{Sets}$ the functor given by
$h(C) = \Mor_A(B[N], C)$. Similarly for $h_1$ and $h_2$.
The map $B[N] \to B[N_2]$ corresponding to the surjection $N \to N_2$
is surjective. It corresponds to a map $h_2 \to h$ such that
$h_2(C) \to h(C)$ is injective for all $A$-algebras $C$. On the other
hand, there are two maps $p, q : h \to h_1$, corresponding to the
zero map $N_1 \to N$ and the injection $N_1 \to N$. Note that
$$
\xymatrix{
h_2 \ar[r] & h \ar@<1ex>[r] \ar@<-1ex>[r] & h_1
}
$$
is an equalizer diagram. Denote $\mathcal{O}_h$ the module-valued functor
$C \mapsto \bigoplus_{h(C)} C$. Similarly for $\mathcal{O}_{h_1}$ and
$\mathcal{O}_{h_2}$. Note that
$$
\Hom_\mathcal{P}(\mathcal{O}_h, F) = F(B[N])
$$
where $\mathcal{P}$ is the category of of module-valued functors on
$\textit{Alg}_A$. We claim there is an equalizer diagram
$$
\xymatrix{