## 56.3 Functors between categories of modules

The following lemma is archetypical of the results in this chapter.

Lemma 56.3.1. Let $A$ and $B$ be rings. Let $F : \text{Mod}_ A \to \text{Mod}_ B$ be a functor. The following are equivalent

$F$ is isomorphic to the functor $M \mapsto M \otimes _ A K$ for some $A \otimes _\mathbf {Z} B$-module $K$,

$F$ is right exact and commutes with all direct sums,

$F$ commutes with all colimits,

$F$ has a right adjoint $G$.

**Proof.**
If (1), then (4) as a right adjoint for $M \mapsto M \otimes _ A K$ is $N \mapsto \mathop{\mathrm{Hom}}\nolimits _ B(K, N)$, see Differential Graded Algebra, Lemma 22.30.3. If (4), then (3) by Categories, Lemma 4.24.5. The implication (3) $\Rightarrow $ (2) is immediate from the definitions.

Assume (2). We will prove (1). By the discussion in Homology, Section 12.7 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.
$\square$

Example 56.3.2. 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

56.3.2.1
\begin{equation} \label{functors-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 56.3.1).

Lemma 56.3.3. Let $R$ be a ring. Let $A$ and $B$ be $R$-algebras. There is an equivalence of categories between

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

the category $\text{Mod}_{A \otimes _ R B}$.

given by sending $K$ to the functor $F$ in (56.3.2.1).

**Proof.**
Let $F$ be an object of the first category. By Lemma 56.3.1 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 (56.3.2.1) is essentially surjective.

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 56.3.1.
$\square$

Lemma 56.3.8. 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 $M \mapsto (M \otimes _ A B) \otimes _ B L$.

**Proof.**
Since an equivalence commutes with all colimits, we see that Lemmas 56.3.1 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 56.3.4 and Lemma 56.3.3 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.

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 15.117.2 and Algebra, Lemma 10.78.2. 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 $F$.

Since $K$ is invertible as a $B$-module we see that $\mathop{\mathrm{Hom}}\nolimits _ B(K, K) = B$. Since $K = F(A)$ the action of $A$ on $K$ defines a ring isomorphism $A \to B$. The lemma follows.
$\square$

Lemma 56.3.9. 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 $M \mapsto (M \otimes _ A B) \otimes _ B L$.

**Proof.**
We get $A \to B$ and $L$ from Lemma 56.3.8. 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.
$\square$

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