The Stacks project

Proof. This follows from Categories, Lemma 4.26.2. To see that the lemma applies observe that finitely presented $A$-modules are categorically compact objects of $\text{Mod}_ A$ by Algebra, Lemma 10.11.4. Also, every $A$-module is a filtered colimit of finitely presented $A$-modules by Algebra, Lemma 10.11.3. $\square$

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 12.7.1. Write $M = \mathop{\mathrm{colim}}\nolimits _ i M_ i$ and $M' = \mathop{\mathrm{colim}}\nolimits _ j M'_ j$ as filtered colimits of finitely presented $A$-modules $M_ i$. Then $F'(M) = \mathop{\mathrm{colim}}\nolimits _ i F(M_ i)$, $F'(M') = \mathop{\mathrm{colim}}\nolimits _ j F(M'_ j)$, and

as desired. To show that $F'$ commutes with direct sums, assume we have $M = \bigoplus _{i \in I} M_ i$. Then $M = \mathop{\mathrm{colim}}\nolimits _{I' \subset I\text{ finite}} \bigoplus _{i \in I'} M_ i$ is a filtered colimit. We obtain

The second equality holds by the additivity of $F'$ already shown. $\square$

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 12.7.1. Hence $F'$ is additive and commutes with direct sums by Lemma 56.2.2. We urge the reader to prove that $F'$ is right exact themselves instead of reading the proof below.

To show that $F'$ is right exact, it suffices to show that $F'$ commutes with coequalizers, see Categories, Lemma 4.23.3. 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

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 $M = \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i$ as a filtered colimit of finitely presented $A$-modules, see Algebra, Lemma 10.11.3. 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 4.14.10 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 $L = \mathop{\mathrm{colim}}\nolimits _{i \in I} L_ i$ 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 $\mathop{\mathrm{Ker}}(L \to M)$ is finite (Algebra, Lemma 10.5.3). Thus we can write $K = \mathop{\mathrm{colim}}\nolimits _{i \in I} K_ i$ as a filtered colimit of finitely presented $A$-modules each surjecting onto $\mathop{\mathrm{Ker}}(L \to M)$. 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. $\square$

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 10.90.4 and 10.90.3. Hence all finite limits exist in $\text{Mod}^{fp}_ A$ and Categories, Definition 4.23.1 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 12.7.1. Hence $F'$ is additive and commutes with direct sums by Lemma 56.2.2. We urge the reader to prove that $F'$ is left exact themselves instead of reading the proof below.

To show that $F'$ is left exact, it suffices to show that $F'$ commutes with equalizers, see Categories, Lemma 4.23.2. 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

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 $M = \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i$ as a filtered colimit of finitely presented $A$-modules, see Algebra, Lemma 10.11.3. 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 $L = \mathop{\mathrm{colim}}\nolimits _{i \in I} L_ i$ 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 $K = \mathop{\mathrm{Ker}}(L \to M)$ 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. $\square$

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

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

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.

For each object $M$ of $\text{Mod}^{fp}_ A$ choose a presentation

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

Having made these choices we can define: for an object $M$ of $\text{Mod}^{fp}_ A$ we set

and for a morphism $f : M \to N$ of $\text{Mod}^{fp}_ A$ we set

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.

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 (56.2.5.1). Namely, say

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 $F(f) = F(f)'$.

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 (56.2.5.1) for $f \circ g$. Hence these induce the correct map which exactly says that $F(f) \circ F(g) = F(f \circ g)$.

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 (56.2.5.1) for $f + g$. Hence these induce the correct map which exactly says that $F(f) + F(g) = F(f + g)$.

Finally, let us prove that $F$ is right exact. It suffices to show that $F$ commutes with coequalizers, see Categories, Lemma 4.23.3. 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

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 $\mathop{\mathrm{Coker}}(F(K) \to F(L)) \to F(M)$. By elementary commutative algebra there exists a commutative diagram

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 $F(M) \to \mathop{\mathrm{Coker}}(F(K) \to F(L))$ which is a right inverse to the map $\mathop{\mathrm{Coker}}(F(K) \to F(L)) \to F(M)$. This first implies that $F(L) \to F(M)$ is an epimorphism always. Next, the above shows we have

where the direct sum decomposition is compatible with both $F(M) \to \mathop{\mathrm{Coker}}(F(K) \to F(L))$ and $\mathop{\mathrm{Coker}}(F(K) \to F(L)) \to F(M)$. 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. $\square$

Proof. Combine Lemmas 56.2.5 and 56.2.3. $\square$