Lemma 56.4.1. 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.

## 56.4 Extending functors on categories of modules

For a ring $A$ let us denote $\text{Mod}^{fp}_ A$ the category of finitely presented $A$-modules.

**Proof.**
Special case of Lemma 56.2.1.
$\square$

Remark 56.4.2. With $A$, $B$, $F$, and $F'$ as in Lemma 56.4.1. Observe that the tensor product of two finitely presented modules is finitely presented, see Algebra, Lemma 10.12.14. 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.

Lemma 56.4.3. With $A$, $B$, $F$, and $F'$ as in Lemma 56.4.1.

If $F$ is additive, then $F'$ is additive and commutes with arbitrary direct sums, and

if $F$ is right exact, then $F'$ is right exact.

Remark 56.4.4. Combining Remarks 56.3.10 and 56.4.2 and Lemma 56.4.3 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.

Lemma 56.4.5. With $A$, $B$, $F$, and $F'$ as in Lemma 56.4.1. Assume $A$ is a coherent ring (Algebra, Definition 10.90.1). If $F$ is left exact, then $F'$ is left exact.

**Proof.**
Special case of Lemma 56.2.4.
$\square$

For a ring $A$ let us denote $\text{Mod}^{fg}_ A$ the category of finitely generated $A$-modules (AKA finite $A$-modules).

Lemma 56.4.6. 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 $F'$.

**Proof.**
See Lemmas 56.4.3 and 56.4.5. Also, use the finite $A$-modules are finitely presented $A$-modules, see Algebra, Lemma 10.31.4, and use that Noetherian rings are coherent, see Algebra, Lemma 10.90.5.
$\square$

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