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functors.tex
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\input{preamble}
% OK, start here.
%
\begin{document}
\title{Functors and Morphisms}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
Let $X$ and $Y$ be schemes. This chapter circles around the relationship
between functors $\QCoh(\mathcal{O}_Y) \to \QCoh(\mathcal{O}_X)$ and
morphisms of schemes $X \to Y$. More broadly speaking we study the
relationship between $\QCoh(\mathcal{O}_X)$ and $X$ or, if $X$ is Noetherian,
the relationship between $\textit{Coh}(\mathcal{O}_X)$ and $X$.
This relationship was studied in \cite{Gabriel}.
\section{Functors on module categories}
\label{section-preliminary}
\noindent
For a ring $A$ let us denote $\text{Mod}^{fp}_A$ the category of
finitely presented $A$-modules.
\begin{lemma}
\label{lemma-functor-on-fp-modules}
Let $A$ be a ring. Let $\mathcal{B}$ be a category having filtered
colimits. Let $F : \text{Mod}^{fp}_A \to \mathcal{B}$ be a functor. Then $F$
extends uniquely to a functor $F' : \text{Mod}_A \to \mathcal{B}$
which commutes with filtered colimits.
\end{lemma}
\begin{proof}
This follows from
Categories, Lemma \ref{categories-lemma-extend-functor-by-colim}.
To see that the lemma applies observe that
finitely presented $A$-modules are
categorically compact objects of $\text{Mod}_A$ by
Algebra, Lemma \ref{algebra-lemma-characterize-finitely-presented-module-hom}.
Also, every $A$-module is a filtered colimit
of finitely presented $A$-modules by
Algebra, Lemma \ref{algebra-lemma-module-colimit-fp}.
\end{proof}
\noindent
If a category $\mathcal{B}$ is additive and has filtered colimits,
then $\mathcal{B}$ has arbitrary direct sums: any direct sum can be written
as a filtered colimit of finite direct sums.
\begin{lemma}
\label{lemma-functor-on-fp-modules-additive}
Let $A$, $\mathcal{B}$, $F$ be as in Lemma \ref{lemma-functor-on-fp-modules}.
Assume $\mathcal{B}$ is additive and $F$ is additive. Then
$F'$ is additive and commutes with arbitrary direct sums.
\end{lemma}
\begin{proof}
To show that $F'$ is additive it suffices to show
that $F'(M) \oplus F'(M') \to F'(M \oplus M')$ is an isomorphism for
any $A$-modules $M$, $M'$, see
Homology, Lemma \ref{homology-lemma-additive-functor}.
Write $M = \colim_i M_i$ and $M' = \colim_j M'_j$ as filtered colimits
of finitely presented $A$-modules $M_i$. Then
$F'(M) = \colim_i F(M_i)$, $F'(M') = \colim_j F(M'_j)$, and
\begin{align*}
F'(M \oplus M')
& =
F'(\colim_{i, j} M_i \oplus M'_j) \\
& =
\colim_{i, j} F(M_i \oplus M'_j) \\
& =
\colim_{i, j} F(M_i) \oplus F(M'_j) \\
& =
F'(M) \oplus F'(M')
\end{align*}
as desired. To show that $F'$ commutes with direct sums, assume
we have $M = \bigoplus_{i \in I} M_i$. Then
$M = \colim_{I' \subset I\text{ finite}} \bigoplus_{i \in I'} M_i$
is a filtered colimit. We obtain
\begin{align*}
F'(M)
& =
\colim_{I' \subset I\text{ finite}}
F'(\bigoplus\nolimits_{i \in I'} M_i) \\
& =
\colim_{I' \subset I\text{ finite}}
\bigoplus\nolimits_{i \in I'} F'(M_i) \\
& =
\bigoplus\nolimits_{i \in I} F'(M_i)
\end{align*}
The second equality holds by the additivity of $F'$ already shown.
\end{proof}
\noindent
If a category $\mathcal{B}$ is additive, has filtered colimits, and
has cokernels, then $\mathcal{B}$ has arbitrary colimits, see
discussion above and Categories, Lemma
\ref{categories-lemma-colimits-coproducts-coequalizers}.
\begin{lemma}
\label{lemma-functor-on-fp-modules-right-exact}
Let $A$, $\mathcal{B}$, $F$ be as in Lemma \ref{lemma-functor-on-fp-modules}.
Assume $\mathcal{B}$ is additive, has cokernels, and $F$ is right exact. Then
$F'$ is additive, right exact, and commutes with arbitrary direct sums.
\end{lemma}
\begin{proof}
Since $F$ is right exact, $F$ commutes with coproducts of pairs, which are
represented by direct sums. Hence $F$ is additive by
Homology, Lemma \ref{homology-lemma-additive-functor}.
Hence $F'$ is additive and commutes with direct sums by
Lemma \ref{lemma-functor-on-fp-modules-additive}.
We urge the reader to prove that $F'$ is right exact themselves
instead of reading the proof below.
\medskip\noindent
To show that $F'$ is right exact, it suffices to show that $F'$ commutes
with coequalizers, see
Categories, Lemma \ref{categories-lemma-characterize-right-exact}.
Now, if $a, b : K \to L$ are maps of $A$-modules, then the
coequalizer of $a$ and $b$ is the cokernel of $a - b : K \to L$.
Thus let $K \to L \to M \to 0$ be an exact sequence
of $A$-modules. We have to show that in
$$
F'(K) \to F'(L) \to F'(M) \to 0
$$
the second arrow is a cokernel for the first arrow in $\mathcal{B}$
(if $\mathcal{B}$ were abelian we would say that the displayed sequence
is exact).
Write $M = \colim_{i \in I} M_i$ as a filtered colimit of finitely
presented $A$-modules, see
Algebra, Lemma \ref{algebra-lemma-module-colimit-fp}.
Let $L_i = L \times_M M_i$.
We obtain a system of exact sequences $K \to L_i \to M_i \to 0$ over $I$.
Since colimits commute with colimits by
Categories, Lemma \ref{categories-lemma-colimits-commute}
and since cokernels are a type of coequalizer,
it suffices to show that $F'(L_i) \to F(M_i)$ is a cokernel of
$F'(K) \to F'(L_i)$ in $\mathcal{B}$ for all $i \in I$. In other words, we may
assume $M$ is finitely presented. Write $L = \colim_{i \in I} L_i$
as a filtered colimit of finitely presented $A$-modules
with the property that each $L_i$ surjects onto $M$.
Let $K_i = K \times_L L_i$. We obtain a system of short exact sequences
$K_i \to L_i \to M \to 0$ over $I$.
Repeating the argument already given, we reduce to showing
$F(L_i) \to F(M_i)$ is a cokernel of
$F'(K) \to F(L_i)$ in $\mathcal{B}$ for all $i \in I$.
In other words, we may assume both
$L$ and $M$ are finitely presented $A$-modules.
In this case the module $\Ker(L \to M)$ is finite
(Algebra, Lemma \ref{algebra-lemma-extension}).
Thus we can write $K = \colim_{i \in I} K_i$ as a filtered colimit
of finitely presented $A$-modules each surjecting onto $\Ker(L \to M)$.
We obtain a system of short exact sequences
$K_i \to L \to M \to 0$ over $I$.
Repeating the argument already given, we reduce to showing
$F(L) \to F(M)$ is a cokernel of
$F(K_i) \to F(L)$ in $\mathcal{B}$ for all $i \in I$.
In other words, we may assume $K$, $L$, and $M$
are finitely presented $A$-modules. This final case follows
from the assumption that $F$ is right exact.
\end{proof}
\noindent
If a category $\mathcal{B}$ is additive and has kernels,
then $\mathcal{B}$ has finite limits. Namely, finite products
are direct sums which exist and the equalizer of $a, b : L \to M$
is the kernel of $a - b : K \to L$ which exists. Thus all finite
limits exist by Categories, Lemma \ref{categories-lemma-finite-limits-exist}.
\begin{lemma}
\label{lemma-functor-on-fp-modules-left-exact}
Let $A$, $\mathcal{B}$, $F$ be as in Lemma \ref{lemma-functor-on-fp-modules}.
Assume $A$ is a coherent ring (Algebra, Definition
\ref{algebra-definition-coherent}), $\mathcal{B}$ is additive, has kernels,
filtered colimits commute with taking kernels, and $F$ is left exact. Then
$F'$ is additive, left exact, and commutes with arbitrary direct sums.
\end{lemma}
\begin{proof}
Since $A$ is coherent, the category $\text{Mod}^{fp}_A$ is abelian
with same kernels and cokernels as in $\text{Mod}_A$, see
Algebra, Lemmas \ref{algebra-lemma-coherent-ring} and
\ref{algebra-lemma-coherent}. Hence all finite limits exist in
$\text{Mod}^{fp}_A$ and
Categories, Definition \ref{categories-definition-exact} applies.
Since $F$ is left exact, $F$ commutes with products of pairs, which are
represented by direct sums. Hence $F$ is additive by
Homology, Lemma \ref{homology-lemma-additive-functor}.
Hence $F'$ is additive and commutes with direct sums by
Lemma \ref{lemma-functor-on-fp-modules-additive}.
We urge the reader to prove that $F'$ is left exact themselves
instead of reading the proof below.
\medskip\noindent
To show that $F'$ is left exact, it suffices to show that $F'$ commutes
with equalizers, see
Categories, Lemma \ref{categories-lemma-characterize-left-exact}.
Now, if $a, b : L \to M$ are maps of $A$-modules, then the
equalizer of $a$ and $b$ is the kernel of $a - b : L \to M$.
Thus let $0 \to K \to L \to M$ be an exact sequence
of $A$-modules. We have to show that in
$$
0 \to F'(K) \to F'(L) \to F'(M)
$$
the arrow $F'(K) \to F'(L)$ is a kernel for $F'(L) \to F'(M)$ in $\mathcal{B}$
(if $\mathcal{B}$ were abelian we would say that the displayed sequence
is exact).
Write $M = \colim_{i \in I} M_i$ as a filtered colimit of finitely
presented $A$-modules, see
Algebra, Lemma \ref{algebra-lemma-module-colimit-fp}.
Let $L_i = L \times_M M_i$.
We obtain a system of exact sequences $0 \to K \to L_i \to M_i$
over $I$. Since filtered colimits commute with taking kernels
in $\mathcal{B}$ by assumption,
it suffices to show that $F'(K) \to F'(L_i)$ is a kernel of
$F'(L_i) \to F(M_i)$ in $\mathcal{B}$ for all $i \in I$. In other words, we may
assume $M$ is finitely presented. Write $L = \colim_{i \in I} L_i$
as a filtered colimit of finitely presented $A$-modules.
Let $K_i = K \times_L L_i$. We obtain a system of short exact sequences
$0 \to K_i \to L_i \to M$ over $I$.
Repeating the argument already given, we reduce to showing
$F'(K_i) \to F(L_i)$ is a kernel of
$F(L_i) \to F(M)$ in $\mathcal{B}$ for all $i \in I$.
In other words, we may assume both
$L$ and $M$ are finitely presented $A$-modules.
Since $A$ is coherent, the $A$-module $K = \Ker(L \to M)$ is
of finite presentation as the category of finitely presented
$A$-modules is abelian (see references given above).
In other words, all three modules $K$, $L$, and $M$
are finitely presented $A$-modules. This final case follows
from the assumption that $F$ is left exact.
\end{proof}
\noindent
If a category $\mathcal{B}$ is additive and has cokernels,
then $\mathcal{B}$ has finite colimits. Namely, finite coproducts
are direct sums which exist and the coequalizer of $a, b : K \to L$
is the cokernel of $a - b : K \to L$ which exists. Thus all finite
colimits exist by Categories, Lemma \ref{categories-lemma-colimits-exist}.
\begin{lemma}
\label{lemma-functor-on-modules-fp}
Let $A$ be a ring. Let $\mathcal{B}$ be an additive category
with cokernels. There is an equivalence of categories between
\begin{enumerate}
\item the category of functors $F : \text{Mod}^{fp}_A \to \mathcal{B}$
which are right exact, and
\item the category of pairs $(K, \kappa)$ where $K \in \Ob(\mathcal{B})$
and $\kappa : A \to \text{End}_\mathcal{B}(K)$ is a ring homomorphism
\end{enumerate}
given by the rule sending $F$ to $F(A)$ with its natural $A$-action.
\end{lemma}
\begin{proof}
Let $(K, \kappa)$ be as in (2). We will construct a functor
$F : \text{Mod}^{fp}_A \to \mathcal{B}$ such that $F(A) = K$
endowed with the given $A$-action $\kappa$. Namely, given an
integer $n \geq 0$ let us set
$$
F(A^{\oplus n}) = K^{\oplus n}
$$
Given an $A$-linear map $\varphi : A^{\oplus m} \to A^{\oplus n}$
with matrix $(a_{ij}) \in \text{Mat}(n \times m, A)$ we define
$$
F(\varphi) :
F(A^{\oplus m}) = K^{\oplus m}
\longrightarrow
K^{\oplus n} = F(A^{\oplus n})
$$
to be the map with matrix $(\kappa(a_{ij}))$. This defines an additive
functor $F$ from the full subcategory of
$\text{Mod}^{fp}_A$ with objects $0$, $A$, $A^{\oplus 2}$, $\ldots$
to $\mathcal{B}$; we omit the verification.
\medskip\noindent
For each object $M$ of $\text{Mod}^{fp}_A$ choose a presentation
$$
A^{\oplus m_M} \xrightarrow{\varphi_M} A^{\oplus n_M} \to M \to 0
$$
of $M$ as an $A$-module. Let us use the trivial presentation
$0 \to A^{\oplus n} \xrightarrow{1} A^{\oplus n} \to 0$ if $M = A^{\oplus n}$
(this isn't necessary but simplifies the exposition).
For each morphism $f : M \to N$ of
$\text{Mod}^{fp}_A$ we can choose a commutative diagram
\begin{equation}
\label{equation-map}
\vcenter{
\xymatrix{
A^{\oplus m_M} \ar[r]_{\varphi_M} \ar[d]_{\psi_f} &
A^{\oplus n_M} \ar[r] \ar[d]_{\chi_f} &
M \ar[r] \ar[d]_f & 0 \\
A^{\oplus m_N} \ar[r]^{\varphi_N} &
A^{\oplus n_N} \ar[r] &
N \ar[r] & 0
}
}
\end{equation}
Having made these choices we can define: for an object
$M$ of $\text{Mod}^{fp}_A$ we set
$$
F(M) = \Coker(F(\varphi_M) : F(A^{\oplus m_M}) \to F(A^{\oplus n_M}))
$$
and for a morphism $f : M \to N$ of $\text{Mod}^{fp}_A$ we set
$$
F(f) = \text{the map }F(M) \to F(N)\text{ induced by }
F(\psi_f)\text{ and }F(\chi_f)\text{ on cokernels}
$$
Note that this rule extends the given functor $F$ on
the full subcategory consisting of the free modules $A^{\oplus n}$.
We still have to show that $F$ is a functor, that $F$ is additive,
and that $F$ is right exact.
\medskip\noindent
Let $f : M \to N$ be a morphism $\text{Mod}^{fp}_A$. We claim that the map
$F(f)$ defined above is independent of the choices of $\psi_f$ and $\chi_f$
in (\ref{equation-map}). Namely, say
$$
\xymatrix{
A^{\oplus m_M} \ar[r]_{\varphi_M} \ar[d]_\psi &
A^{\oplus n_M} \ar[r] \ar[d]_\chi &
M \ar[r] \ar[d]_f & 0 \\
A^{\oplus m_N} \ar[r]^{\varphi_N} &
A^{\oplus n_N} \ar[r] &
N \ar[r] & 0
}
$$
is also commutative. Denote $F(f)' : F(M) \to F(N)$ the map
induced by $F(\psi)$ and $F(\chi)$. Looking at the commutative
diagrams, by elementary commutative algebra there exists a map
$\omega : A^{\oplus n_M} \to A^{\oplus m_N}$ such that
$\chi = \chi_f + \varphi_N \circ \omega$. Applying $F$ we
find that $F(\chi) = F(\chi_f) + F(\varphi_N) \circ F(\omega)$.
As $F(N)$ is the cokernel of $F(\varphi_N)$ we find
that the map $F(A^{\oplus n_M}) \to F(M)$ equalizes $F(f)$ and $F(f)'$.
Since a cokernel is an epimorphism, we conclude that $F(f) = F(f)'$.
\medskip\noindent
Let us prove $F$ is a functor. First, observe that
$F(\text{id}_M) = \text{id}_{F(M)}$ because we may pick
the identities for $\psi_f$ and $\chi_f$ in the diagram above
in case $f = \text{id}_M$. Second, suppose we have
$f : M \to N$ and $g : L \to M$. Then we see that
$\psi = \psi_f \circ \psi_g$ and $\chi = \chi_f \circ \chi_g$
fit into (\ref{equation-map}) for $f \circ g$.
Hence these induce the correct map which exactly
says that $F(f) \circ F(g) = F(f \circ g)$.
\medskip\noindent
Let us prove that $F$ is additive. Namely, suppose we have
$f, g : M \to N$. Then we see that $\psi = \psi_f + \psi_g$ and
$\chi = \chi_f + \chi_g$ fit into (\ref{equation-map}) for $f + g$.
Hence these induce the correct map which exactly
says that $F(f) + F(g) = F(f + g)$.
\medskip\noindent
Finally, let us prove that $F$ is right exact. It suffices to show that $F$
commutes with coequalizers, see
Categories, Lemma \ref{categories-lemma-characterize-right-exact}.
For this, it suffices to prove that $F$ commutes with cokernels.
Let $K \to L \to M \to 0$ be an exact sequence of $A$-modules
with $K$, $L$, $M$ finitely presented. Since $F$ is an additive
functor, this certainly gives a complex
$$
F(K) \to F(L) \to F(M) \to 0
$$
and we have to show that the second arrow is the cokernel of the first
in $\mathcal{B}$. In any case, we obtain a map
$\Coker(F(K) \to F(L)) \to F(M)$.
By elementary commutative algebra there exists a commutative diagram
$$
\xymatrix{
A^{\oplus m_M} \ar[r]_{\varphi_M} \ar[d]_\psi &
A^{\oplus n_M} \ar[r] \ar[d]_\chi &
M \ar[r] \ar[d]_1 & 0 \\
K \ar[r] &
L \ar[r] &
M \ar[r] & 0
}
$$
Applying $F$ to this diagram and using the construction of $F(M)$ as
the cokernel of $F(\varphi_M)$ we find there exists a map
$F(M) \to \Coker(F(K) \to F(L))$ which is a right inverse
to the map $\Coker(F(K) \to F(L)) \to F(M)$. This first implies
that $F(L) \to F(M)$ is an epimorphism always. Next, the above shows
we have
$$
\Coker(F(K) \to F(L)) = F(M) \oplus E
$$
where the direct sum decomposition is compatible with both
$F(M) \to \Coker(F(K) \to F(L))$ and $\Coker(F(K) \to F(L)) \to F(M)$.
However, then the epimorphism $p : F(L) \to E$ becomes zero both
after composition with $F(K) \to F(L)$ and after composition
with $F(A^{n_M}) \to F(L)$. However, since $K \oplus A^{n_M} \to L$
is surjective (algebra argument omitted), we conclude that
$F(K \oplus A^{n_M}) \to F(L)$ is an epimorphism (by the above)
whence $E = 0$. This finishes the proof.
\end{proof}
\begin{lemma}
\label{lemma-functor-on-modules}
Let $A$ be a ring. Let $\mathcal{B}$ be an additive category
with arbitrary direct sums and cokernels. There is an equivalence
of categories between
\begin{enumerate}
\item the category of functors $F : \text{Mod}_A \to \mathcal{B}$
which are right exact and commute with arbitrary direct sums, and
\item the category of pairs $(K, \kappa)$ where $K \in \Ob(\mathcal{B})$
and $\kappa : A \to \text{End}_\mathcal{B}(K)$ is a ring homomorphism
\end{enumerate}
given by the rule sending $F$ to $F(A)$ with its natural $A$-action.
\end{lemma}
\begin{proof}
Combine Lemmas \ref{lemma-functor-on-modules-fp} and
\ref{lemma-functor-on-fp-modules-right-exact}.
\end{proof}
\section{Functors between categories of modules}
\label{section-functors}
\noindent
The following lemma is archetypical of the results in this chapter.
\begin{lemma}
\label{lemma-functor}
Let $A$ and $B$ be rings. Let $F : \text{Mod}_A \to \text{Mod}_B$
be a functor. The following are equivalent
\begin{enumerate}
\item $F$ is isomorphic to the functor $M \mapsto M \otimes_A K$
for some $A \otimes_\mathbf{Z} B$-module $K$,
\item $F$ is right exact and commutes with all direct sums,
\item $F$ commutes with all colimits,
\item $F$ has a right adjoint $G$.
\end{enumerate}
\end{lemma}
\begin{proof}
If (1), then (4) as a right adjoint for $M \mapsto M \otimes_A K$
is $N \mapsto \Hom_B(K, N)$, see
Differential Graded Algebra, Lemma \ref{dga-lemma-tensor-hom-adjunction}.
If (4), then (3) by Categories, Lemma \ref{categories-lemma-adjoint-exact}.
The implication (3) $\Rightarrow$ (2) is immediate from the definitions.
\medskip\noindent
Assume (2). We will prove (1). By the discussion in
Homology, Section \ref{homology-section-functors}
the functor $F$ is additive. Hence $F$ induces
a ring map $A \to \text{End}_B(F(M))$, $a \mapsto F(a \cdot \text{id}_M)$
for every $A$-module $M$. We conclude that $F(M)$ is an
$A \otimes_\mathbf{Z} B$-module functorially in $M$.
Set $K = F(A)$. Define
$$
M \otimes_A K = M \otimes_A F(A) \longrightarrow F(M),
\quad m \otimes k \longmapsto F(\varphi_m)(k)
$$
Here $\varphi_m : A \to M$ sends $a \to am$. The rule
$(m, k) \mapsto F(\varphi_m)(k)$ is $A$-bilinear (and $B$-linear
on the right) as required to obtain the displayed
$A \otimes_\mathbf{Z} B$-linear map.
This construction is functorial in $M$, hence defines a transformation
of functors $- \otimes_A K \to F(-)$ which is an isomorphism when
evaluated on $A$. For every $A$-module $M$ we can choose an exact sequence
$$
\bigoplus\nolimits_{j \in J} A \to
\bigoplus\nolimits_{i \in I} A \to
M \to 0
$$
Using the maps constructed above we find a commutative diagram
$$
\xymatrix{
(\bigoplus\nolimits_{j \in J} A) \otimes_A K \ar[r] \ar[d] &
(\bigoplus\nolimits_{i \in I} A) \otimes_A K \ar[r] \ar[d] &
M \otimes_A K \ar[r] \ar[d] &
0 \\
F(\bigoplus\nolimits_{j \in J} A) \ar[r] &
F(\bigoplus\nolimits_{i \in I} A) \ar[r] &
F(M) \ar[r] & 0
}
$$
The lower row is exact as $F$ is right exact.
The upper row is exact as tensor product with $K$ is right exact.
Since $F$ commutes with direct sums the left two vertical arrows
are bijections. Hence we conclude.
\end{proof}
\begin{example}
\label{example-functor-modules}
Let $R$ be a ring. Let $A$ and $B$ be $R$-algebras. Let $K$ be a
$A \otimes_R B$-module. Then we can consider the functor
\begin{equation}
\label{equation-FM-modules}
F : \text{Mod}_A \longrightarrow \text{Mod}_B,\quad
M \longmapsto M \otimes_A K
\end{equation}
This functor is $R$-linear, right exact,
commutes with arbitrary direct sums, commutes
with all colimits, has a right adjoint (Lemma \ref{lemma-functor}).
\end{example}
\begin{lemma}
\label{lemma-functor-modules}
Let $R$ be a ring. Let $A$ and $B$ be $R$-algebras. There is an
equivalence of categories between
\begin{enumerate}
\item the category of $R$-linear functors
$F : \text{Mod}_A \to \text{Mod}_B$ which
are right exact and commute with arbitrary direct sums, and
\item the category $\text{Mod}_{A \otimes_R B}$.
\end{enumerate}
given by sending $K$ to the functor $F$ in (\ref{equation-FM-modules}).
\end{lemma}
\begin{proof}
Let $F$ be an object of the first category. By
Lemma \ref{lemma-functor} we may assume $F(M) = M \otimes_A K$
functorially in $M$ for some $A \otimes_\mathbf{Z} B$-module $K$.
The $R$-linearity of $F$ immediately implies that the
$A \otimes_\mathbf{Z} B$-module structure on $K$ comes
from a (unique) $A \otimes_R B$-module structure on $K$.
Thus we see that sending $K$ to $F$ as in (\ref{equation-FM-modules})
is essentially surjective.
\medskip\noindent
To prove that our functor is fully faithful, we have to show that
given $A \otimes_R B$-modules $K$ and $K'$ any transformation
$t : F \to F'$ between the corresponding functors, comes from
a unique $\varphi : K \to K'$. Since $K = F(A)$ and $K' = F'(A)$
we can take $\varphi$ to be the value $t_A : F(A) \to F'(A)$
of $t$ at $A$. This maps is $A \otimes_R B$-linear by the
definition of the $A \otimes B$-module structure on $F(A)$
and $F'(A)$ given in the proof of Lemma \ref{lemma-functor}.
\end{proof}
\begin{remark}
\label{remark-composition}
Let $R$ be a ring. Let $A$, $B$, $C$ be $R$-algebras.
Let $F : \text{Mod}_A \to \text{Mod}_B$ and
$F' : \text{Mod}_B \to \text{Mod}_C$ be
$R$-linear, right exact functors which commute with arbitrary direct sums.
If by the equivalence of Lemma \ref{lemma-functor-modules} the object
$K$ in $\text{Mod}_{A \otimes_R B}$ corresponds to $F$ and the object
$K'$ in $\text{Mod}_{B \otimes_R C}$ corresponds to $F'$, then
$K \otimes_B K'$ viewed as an object of
$\text{Mod}_{A \otimes_R C}$ corresponds to $F' \circ F$.
\end{remark}
\begin{remark}
\label{remark-exact-flat}
In the situation of Lemma \ref{lemma-functor-modules}
suppose that $F$ corresponds to $K$. Then
$F$ is exact $\Leftrightarrow$ $K$ is flat over $A$.
\end{remark}
\begin{remark}
\label{remark-finite}
In the situation of Lemma \ref{lemma-functor-modules}
suppose that $F$ corresponds to $K$. Then
$F$ sends finite $A$-modules to finite $B$-modules
$\Leftrightarrow$ $K$ is finite as a $B$-module.
\end{remark}
\begin{remark}
\label{remark-finite-presentation}
In the situation of Lemma \ref{lemma-functor-modules}
suppose that $F$ corresponds to $K$. Then
$F$ sends finitely presented $A$-modules to finitely presented $B$-modules
$\Leftrightarrow$ $K$ is finitely presented as a $B$-module.
\end{remark}
\begin{lemma}
\label{lemma-functor-equivalence}
Let $A$ and $B$ be rings. If
$$
F : \text{Mod}_A \longrightarrow \text{Mod}_B
$$
is an equivalence of categories, then there exists an isomorphism
$A \to B$ of rings and an invertible $B$-module $L$ such that
$F$ is isomorphic to the functor $M \mapsto (M \otimes_A B) \otimes_B L$.
\end{lemma}
\begin{proof}
Since an equivalence commutes with all colimits, we see that
Lemmas \ref{lemma-functor} applies. Let $K$ be the
$A \otimes_\mathbf{Z} B$-module such that $F$ is
isomorphic to the functor $M \mapsto M \otimes_A K$.
Let $K'$ be the $B \otimes_\mathbf{Z} A$-module such that
a quasi-inverse of $F$ is
isomorphic to the functor $N \mapsto N \otimes_B K'$.
By Remark \ref{remark-composition} and
Lemma \ref{lemma-functor-modules} we have an isomorphism
$$
\psi : K \otimes_B K' \longrightarrow A
$$
of $A \otimes_\mathbf{Z} A$-modules.
Similarly, we have an isomorphism
$$
\psi' : K' \otimes_A K \longrightarrow B
$$
of $B \otimes_\mathbf{Z} B$-modules. Choose an element
$\xi = \sum_{i = 1, \ldots, n} x_i \otimes y_i \in K \otimes_B K'$
such that $\psi(\xi) = 1$. Consider the isomorphisms
$$
K \xrightarrow{\psi^{-1} \otimes \text{id}_K}
K \otimes_B K' \otimes_A K \xrightarrow{\text{id}_K \otimes \psi'} K
$$
The composition is an isomorphism and given by
$$
k \longmapsto \sum x_i \psi'(y_i \otimes k)
$$
We conclude this automorphism factors as
$$
K \to B^{\oplus n} \to K
$$
as a map of $B$-modules. It follows that $K$ is finite
projective as a $B$-module.
\medskip\noindent
We claim that $K$ is invertible as a $B$-module. This is equivalent
to asking the rank of $K$ as a $B$-module to have the constant value $1$,
see More on Algebra, Lemma \ref{more-algebra-lemma-invertible} and
Algebra, Lemma \ref{algebra-lemma-finite-projective}.
If not, then there exists a maximal ideal $\mathfrak m \subset B$
such that either (a) $K \otimes_B B/\mathfrak m = 0$ or
(b) there is a surjection $K \to (B/\mathfrak m)^{\oplus 2}$ of
$B$-modules. Case (a) is absurd as $K' \otimes_A K \otimes_B N = N$
for all $B$-modules $N$. Case (b) would imply we get a surjection
$$
A = K \otimes_B K' \longrightarrow (B/\mathfrak m \otimes_B K')^{\oplus 2}
$$
of (right) $A$-modules. This is impossible as the target is an $A$-module
which needs at least two generators: $B/\mathfrak m \otimes_B K'$
is nonzero as the image of the nonzero module $B/\mathfrak m$ under
the quasi-inverse of $F$.
\medskip\noindent
Since $K$ is invertible as a $B$-module we see that $\Hom_B(K, K) = B$.
Since $K = F(A)$ the action of $A$ on $K$ defines a ring isomorphism
$A \to B$. The lemma follows.
\end{proof}
\begin{lemma}
\label{lemma-functor-equivalence-linear}
Let $R$ be a ring. Let $A$ and $B$ be $R$-algebras. If
$$
F : \text{Mod}_A \longrightarrow \text{Mod}_B
$$
is an $R$-linear equivalence of categories, then there exists an isomorphism
$A \to B$ of $R$-algebras and an invertible $B$-module $L$ such that
$F$ is isomorphic to the functor $M \mapsto (M \otimes_A B) \otimes_B L$.
\end{lemma}
\begin{proof}
We get $A \to B$ and $L$ from Lemma \ref{lemma-functor-equivalence}.
To finish the proof, we need to show that the $R$-linearity
of $F$ forces $A \to B$ to be an $R$-algebra map. We omit the details.
\end{proof}
\begin{remark}
\label{remark-monoidal}
Let $A$ and $B$ be rings. Let us endow $\text{Mod}_A$ and $\text{Mod}_B$
with the usual monoidal structure given by tensor products of modules.
Let $F : \text{Mod}_A \to \text{Mod}_B$ be a functor of
monoidal categories, see
Categories, Definition \ref{categories-definition-functor-monoidal-categories}.
Here are some comments:
\begin{enumerate}
\item Since $F(A)$ is a unit (by our definitions) we have $F(A) = B$.
\item We obtain a multiplicative map $\varphi : A \to B$
by sending $a \in A$ to its action on $F(A) = B$.
\item Take $A = B$ and $F(M) = M \otimes_A M$. In this case $\varphi(a) = a^2$.
\item If $F$ is additive, then $\varphi$ is a ring map.
\item Take $A = B = \mathbf{Z}$ and $F(M) = M/\text{torsion}$. Then
$\varphi = \text{id}_\mathbf{Z}$ but $F$ is not the identity functor.
\item If $F$ is right exact and commutes with direct sums,
then $F(M) = M \otimes_{A, \varphi} B$ by Lemma \ref{lemma-functor}.
\end{enumerate}
In other words, ring maps $A \to B$ are in bijection with isomorphism classes
of functors of monoidal categories $\text{Mod}_A \to \text{Mod}_B$
which commute with all colimits.
\end{remark}
\section{Extending functors on categories of modules}
\label{section-functors-extend}
\noindent
For a ring $A$ let us denote $\text{Mod}^{fp}_A$ the category of
finitely presented $A$-modules.
\begin{lemma}
\label{lemma-functor-fp-modules}
Let $A$ and $B$ be rings. Let
$F : \text{Mod}^{fp}_A \to \text{Mod}^{fp}_B$ be a functor.
Then $F$ extends uniquely to a functor
$F' : \text{Mod}_A \to \text{Mod}_B$
which commutes with filtered colimits.
\end{lemma}
\begin{proof}
Special case of Lemma \ref{lemma-functor-on-fp-modules}.
\end{proof}
\begin{remark}
\label{remark-monoidal-extension}
With $A$, $B$, $F$, and $F'$ as in Lemma \ref{lemma-functor-fp-modules}.
Observe that the tensor product of two finitely presented modules is
finitely presented, see Algebra, Lemma \ref{algebra-lemma-tensor-finiteness}.
Thus we may endow $\text{Mod}^{fp}_A$, $\text{Mod}^{fp}_B$,
$\text{Mod}_A$, and $\text{Mod}_B$ with the usual monoidal structure
given by tensor products of modules. In this case, if $F$ is
a functor of monoidal categories, so is $F'$. This follows immediately
from the fact that tensor products of modules commutes with filtered
colimits.
\end{remark}
\begin{lemma}
\label{lemma-functor-fp-modules-exact}
With $A$, $B$, $F$, and $F'$ as in Lemma \ref{lemma-functor-fp-modules}.
\begin{enumerate}
\item If $F$ is additive, then $F'$ is additive and
commutes with arbitrary direct sums, and
\item if $F$ is right exact, then $F'$ is right exact.
\end{enumerate}
\end{lemma}
\begin{proof}
Follows from Lemmas \ref{lemma-functor-on-fp-modules-additive} and
\ref{lemma-functor-on-fp-modules-right-exact}.
\end{proof}
\begin{remark}
\label{remark-monoidal-extension-exact}
Combining Remarks \ref{remark-monoidal} and \ref{remark-monoidal-extension}
and Lemma \ref{lemma-functor-fp-modules-exact}
we find the following. Given rings $A$ and $B$ the set of ring maps $A \to B$
is in bijection with the set of isomorphism classes
of functors of monoidal categories $\text{Mod}^{fp}_A \to \text{Mod}^{fp}_B$
which are right exact.
\end{remark}
\begin{lemma}
\label{lemma-functor-fp-modules-left-exact}
With $A$, $B$, $F$, and $F'$ as in Lemma \ref{lemma-functor-fp-modules}.
Assume $A$ is a coherent ring
(Algebra, Definition \ref{algebra-definition-coherent}).
If $F$ is left exact, then $F'$ is left exact.
\end{lemma}
\begin{proof}
Special case of Lemma \ref{lemma-functor-on-fp-modules-left-exact}.
\end{proof}
\noindent
For a ring $A$ let us denote $\text{Mod}^{fg}_A$ the category of
finitely generated $A$-modules (AKA finite $A$-modules).
\begin{lemma}
\label{lemma-functor-finite-modules}
Let $A$ and $B$ be Noetherian rings. Let
$F : \text{Mod}^{fg}_A \to \text{Mod}^{fg}_B$ be a functor.
Then $F$ extends uniquely to a functor $F' : \text{Mod}_A \to \text{Mod}_B$
which commutes with filtered colimits. If $F$ is additive, then
$F'$ is additive and commutes with arbitrary direct sums.
If $F$ is exact, left exact, or right exact, so is $F'$.
\end{lemma}
\begin{proof}
See Lemmas \ref{lemma-functor-fp-modules-exact} and
\ref{lemma-functor-fp-modules-left-exact}.
Also, use the finite $A$-modules are finitely presented $A$-modules,
see Algebra, Lemma
\ref{algebra-lemma-Noetherian-finite-type-is-finite-presentation},
and use that Noetherian rings are coherent, see
Algebra, Lemma \ref{algebra-lemma-Noetherian-coherent}.
\end{proof}
\section{Functors between categories of quasi-coherent modules}
\label{section-functor-quasi-coherent}
\noindent
In this section we briefly study functors between categories of
quasi-coherent modules.
\begin{example}
\label{example-functor-quasi-coherent}
Let $R$ be a ring. Let $X$ and $Y$ be
schemes over $R$ with $X$ quasi-compact and quasi-separated.
Let $\mathcal{K}$ be a quasi-coherent $\mathcal{O}_{X \times_R Y}$-module.
Then we can consider the functor
\begin{equation}
\label{equation-FM-QCoh}
F : \QCoh(\mathcal{O}_X) \longrightarrow \QCoh(\mathcal{O}_Y),\quad
\mathcal{F} \longmapsto
\text{pr}_{2, *}(\text{pr}_1^*\mathcal{F}
\otimes_{\mathcal{O}_{X \times_R Y}} \mathcal{K})
\end{equation}
The morphism $\text{pr}_2$ is quasi-compact and quasi-separated
(Schemes, Lemmas \ref{schemes-lemma-quasi-compact-preserved-base-change}
and \ref{schemes-lemma-separated-permanence}). Hence pushforward along
this morphism preserves quasi-coherent modules, see
Schemes, Lemma \ref{schemes-lemma-push-forward-quasi-coherent}.
Moreover, our functor is $R$-linear and commutes with arbitrary direct sums,
see Cohomology of Schemes, Lemma \ref{coherent-lemma-colimit-cohomology}.
\end{example}
\noindent
The following lemma is a natural generalization of
Lemma \ref{lemma-functor-modules}.
\begin{lemma}
\label{lemma-functor-quasi-coherent-from-affine}
Let $R$ be a ring. Let $X$ and $Y$ be schemes over $R$ with $X$ affine.
There is an equivalence of categories between
\begin{enumerate}
\item the category of $R$-linear functors
$F : \QCoh(\mathcal{O}_X) \to \QCoh(\mathcal{O}_Y)$
which are right exact and commute with arbitrary direct sums, and
\item the category $\QCoh(\mathcal{O}_{X \times_R Y})$
\end{enumerate}
given by sending $\mathcal{K}$ to the functor $F$ in (\ref{equation-FM-QCoh}).
\end{lemma}
\begin{proof}
Let $\mathcal{K}$ be an object of $\QCoh(\mathcal{O}_{X \times_R Y})$
and $F_\mathcal{K}$ the functor (\ref{equation-FM-QCoh}). By the discussion in
Example \ref{example-functor-quasi-coherent} we already know that
$F$ is $R$-linear and commutes with arbitrary direct sums.
Since $\text{pr}_2 : X \times_R Y \to Y$ is affine
(Morphisms, Lemma \ref{morphisms-lemma-base-change-affine}) the functor
$\text{pr}_{2, *}$ is exact, see Cohomology of Schemes, Lemma
\ref{coherent-lemma-relative-affine-vanishing}.
Hence $F$ is right exact as well, in other words $F$ is as in (1).
\medskip\noindent
Let $F$ be as in (1). Say $X = \Spec(A)$. Consider the quasi-coherent
$\mathcal{O}_Y$-module $\mathcal{G} = F(\mathcal{O}_X)$.
The functor $F$ induces an $R$-linear map
$A \to \text{End}_{\mathcal{O}_Y}(\mathcal{G})$,
$a \mapsto F(a \cdot \text{id})$. Thus $\mathcal{G}$ is a sheaf of modules over
$$
A \otimes_R \mathcal{O}_Y = \text{pr}_{2, *}\mathcal{O}_{X \times_R Y}
$$
By Morphisms, Lemma \ref{morphisms-lemma-affine-equivalence-modules}
we find that there is a unique quasi-coherent module $\mathcal{K}$
on $X \times_R Y$ such that $F(\mathcal{O}_X) = \mathcal{G} =
\text{pr}_{2, *}\mathcal{K}$ compatible with action
of $A$ and $\mathcal{O}_Y$. Denote $F_\mathcal{K}$ the functor
given by (\ref{equation-FM-QCoh}). There is an equivalence
$\text{Mod}_A \to \QCoh(\mathcal{O}_X)$ sending $A$ to $\mathcal{O}_X$, see
Schemes, Lemma \ref{schemes-lemma-equivalence-quasi-coherent}.
Hence we find an isomorphism $F \cong F_\mathcal{K}$ by
Lemma \ref{lemma-functor-on-modules} because we have an isomorphism
$F(\mathcal{O}_X) \cong F_\mathcal{K}(\mathcal{O}_X)$ compatible with
$A$-action by construction.
\medskip\noindent
This shows that the functor sending $\mathcal{K}$ to $F_\mathcal{K}$
is essentially surjective. We omit the verification of fully faithfulness.
\end{proof}
\begin{remark}
\label{remark-affine-morphism}
Below we will use that for an affine morphism
$h : T \to S$ we have $h_*\mathcal{G} \otimes_{\mathcal{O}_S} \mathcal{H} =
h_*(\mathcal{G} \otimes_{\mathcal{O}_T} h^*\mathcal{H})$ for
$\mathcal{G} \in \QCoh(\mathcal{O}_T)$ and
$\mathcal{H} \in \QCoh(\mathcal{O}_S)$. This follows
immediately on translating into algebra.
\end{remark}
\begin{lemma}
\label{lemma-functor-quasi-coherent-from-affine-compose}
In Lemma \ref{lemma-functor-quasi-coherent-from-affine} let $F$
correspond to $\mathcal{K}$ in $\QCoh(\mathcal{O}_{X \times_R Y})$.
We have
\begin{enumerate}
\item If $f : X' \to X$ is an affine morphism, then $F \circ f_*$
corresponds to $(f \times \text{id}_Y)^*\mathcal{K}$.
\item If $g : Y' \to Y$ is a flat morphism, then $g^* \circ F$ corresponds to
$(\text{id}_X \times g)^*\mathcal{K}$.
\item If $j : V \to Y$ is an open immersion, then $j^* \circ F$
corresponds to $\mathcal{K}|_{X \times_R V}$.
\end{enumerate}
\end{lemma}
\begin{proof}
Proof of (1). Consider the commutative diagram
$$
\xymatrix{
X' \times_R Y \ar[rrd]^{\text{pr}'_2} \ar[rd]_{f \times \text{id}_Y}
\ar[dd]_{\text{pr}'_1} \\
& X \times_R Y \ar[r]_{\text{pr}_2} \ar[d]_{\text{pr}_1} & Y \\
X' \ar[r]^f & X
}
$$
Let $\mathcal{F}'$ be a quasi-coherent module on $X'$. We have
\begin{align*}
\text{pr}_{2, *}(\text{pr}_1^*f_*\mathcal{F}'
\otimes_{\mathcal{O}_{X \times_R Y}} \mathcal{K})
& =
\text{pr}_{2, *}((f \times \text{id}_Y)_*
(\text{pr}'_1)^*\mathcal{F}'
\otimes_{\mathcal{O}_{X \times_R Y}} \mathcal{K}) \\
& =
\text{pr}_{2, *}(f \times \text{id}_Y)_*
\left((\text{pr}'_1)^*\mathcal{F}'
\otimes_{\mathcal{O}_{X' \times_R Y}}
(f \times \text{id}_Y)^*\mathcal{K})\right) \\
& =
\text{pr}'_{2, *}((\text{pr}'_1)^*\mathcal{F}'
\otimes_{\mathcal{O}_{X' \times_R Y}} (f \times \text{id}_Y)^*\mathcal{K})
\end{align*}
Here the first equality is affine base change for the left hand
square in the diagram, see
Cohomology of Schemes, Lemma \ref{coherent-lemma-affine-base-change}.
The second equality hold by Remark \ref{remark-affine-morphism}.
The third equality is functoriality of pushforwards for modules.
This proves (1).
\medskip\noindent
Proof of (2). Consider the commutative diagram
$$
\xymatrix{
X \times_R Y' \ar[rr]_-{\text{pr}'_2} \ar[rd]^{\text{id}_X \times g}
\ar[rdd]_{\text{pr}'_1} & & Y' \ar[d]^g \\
& X \times_R Y \ar[r]_-{\text{pr}_2} \ar[d]^{\text{pr}_1} & Y \\
& X
}
$$
We have
\begin{align*}
g^*\text{pr}_{2, *}(\text{pr}_1^*\mathcal{F}
\otimes_{\mathcal{O}_{X \times_R Y}} \mathcal{K})
& =
\text{pr}'_{2, *}(
(\text{id}_X \times g)^*(
\text{pr}_1^*\mathcal{F} \otimes_{\mathcal{O}_{X \times_R Y}} \mathcal{K})) \\
& =
\text{pr}'_{2, *}((\text{pr}'_1)^*\mathcal{F}
\otimes_{\mathcal{O}_{X \times_R Y'}}
(\text{id}_X \times g)^*\mathcal{K})
\end{align*}
The first equality by flat base change for the square in the diagram, see
Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}.
The second equality by functoriality of pullback and the fact that
a pullback of tensor products it the tensor product of the pullbacks.
\medskip\noindent
Part (3) is a special case of (2).
\end{proof}
\begin{lemma}
\label{lemma-functor-quasi-coherent-from-affine-diagonal-pre}
Let $R$ be a ring. Let $X$ and $Y$ be schemes over $R$. Assume $X$
is quasi-compact with affine diagonal. Let
$F : \QCoh(\mathcal{O}_X) \to \QCoh(\mathcal{O}_Y)$
be an $R$-linear, right exact functor which commutes
with arbitrary direct sums. Then we can construct
\begin{enumerate}
\item a quasi-coherent module $\mathcal{K}$ on $X \times_R Y$, and
\item a natural transformation $t : F \to F_\mathcal{K}$
where $F_\mathcal{K}$ denotes the functor (\ref{equation-FM-QCoh})
\end{enumerate}
such that $t : F \circ f_* \to F_\mathcal{K} \circ f_*$ is an isomorphism
for every morphism $f : X' \to X$ whose source is an affine scheme.
\end{lemma}
\begin{proof}
Consider a morphism $f' : X' \to X$ with $X'$ affine. Since the