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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0GNR. Beware of the difference between the letter 'O' and the digit '0'.