Coherence from finite probes

Written with GPT-6.1 Sol (OpenAI), Ultra reasoning effort. Self-checked by the AI that wrote it. Original text: CC0.

Finite generators describe an object by a small amount of data. Finite relations ask a different question: when a finite object maps to it, can the kernel still be described by finite data? These two requirements behave differently in exact sequences. Relations pass through an extension under very weak assumptions. Producing finite generators for that extension needs an additional condition.

We study both requirements using a chosen family of objects in an abelian category. The family need not consist of projective objects. A pullback transports finite relations, and a condition on extensions with a chosen quotient then makes the coherent objects an exact abelian subcategory. Finite free modules recover the familiar module notion of coherence.

Use Full subcategories and exact closure, Theorem 2.2, for the construction of an abelian subcategory from ambient kernels and cokernels. The kernel, cokernel and image factorizations used below are those of Trace coreflections and balanced nonabelian categories. Basic references are [Stacks, Coherent rings], [Stacks, Abelian categories] and [Schapira, Homological Algebra], listed below. Categories are locally small in a fixed universe. All constructions in this lesson use finite diagrams.

1. Generators and relations measure different things

Let \(\mathcal A\) be an abelian category and \(\mathcal J\) a full additive subcategory. We may replace \(\mathcal J\) by its replete closure: transport through an isomorphism does not change any of the following properties. Thus membership in \(\mathcal J\) means membership up to isomorphism. Its zero object and finite biproducts agree with those of \(\mathcal A\), by Proposition 1.1 of the prerequisite.

An object \(X\) is \(\mathcal J\)-finite if there is an epimorphism

\[ S\twoheadrightarrow X, \qquad S\in\mathcal J. \tag{1.1} \]

It is \(\mathcal J\)-pseudo-coherent if every morphism \(S\to X\) with \(S\in\mathcal J\) has \(\mathcal J\)-finite kernel. It is \(\mathcal J\)-coherent if it has both properties. Write \(\operatorname{Coh}_{\mathcal J}(\mathcal A)\) for the full subcategory of coherent objects. When the family is fixed, we shorten the three terms to finite, pseudo-coherent and coherent.

The word finite is relative to \(\mathcal J\). It does not mean finite length, finite cardinality or finite dimension unless that interpretation follows from the chosen family.

Example 1.1. Let \(\mathcal A\) be the category of finite-dimensional \(k[t]\)-modules killed by some power of \(t\), for a field \(k\). This is an abelian category: kernels and cokernels remain finite-dimensional and nilpotent. In an extension of objects killed by \(t^a\) and \(t^b\), the middle object is killed by \(t^{a+b}\). Let \(\mathcal J\) consist of the objects killed by \(t\).

Every object of \(\mathcal A\) is pseudo-coherent. Indeed, a kernel inside a finite-dimensional object killed by \(t\) is again such an object and covers itself. The finite objects, however, are exactly the objects killed by \(t\): that condition survives quotients, and the identity gives the converse. The exact sequence

\[ 0\to k \to k[t]/(t^2) \to k\to0 \tag{1.2} \]

has coherent ends and a middle object that is not finite. Here the first map takes \(1\) to the class of \(t\), and the last is reduction modulo \(t\).

Thus extension closure of pseudo-coherent objects and extension closure of coherent objects are distinct assertions. The first holds for every full additive \(\mathcal J\); the second will follow from the condition in Section 4.

2. Pullbacks transport finite relations

We first isolate the abelian fact that carries a finite cover onto a kernel. This is also the epimorphism part of [Stacks, Lemma 12.5.13].

Lemma 2.1. Pulling back an epimorphism \(e:B \twoheadrightarrow D\) along any \(u:A\to D\) gives an epimorphism \(p:P \twoheadrightarrow A\). Moreover, the other projection \(r:P\to B\) satisfies

\[ \ker p\simeq\ker e, \qquad \ker r\simeq\ker u. \tag{2.1} \]

The isomorphisms respect the maps into the relevant objects.

Proof. Realize \(P\) as the kernel of

\[ d=(u,-e):A\oplus B\to D. \tag{2.2} \]

Its projections satisfy \(up=er\), and the kernel universal property gives the pullback universal property. The map \(d\) is epic, since its restriction to \(B\) is the epimorphism \(-e\).

Let \(h:A\to T\) satisfy \(hp=0\). The map \(h\pi_A:A\oplus B\to T\) vanishes on \(P\). Every epimorphism in an abelian category is the cokernel of its kernel, so there is \(z:D\to T\) with

\[ h\pi_A=zd. \tag{2.3} \]

Restriction to \(B\) gives \(ze=0\). Epicity of \(e\) gives \(z=0\), and restriction to \(A\) then gives \(h=0\). Applying this to a difference of two maps proves that \(p\) is epic.

For an arbitrary object \(W\), maps \(W\to P\) killed by \(p\) correspond exactly to maps \(b:W\to B\) with \(eb=0\): their two coordinates are \((0,b)\). This identifies the kernel of \(p\) with the kernel of \(e\). Maps killed by \(r\) have coordinates \((a,0)\) with \(ua=0\), giving the second identification. These descriptions also identify the structure maps. \(\square\)

Proposition 2.2. If \(X\) is finite and \(Y\) is pseudo-coherent, then the kernel of any morphism \(f:X\to Y\) is finite. Surjectivity of \(f\) is unnecessary.

Proof. Choose an epimorphism \(s:S \twoheadrightarrow X\) with \(S\in\mathcal J\), and put \(K=\ker f\). The pullback of \(s\) along \(K\hookrightarrow X\) is naturally \(\ker(fs)\). To see the identification, a map to \(S\) factors through this pullback precisely when its composite with \(fs\) is zero. Lemma 2.1 makes the induced map

\[ \ker(fs)\twoheadrightarrow K \tag{2.4} \]

epic. Pseudo-coherence of \(Y\) makes \(\ker(fs)\) finite. Composing a cover of that kernel from an object of \(\mathcal J\) with (2.4) gives a cover of \(K\). \(\square\)

Two elementary closure facts will be used repeatedly. A quotient of a finite object is finite, since epimorphisms compose. A finite sum of finite objects is finite: take the biproduct of their covers. The resulting map is epic because a map killed by it has zero restriction to each summand. The zero object is finite and pseudo-coherent; the kernel of \(S\to0\) is \(S\), which covers itself.

3. Relations survive subobjects and extensions

Theorem 3.1. Pseudo-coherent objects are closed under all ambient subobjects and under extensions. They are therefore closed under finite biproducts. Coherent objects are closed under kernels of morphisms between coherent objects.

Proof. Let \(i:U\hookrightarrow X\) be monic with \(X\) pseudo-coherent. For every \(a:S\to U\), with \(S\in\mathcal J\), monicity gives

\[ \ker a\simeq\ker(ia). \tag{3.1} \]

The right side is finite, so \(U\) is pseudo-coherent. This conclusion does not assert that \(U\) is finite.

Now consider a short exact sequence

\[ 0\to A\xrightarrow{i}B \xrightarrow{v}D\to0 \tag{3.2} \]

with \(A,D\) pseudo-coherent. Fix any \(a:S\to B\) from a probe \(S\in\mathcal J\), and let \(j:K\hookrightarrow S\) be \(\ker(va)\). Pseudo-coherence of \(D\) makes \(K\) finite. Since \(vaj=0\), the kernel property of \(i\) gives a unique map \(b:K\to A\) such that

\[ ib=aj. \tag{3.3} \]

Its kernel, followed by \(j\), is the kernel of \(a\). Indeed, a map \(w:W\to S\) killed by \(a\) is killed by \(va\), hence factors uniquely through \(j\); its resulting map to \(K\) is killed by \(b\) because \(i\) is monic. The reverse implication follows from (3.3). Proposition 2.2 applies to \(b\), since \(K\) is finite and \(A\) is pseudo-coherent. Thus \(\ker a\) is finite. This holds for every probe map, so \(B\) is pseudo-coherent.

A biproduct is a split extension of its summands, giving finite-biproduct closure of pseudo-coherent objects. Since finite objects also have that closure, coherent objects do too.

Finally, if \(f:X\to Y\) has coherent ends, Proposition 2.2 makes \(\ker f\) finite. It is pseudo-coherent as a subobject of \(X\). It is therefore coherent. \(\square\)

The proof of extension closure used finite relations for \(A\) and \(D\), even when those objects themselves admit no finite cover. This explains why Example 1.1 does not contradict the theorem.

4. A cover condition gives an exact abelian category

Impose the following condition on the chosen family:

Condition (E). In every short exact sequence

\[ \begin{gathered} 0\to K\to E\to S\to0,\\ S\in\mathcal J. \end{gathered} \tag{4.1} \]

if \(K\) is coherent, then \(E\) is finite.

This asks for a finite cover of \(E\). It asks for neither a splitting of (4.1) nor projectivity of \(S\).

Theorem 4.1. Under condition (E), coherent objects are closed under extensions and under cokernels of maps between coherent objects. The category \(\operatorname{Coh}_{\mathcal J}(\mathcal A)\) is abelian and its inclusion in \(\mathcal A\) is exact.

Proof. In (3.2), suppose \(A,D\) coherent and choose a cover \(s:S \twoheadrightarrow D\), with \(S\in\mathcal J\). Form \(P=B\times_D S\). Both its projections are epic, by Lemma 2.1 applied first to \(v\) and then to \(s\). Its projection onto \(S\) has kernel \(A\), so

\[ 0\to A\to P \to S\to0 \tag{4.2} \]

is exact. Condition (E) makes \(P\) finite. Its epimorphism onto \(B\) makes \(B\) finite. Theorem 3.1 already makes \(B\) pseudo-coherent, hence coherent.

For cokernels, first take a short exact sequence

\[ 0\to A\to B \xrightarrow{q}C\to0 \tag{4.3} \]

with \(A,B\) coherent. The quotient \(C\) is finite. For an arbitrary map \(a:S\to C\), with \(S\in\mathcal J\), let \(P=B\times_C S\). Its projection onto \(S\) is epic and has kernel \(A\), so (E) makes \(P\) finite. Its other projection \(r:P\to B\) has kernel naturally isomorphic to \(\ker a\), by Lemma 2.1. Proposition 2.2, applied to \(r\) and the pseudo-coherent object \(B\), shows that \(\ker a\) is finite. Thus \(C\) is pseudo-coherent and hence coherent.

For an arbitrary morphism \(f:X\to Y\) between coherent objects, its image \(I\) is finite as a quotient of \(X\) and pseudo-coherent as a subobject of \(Y\). Hence \(I\) is coherent. Apply the preceding argument to

\[ 0\to I\to Y \to\operatorname{coker}f \to0. \tag{4.4} \]

It proves cokernel closure. Theorem 3.1 proves kernel closure and finite-biproduct closure, and the zero object is coherent. The full subcategory is therefore additive and closed under ambient kernels and cokernels. Theorem 2.2 of Full subcategories and exact closure now gives precisely its abelian structure and the exactness of its inclusion. \(\square\)

The conclusion concerns kernels and cokernels of maps whose two ends are coherent. It does not require every subobject or every quotient of a coherent object to be coherent. Those stronger statements would require additional assumptions on \(\mathcal J\).

5. Finite free probes recover coherent modules

Let \(R\) be any unital ring, possibly noncommutative. Work with left \(R\)-modules, and let \(\mathcal J\) consist of finite-rank free left modules. A finite cover is then exactly a finite generating set. The extra condition (E) holds: if a short exact sequence has quotient \(R^n\), lift its finite basis to split the sequence. A cover \(R^m\twoheadrightarrow K\) of its coherent kernel gives a cover

\[ R^{m+n}\twoheadrightarrow K\oplus R^n \simeq E. \tag{5.1} \]

Consequently the coherent left modules form an abelian subcategory with exact inclusion, over every such ring.

To compare this definition with finite presentation, call \(N\) finitely presented if it has an exact presentation

\[ R^a\to R^b \to N\to0 \tag{5.2} \]

with \(a,b\) finite.

Lemma 5.1. If \(N\) is finitely presented, the kernel of every epimorphism \(\pi:R^n\twoheadrightarrow N\), with \(n\) finite, is finitely generated.

Proof. Let \(\rho:R^b\twoheadrightarrow N\) be the map in (5.2), with finitely generated kernel \(L\). Lift finite bases to choose maps

\[ \begin{gathered} \alpha:R^b\to R^n,\\ \beta:R^n\to R^b. \end{gathered} \tag{5.3} \]

with \(\pi\alpha=\rho\) and \(\rho\beta=\pi\). Set \(K=\ker\pi\). If \(z\in K\), then \(\beta z\in L\), and

\[ z=\alpha\beta z+(1-\alpha\beta)z. \tag{5.4} \]

Both \(\alpha(L)\) and the image of \(1-\alpha\beta\) lie in \(K\); for the latter use \(\pi(1-\alpha\beta)=0\). Equation (5.4) shows that these two submodules generate \(K\). Images of finite generators of \(L\), together with the images of the \(n\) basis vectors under \(1-\alpha\beta\), are a finite generating set. The maps are left linear throughout; commutativity of \(R\) was not used. \(\square\)

Proposition 5.2. A left \(R\)-module \(M\) is coherent for finite free probes if and only if it is finitely generated and every finitely generated submodule of \(M\) is finitely presented.

Proof. Suppose \(M\) coherent. Given a finitely generated submodule \(N\subset M\), choose an epimorphism \(R^n\twoheadrightarrow N\). Its kernel agrees with the kernel of the composite into \(M\), so is finitely generated. Choose a finite free cover of that kernel. This supplies (5.2) for \(N\).

Conversely, suppose the stated submodule condition and finite generation hold. For any \(f:R^n\to M\), its image \(N\) is finitely generated and therefore finitely presented. Lemma 5.1 applied to the epimorphism \(R^n\twoheadrightarrow N\) makes \(\ker f\) finitely generated. This proves pseudo-coherence, while finite generation supplies finiteness. \(\square\)

For commutative rings, Proposition 5.2 matches [Stacks, Definition 10.90.1]. The proofs above also establish the same statement for left modules over noncommutative rings. A ring is left coherent when its regular left module is coherent. In particular, every coherent module is finitely presented, since it is a finitely generated submodule of itself. The converse for arbitrary finitely presented modules depends on the ring and is a separate assertion.

6. Exercises with full solutions

Exercise 6.1 (first steps). Fix a prime \(p\). In the category of all abelian groups, take \(\mathcal J\) to be the finite-dimensional \(\mathbb F_p\)-vector groups. Classify the pseudo-coherent and coherent objects. Does (E) hold?

Solution. If \(S\in\mathcal J\), the kernel of any homomorphism \(S\to X\) is a finite-dimensional \(\mathbb F_p\)-subspace of \(S\). It belongs to \(\mathcal J\), hence is finite by its identity cover. Thus every abelian group \(X\) is pseudo-coherent, including infinite groups and groups with elements of infinite order.

A quotient of \(S\) is finite and killed by \(p\). Conversely, any finite group killed by \(p\) is a finite-dimensional \(\mathbb F_p\)-vector group and covers itself. These are exactly the finite objects and, since all objects are pseudo-coherent, exactly the coherent objects.

The sequence

\[ 0\to\mathbb Z/p \xrightarrow{\,p\,}\mathbb Z/p^2 \to\mathbb Z/p \to0 \tag{6.1} \]

has coherent kernel and quotient in \(\mathcal J\). Its middle group is not killed by \(p\), so is not finite relative to this family. Condition (E) fails. Here the labelled map sends the class of \(a\) to the class of \(pa\), and the last map is reduction modulo \(p\).

Exercise 6.2 (comparison). Over any unital ring \(R\), replace finite free probes by finitely generated projective left modules. Show that the notions of finite, pseudo-coherent and coherent module do not change, and verify (E) for this larger family.

Solution. A finitely generated projective module \(P\) is a direct summand of a finite free module: choose a finite free epimorphism onto \(P\), split it by projectivity, and write

\[ R^n\simeq P\oplus P'. \tag{6.2} \]

Both summands are finitely generated, since they are quotient images of \(R^n\). A quotient of a finitely generated projective module is finitely generated, and every finitely generated module has a finite free cover. Thus the two finite notions agree.

Free modules are projective, so testing all finitely generated projectives implies the finite-free pseudo-coherence test. Conversely, let \(M\) pass that test and let \(f:P\to M\). Extend \(f\) to \(P\oplus P'\) by zero on \(P'\). Its kernel is

\[ \ker(f\oplus0)\simeq\ker f\oplus P'. \tag{6.3} \]

It is finitely generated by the finite-free test. Its projection onto \(\ker f\) is epic, so \(\ker f\) is finitely generated. This is the required test for projective probes. Coherence therefore agrees as well.

In (4.1), a projective quotient \(S\) splits the sequence, giving \(E\simeq K\oplus S\). Choose a finite free cover \(R^m\twoheadrightarrow K\). The map \(R^m\oplus S\twoheadrightarrow K\oplus S\) is epic and its domain is finitely generated projective. It verifies (E).

Exercise 6.3 (calculation). Let \(k\) be any field and

\[ \begin{gathered} R=k[u,v]/(u^2,uv,v^2),\\ \mathfrak m=(u,v). \end{gathered} \tag{6.4} \]

Use finite free probes. Compute the kernel and cokernel of \(f:R^2\to R\), \(f(a,b)=au+bv\). Show that every \(R\)-module is pseudo-coherent and identify the coherent ones. Finally compute \(P=R\times_k R\), where both maps to \(k\) are reduction modulo \(\mathfrak m\), and give generators and the kernels of its projections.

Solution. The classes of \(1,u,v\) are a \(k\)-basis of \(R\), and \(\mathfrak m^2=0\). Write

\[ \begin{gathered} a=a_0+a_1u+a_2v,\\ b=b_0+b_1u+b_2v. \end{gathered} \tag{6.5} \]

Then \(f(a,b)=a_0u+b_0v\). Linear independence of \(u,v\) gives

\[ \ker f=\mathfrak m\oplus\mathfrak m, \qquad\operatorname{coker}f\simeq k. \tag{6.6} \]

The kernel has \(k\)-dimension four and is generated over \(R\) by \((u,0),(v,0),(0,u),(0,v)\). The image is exactly \(\mathfrak m\).

For any map \(R^n\to M\), its kernel is a \(k\)-subspace of the finite-dimensional vector space \(R^n\). A finite \(k\)-basis of the kernel also generates it over \(R\), because \(k\) acts through its subring of scalars. Hence every module is pseudo-coherent. A finitely generated module is finite-dimensional over \(k\), since it is a quotient of some \(R^n\); a finite-dimensional module is finitely generated, using a \(k\)-basis. The coherent modules are precisely those of finite \(k\)-dimension.

The pullback consists of pairs \((a,b)\) with \(b-a\in\mathfrak m\). The map

\[ \begin{gathered} R\oplus\mathfrak m\to P,\\ (a,c)\longmapsto(a,a+c). \end{gathered} \tag{6.7} \]

is an \(R\)-linear isomorphism, with inverse \((a,b)\mapsto(a,b-a)\). Therefore \(P\) is generated by \((1,1),(0,u),(0,v)\), and has \(k\)-dimension five. The first projection has kernel \(0\oplus\mathfrak m\), and the second has kernel \(\mathfrak m\oplus0\), as submodules of \(P\). Both projections are surjective: the diagonal pair \((a,a)\) maps to \(a\) under either one.

Exercise 6.4 (synthesis). In all abelian groups, take \(\mathcal J\) to consist of all finite abelian groups. Classify the pseudo-coherent and coherent objects, verify (E), and show that (E) does not force every object of \(\mathcal J\) to be projective.

Solution. Every kernel of a map from a finite group is finite and belongs to \(\mathcal J\). Hence every abelian group is pseudo-coherent. A quotient of a finite group is finite, and a finite group covers itself. Thus coherent objects are exactly finite abelian groups.

In (4.1), both \(K\) and \(S\) are finite. Each fibre of \(E\twoheadrightarrow S\) is a coset of \(K\), so

\[ |E|=|K|\,|S|<\infty. \tag{6.8} \]

It follows that \(E\in\mathcal J\) and its identity supplies the cover required by (E). Theorem 4.1 therefore gives the exact abelian inclusion of finite abelian groups into all abelian groups.

For any prime \(p\), the object \(\mathbb Z/p\) of \(\mathcal J\) is not projective in the ambient category. If reduction \(\mathbb Z/p^2\twoheadrightarrow\mathbb Z/p\) had a section, the section would take \(1\) to a class congruent to \(1\) modulo \(p\). That class is not killed by \(p\), whereas every image of a homomorphism from \(\mathbb Z/p\) must be killed by \(p\). This contradiction proves the assertion. Condition (E) ensures finite covers of the relevant extensions while allowing such nonsplit extensions.

References