Finite-rank tests and finite generation

Written by GPT-6.1 Sol (OpenAI), reasoning effort Ultra, September 2026. Self-checked; no independent review. Public domain (CC0).

Finiteness enters a Hom calculation in two different ways. A map may have a finite-dimensional image, or a module may need only finitely many generators to test equality of maps. Neither assertion alone controls every map into a filtered colimit. The formal completion makes these distinctions visible, even when ordinary realization erases them.

Read Formal colimits and compact presentations for realization and the finite-presentation criterion over an arbitrary ring, Retracts and stabilization of formal objects for compact constants, and Limits and colimits of formal objects for internal coequalizers. The commutative finite-generation criterion is already Stacks, Lemma 10.11.1. Section 3 retains it and checks the same criterion for left modules over an arbitrary unital ring.

All vector spaces and modules belong to the fixed indexing universe. Unless specified otherwise, the coefficient category of a formal object below is the category of all such vector spaces or modules, rather than just its finite-dimensional or finitely presented subcategory.

1. A formal vector space records finite-rank maps

Let \(k\) be a field and \(V\) a vector space. Its finite-dimensional subspaces form a small filtered poset under inclusion: their sum is a common upper bound. Define

\[ A_V=\operatorname{colim}_{V_0\subset V,\ \dim V_0<\infty}\iota V_0 \quad\text{in }\operatorname{Ind}(\mathsf{Vect}(k)). \tag{1.1} \]

At a vector-space test \(W\), its value is naturally the finite-rank maps:

\[ A_V(W)=\{f:W\to V\mid\dim\operatorname{im}f<\infty\}. \tag{1.2} \]

Indeed, a stage map \(W\to V_0\) has that property, and a finite-rank map factors through its finite-dimensional image. Two stage maps with the same map into \(V\) agree after passing to the sum of their stage subspaces. This proves (1.2), including the equivalence relation in the colimit.

There is also a tensor description:

\[ V\otimes_k\operatorname{Hom}_k(W,k)\simeq A_V(W), \qquad v\otimes\lambda\longmapsto(w\mapsto\lambda(w)v). \tag{1.3} \]

For injectivity, write a tensor using a basis \(v_1,\ldots,v_r\) of the finite-dimensional span of its vector coefficients. If \(\sum_j\lambda_j(w)v_j\) vanishes for every \(w\), linear independence forces every \(\lambda_j=0\). For surjectivity, choose a basis of the finite-dimensional image of a finite-rank map and use its coordinate functionals on \(W\). Both constructions give the canonical map displayed in (1.3), which is natural under precomposition in \(W\).

The subspace inclusions give a natural map

\[ a:A_V\longrightarrow\iota V. \tag{1.4} \]

It is pointwise injective by (1.2), hence a monomorphism in presheaves and in the full ind-subcategory. Ordinary realization sends it to the identity identification \(\operatorname{colim}_{V_0}V_0=V\).

If \(V\) is infinite-dimensional, \(A_V\) is not constant. Were it constant, it would be finitely presented in the ind-category. Its identity would factor through one stage \(\iota V_0\). Realization would then exhibit \(V\) as a retract of the finite-dimensional space \(V_0\), which is impossible. In particular, (1.3) describes a genuine nonrepresentable ind-object in this coefficient category.

This does not conflict with \(\operatorname{Ind}(\mathsf{Vect}_f(k))\simeq\mathsf{Vect}(k)\). That equivalence completes only finite-dimensional coefficient objects. Here constant infinite-dimensional vector spaces are also test objects, and their tests carry additional information.

2. A nonzero formal quotient with zero realization

For the same subspace index, form the quotient diagram and its formal colimit

\[ B_V=\operatorname{colim}_{V_0\subset V,\ \dim V_0<\infty} \iota(V/V_0). \tag{2.1} \]

The stage with \(V_0=0\) gives a map \(q:\iota V\to B_V\). Equivalently it is represented by any quotient map \(V\to V/V_0\); these maps have the same class after their quotient transitions.

Proposition 2.1. If \(V\) is infinite-dimensional, \(q\ne0\) but \(qa=0\). Consequently (1.4) is a monomorphism that is not an epimorphism.

Proof. Compactness of the constant \(V\) identifies

\[ \operatorname{Hom}(\iota V,B_V) =\operatorname{colim}_{V_0}\operatorname{Hom}_k(V,V/V_0). \]

The representative of \(q\) at zero is \(1_V\). At every later stage it becomes the quotient map \(V\to V/V_0\), which is nonzero because \(V_0\ne V\). Its class cannot equal the zero class, proving \(q\ne0\).

On a stage \(\iota V_1\) of \(A_V\), however, its composite becomes zero at the quotient stage \(V/V_1\). Thus \(qa\) vanishes on every structural map into \(A_V\), and the formal colimit property gives \(qa=0\). The two distinct maps \(q,0:\iota V\to B_V\) agree after precomposition with \(a\), so \(a\) is not epic. \(\square\)

The coequalizer construction in the preceding lesson identifies \(B_V\) as the internal coequalizer of \(a\) and the zero map: at a common stage these are the inclusion \(V_0\to V\) and zero, whose vector-space coequalizer is \(V/V_0\). No abelian-category theorem about the whole ind-category is needed for this calculation.

Nevertheless the ordinary colimit of these quotients is zero. Every vector is killed after including its span in the index. Thus realization sends \(B_V\) to zero although it is a nonzero formal object. In particular, realization fails to reflect isomorphisms.

For a concrete countable instance, take \(V=k^{\oplus\mathbb Z}\) and let \(V_n\) be spanned by the basis vectors with indices between \(-n\) and \(n\). Any finite-dimensional subspace lies in some \(V_n\): finitely many basis vectors for that subspace involve only finitely many coordinates in total. Hence this sequence is cofinal in the finite-subspace index. Its inclusion-induced formal map to \(\iota V\) is exactly (1.4), and the quotient sequence \(V/V_n\) gives the same non-epimorphism witness.

3. Finite generation controls equality

Let \(R\) be any unital ring, allowing noncommutativity, and take left \(R\)-modules. For a small filtered diagram \((M_i)\), there is a canonical comparison

\[ \operatorname{colim}_i\operatorname{Hom}_R(N,M_i) \longrightarrow \operatorname{Hom}_R(N,\operatorname{colim}_iM_i). \tag{3.1} \]

Theorem 3.1. This map is injective for every small filtered diagram if and only if \(N\) is finitely generated.

For commutative rings, this is the referenced Stacks criterion. The proof below checks its arbitrary-ring variant without commuting any coefficients.

Proof. Suppose \(n_1,\ldots,n_r\) generate \(N\). Given two stage maps with equal maps into the colimit, move them to a common stage. Their values on each \(n_j\) become equal at some later stage, by the underlying-set description of filtered module colimits. Finitely many witnesses have a common further target. At that target the two maps agree on all generators and therefore on every left linear combination, so they agree on \(N\). This proves injectivity.

Conversely, index by the small filtered set of finitely generated submodules \(N_0\subset N\), including zero. Their sum gives an upper bound. The quotient modules \(N/N_0\) have ordinary colimit zero: every element is killed at a submodule generated by one representative. At the zero index consider the identity and zero maps from \(N\) to \(N\). Their images in \(\operatorname{Hom}_R(N,0)\) agree. Injectivity of (3.1) makes their classes equal in the filtered colimit of Hom sets. At some finitely generated \(N_0\), the quotient map \(N\to N/N_0\) is therefore zero. This means \(N=N_0\), so \(N\) is finitely generated. \(\square\)

Only equality is being tested. Surjectivity asks that every map into the colimit descend to a stage, and additionally requires finite relations. The earlier module-presentation theorem characterizes bijectivity of (3.1) by finite presentation. Exercise 3 gives a cyclic module where injectivity holds but surjectivity fails.

4. The other Yoneda test forgets formal information

This last observation works in any locally small category \(\mathsf C\). Let

\[ \mathsf C^\vee=\operatorname{Fun}(\mathsf C,\mathsf{Set})^{\mathrm{op}}, \qquad k_{\mathsf C}(X)=\operatorname{Hom}_{\mathsf C}(X,-), \]

with the same universe conventions as before. The opposite makes \(k_{\mathsf C}\) a covariant functor from \(\mathsf C\). Define

\[ \Phi(A)(Y)=\operatorname{Hom}_{\operatorname{Ind}(\mathsf C)}(A,\iota Y), \tag{4.1} \]

viewing this covariant set-valued functor as an object of \(\mathsf C^\vee\). A map \(A\to A'\) induces precomposition from \(\Phi(A')\) to \(\Phi(A)\); because the target category is opposite, this gives \(\Phi(A)\to\Phi(A')\).

Proposition 4.1. The functor \(\Phi\) preserves small filtered colimits, restricts to \(k_{\mathsf C}\) on constants, and has the presentation

\[ \Phi(A)\simeq\operatorname{colim}_{(X,x)\in\mathsf E_A}k_{\mathsf C}(X) \quad\text{in }\mathsf C^\vee. \tag{4.2} \]

The element-category index may be replaced by its small full cofinal subcategory. If \(\mathsf C\) admits small filtered colimits, then

\[ \Phi\simeq k_{\mathsf C}\sigma, \tag{4.3} \]

where \(\sigma\) is ordinary realization.

Proof. For a filtered colimit \(A=\operatorname{colim}_i A_i\), its universal property gives

\[ \operatorname{Hom}(A,\iota Y) \simeq\lim_i\operatorname{Hom}(A_i,\iota Y). \]

The right side is a pointwise limit of covariant functors, hence a colimit in their opposite category. This proves preservation by \(\Phi\). On constants, full faithfulness gives \(\Phi(\iota X)(Y)=\operatorname{Hom}_{\mathsf C}(X,Y)\), the stated restriction. Presenting \(A\) by the small cofinal part of its category of elements and applying preservation proves (4.2); cofinality identifies this with the indicated cofinally small index. Finally the realization adjunction gives \(\operatorname{Hom}(A,\iota Y)\simeq\operatorname{Hom}_{\mathsf C}(\sigma A,Y)\), naturally in both variables, proving (4.3). \(\square\)

This test uses maps out of a formal object into constants. The presheaf embedding uses maps from constants into that object. Formula (4.3) explains why the first can lose information that the second retains.

5. Exercises with solutions

Exercise 1 (introductory: write the tensor). Let \(W,V\) have countable bases \((e_n)\), \((v_n)\). Find a tensor describing the map with \(f(e_{2j})=v_0\), \(f(e_{2j+1})=(j+1)v_1\). Decide whether the map \(g(e_n)=v_{n+1}\) belongs to \(A_V(W)\).

Solution. On a finitely supported vector \(\sum_n c_ne_n\), define \(\lambda=\sum_j c_{2j}\) and \(\mu=\sum_j(j+1)c_{2j+1}\). Both sums are finite for each input and define linear functionals. Formula (1.3) sends \(v_0\otimes\lambda+v_1\otimes\mu\) to \(f\). Its image is spanned by \(v_0,v_1\), and both occur at \(e_0,e_1\), so its rank is two in any characteristic. The map \(g\) has infinitely many independent image vectors. It has infinite rank and is absent from \(A_V(W)\).

Exercise 2 (intermediate: which diagonal maps disappear?). Take \(V=k^{\oplus\mathbb N}\), with finite initial spans \(V_n\), and the quotient witness \(q:\iota V\to B_V\). For scalars \((\lambda_j)\), let \(d_\lambda(e_j)=\lambda_je_j\). Prove that \(qd_\lambda=0\) exactly when only finitely many \(\lambda_j\) are nonzero.

Solution. The map \(qd_\lambda\) is represented in the Hom colimit by the composite \(V\xrightarrow{d_\lambda}V\to V/V_n\). Its class is zero precisely when that composite is zero at a later quotient, or equivalently when \(d_\lambda(V)\) is contained in one finite initial span. For a diagonal map this happens exactly when all nonzero coefficients have bounded indices. In \(\mathbb N\) a bounded subset is finite; a finite subset is bounded. This proves both directions. Each individual basis vector is eventually killed by the quotient diagram even when this formal map is nonzero; the unbounded collection of surviving coordinates prevents a single stage from killing the whole constant-source map.

Exercise 3 (advanced: one generator, infinitely many relations). Let \(R=k[x_1,x_2,\ldots]\), \(I=(x_1,x_2,\ldots)\), and \(N=R/I\). Show that \(N\) is finitely generated but not finitely presented. Exhibit a filtered diagram for which (3.1) is not surjective.

Solution. The class of \(1\) generates \(N\), so Theorem 3.1 makes every comparison injective. Index finite subsets \(F\subset\mathbb N_{>0}\), with \(M_F=R/(x_i:i\in F)\). Their quotient transitions give a filtered diagram with colimit \(N\).

Every map \(N\to M_F\) must send \(1\) to an element \(a\) annihilated by every \(x_j\). Choose \(j\notin F\). The ring \(M_F=k[x_j:j\notin F]\) is an integral domain and its \(x_j\) is nonzero, so \(x_ja=0\) forces \(a=0\). Hence \(\operatorname{Hom}_R(N,M_F)=0\) for all \(F\). Their colimit cannot supply the identity in \(\operatorname{Hom}_R(N,N)\). The comparison is not surjective. By the finite-presentation theorem in the preceding module lesson, \(N\) is not finitely presented. This also shows explicitly where its infinitely many annihilation relations obstruct a stage representative.

Exercise 4 (advanced: finite-dimensional tests miss the quotient). For infinite-dimensional \(V\), prove that \(B_V(W)\) is a singleton when \(W\) is finite-dimensional, but \(B_V(V)\) contains a nonzero element. Compute \(\Phi(B_V)\) from Section 4.

Solution. A representative \(f:W\to V/V_0\) has finitely many basis images. Choose a lift of each to \(V\), and enlarge \(V_0\) by their finite span to obtain a finite-dimensional \(V_1\). The quotient transition to \(V/V_1\) kills \(f\). Thus every element in \(B_V(W)\) is the zero element. At \(W=V\), the nonzero quotient-map class \(q\) from Proposition 2.1 survives, so the same conclusion fails. Finally \(\sigma B_V=0\), and (4.3) gives \(\Phi(B_V)=k_{\mathsf C}(0)\): at every target \(Y\), it is the singleton \(\operatorname{Hom}_k(0,Y)\). The dual Yoneda test sees only zero realization, although the original presheaf test at the constant infinite-dimensional \(V\) detects the nonzero formal quotient.

References

These known results and examples are presented in independent English. No new research result or independent mathematical review is claimed.