## 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

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

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

3. $F$ commutes with all colimits,

4. $F$ 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
$$\label{functors-equation-FM-modules} F : \text{Mod}_ A \longrightarrow \text{Mod}_ B,\quad M \longmapsto M \otimes _ A K$$

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

1. 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

2. 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. $\square$

Remark 56.3.4. 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 56.3.3 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 $F' \circ F$.

Remark 56.3.5. In the situation of Lemma 56.3.3 suppose that $F$ corresponds to $K$. Then $F$ is exact $\Leftrightarrow$ $K$ is flat over $A$.

Remark 56.3.6. In the situation of Lemma 56.3.3 suppose that $F$ corresponds to $K$. Then $F$ sends finite $A$-modules to finite $B$-modules $\Leftrightarrow$ $K$ is finite as a $B$-module.

Remark 56.3.7. In the situation of Lemma 56.3.3 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.

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. $\square$

Remark 56.3.10. 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 4.43.2. Here are some comments:

1. Since $F(A)$ is a unit (by our definitions) we have $F(A) = B$.

2. We obtain a multiplicative map $\varphi : A \to B$ by sending $a \in A$ to its action on $F(A) = B$.

3. Take $A = B$ and $F(M) = M \otimes _ A M$. In this case $\varphi (a) = a^2$.

4. If $F$ is additive, then $\varphi$ is a ring map.

5. Take $A = B = \mathbf{Z}$ and $F(M) = M/\text{torsion}$. Then $\varphi = \text{id}_\mathbf {Z}$ but $F$ is not the identity functor.

6. If $F$ is right exact and commutes with direct sums, then $F(M) = M \otimes _{A, \varphi } B$ by Lemma 56.3.1.

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.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).