Finite module tests and Banach quotients

Written and self-checked by GPT-6.1 Sol (OpenAI), Ultra reasoning effort, October 2026. Original text: CC0.

An additive subcategory can lose a kernel because the required object is too large for its finiteness condition. A category of complete normed spaces can instead have every kernel and cokernel while its coimage–image comparison fails. We will compute both boundaries, and prove that the normed-space category still preserves strict quotients under pullback and strict embeddings under pushout.

Work in a fixed universe. Rings are unital, modules are left modules, and Banach spaces are over \(\mathbb C\), with bounded linear maps. Retain the complete general universal and module constructions in Kernels, cokernels and the abelian comparison. Retain Full subcategories and exact closure, §§1–2, Coherence from finite probes, §§2–5, and Coimages, images and composition of quotients, §§1–3. These full proofs remain the foundations; the new arguments check the finite-object and norm interfaces.

1. Finitely generated modules test the ring

Let \(R\) be a unital ring, possibly noncommutative. Call it left Noetherian when every left ideal is finitely generated. Retain the complete left-module proofs in Stafford–Prest, Noncommutative Algebra, Theorems 3.1 and 3.6, Corollaries 3.7–3.8, printed pp. 39–40 and 43–44. They identify the ascending chain condition with the maximal-member condition and finite generation of every submodule, and prove preservation under submodules, quotients and extensions. Consequently every finitely generated left module over a left Noetherian ring has all its submodules finitely generated. The zero module gives the rank-zero case, and every unital module over the zero ring is zero.

Let \(\operatorname{Mod}^f(R)\) be the full category of finitely generated left modules. Then \[ \begin{gathered} \operatorname{Mod}^f(R)\text{ is abelian} \\ \Longleftrightarrow \\ R\text{ is left Noetherian}. \end{gathered} \tag{1.1} \]

Proof. Suppose \(R\) left Noetherian. The retained result makes the ordinary kernel of a map between finitely generated modules finitely generated. Its ordinary cokernel is finitely generated because the images of generators of the target generate the quotient. Zero and finite direct sums are finitely generated. The full category of all modules is abelian by the previous lesson's complete proof. Thus the full-subcategory criterion, Theorem 2.2 of Full subcategories, applies and supplies the abelian structure and exact inclusion.

Conversely, if a left ideal \(I\) is not finitely generated, the quotient \(R\to R/I\) has both ends in \(\operatorname{Mod}^f(R)\), but has no kernel there. This is exactly the full rank-one-probe proof in the preceding lesson, §5, for an arbitrary left ideal. An abelian category must have that kernel. Therefore no such \(I\) exists. This proves (1.1), including the zero ring. \(\square\)

The side matters: (1.1) concerns left ideals and left modules throughout. No right-Noetherian assumption is added.

2. Finite presentation and the coherence convention

A module \(M\) is finitely presented if it has an exact presentation \(R^a\to R^b\to M\to0\), with \(a,b\) finite. Write \(\operatorname{Mod}^{fp}(R)\) for their full category. The categorical convention calls \(R\) left coherent when this category is abelian. We reconcile this with the already proved finite-free-probe convention, which calls the regular left module \(R\) coherent when every map \(R^n\to R\) has finitely generated kernel.

Proposition 2.1. These two conventions agree. When they hold, \(\operatorname{Mod}^{fp}(R)\) is the exact full abelian subcategory of modules coherent for finite free probes.

Proof. First assume the regular module coherent in the probe convention. The full proofs in Coherence, §§2–5, show that coherent modules are closed under finite biproducts and cokernels, and form an exact abelian subcategory. In Section 5 the cover condition is proved for arbitrary unital \(R\) by lifting a finite free basis; commutativity is unnecessary. Consequently finite free modules are coherent, and the cokernel of any map \(R^a\to R^b\) is coherent. Thus every finitely presented module is coherent. Conversely every coherent module is finitely presented by the complete proof of Proposition 5.2 there, applied to the module as its own finitely generated submodule. The two full subcategories agree, so the categorical condition follows, with exact inclusion.

Now suppose only that \(\operatorname{Mod}^{fp}(R)\) is abelian; exactness of its inclusion has not yet been established. Take any \(f:R^n\to R\), and let \(\kappa:K\to R^n\) be its kernel in that category. Both finite free modules and \(R\) are allowed test objects. The rank-one test identifies this intrinsic kernel with the ordinary one. Indeed, if \(\kappa(z)=0\), the two linear maps \(R\to K\) given by \(r\mapsto rz\) and zero factor the same zero map to \(R^n\); kernel uniqueness forces \(z=0\). Every \(x\in\ker_R f\) defines a linear map \(R\to R^n\), \(r\mapsto rx\), killed by \(f\), so factors through \(\kappa\). Evaluation at \(1\) puts \(x\) in its image. Hence \[ \begin{gathered} K\xrightarrow{\kappa}\ker_R f \\ \text{is a module isomorphism}. \end{gathered} \tag{2.1} \] Since \(K\) is finitely presented, it is finitely generated. This makes every such ordinary kernel finitely generated, so the regular module is coherent for finite free probes. The first implication now applies and proves exactness as well. \(\square\)

This argument distinguishes an intrinsic kernel from an ambient kernel before identifying them. It does not assume that an abelian full subcategory automatically has exact inclusion.

3. Closed subspaces supply Banach universal objects

The bounded-map convention includes exactly the continuous linear maps used in the source. A bound proves continuity by \(\|f(x)-f(y)\|\le\|f\|\|x-y\|\). Conversely, continuity at zero gives some \(\delta>0\) for which \(\|x\|<\delta\) implies \(\|f(x)\|<1\). For nonzero \(x\), apply this to \(\delta x/(2\|x\|)\) to get \[ \|f(x)\|\le2\|x\|/\delta. \] The zero vector causes no exception. Thus no continuous morphisms have been excluded.

Closed subspaces of a Banach space are complete: a Cauchy sequence converges in the ambient space, and closedness places its limit in the subspace. For a closed linear subspace \(K\subseteq X\), equip the quotient with \[ \|[x]\|_{X/K}=\inf_{k\in K}\|x+k\|_X. \tag{3.1} \] This is independent of the representative. It is positive on a nonzero coset because distance zero places \(x\) in the closed set \(K\). Homogeneity follows by replacing \(k\) with \(\lambda k\) for \(\lambda\ne0\), and the zero scalar is immediate. Adding representatives whose norms approach the two infima proves the triangle inequality. Thus (3.1) is a norm, and the quotient map is contractive.

The quotient is complete. Given a Cauchy sequence of cosets, choose a subsequence \(c_j\) whose successive differences have norm less than \(2^{-j-1}\). Choose representatives \(v_j\) of \(c_{j+1}-c_j\) with \(\|v_j\|<2^{-j}\). Starting with any representative of \(c_1\), its sum with \(\sum_jv_j\) converges in \(X\): the partial sums are Cauchy by the convergent geometric majorant, and \(X\) is complete. The contractive quotient sends these partial sums to the chosen subsequence. That subsequence converges, and the Cauchy property makes the full original sequence converge to the same coset.

For additivity and finite sums, retain the full proof of Additive localization and projector decompositions, Proposition 1.2 and its full Exercise 1 solution in §5. They prove completeness and both bounded universal properties, and the bounded isomorphism between the maximum and sum norms. We use their sum-norm biproduct \(X\oplus_1Y\), with \(\|(x,y)\|=\|x\|+\|y\|\); the later estimates depend on this chosen norm. These exact stronger norm computations remain in the earlier lesson.

For bounded linear \(f:X\to Y\), its nullspace is closed: if \(x_j\to x\) and \(f(x_j)=0\), boundedness gives \(f(x)=0\). With its subspace norm, inclusion is a kernel. A bounded map \(t:T\to X\) killed by \(f\) has exactly the same bounded element function as its unique factor through that nullspace.

Let \(L=\overline{f(X)}\). This is a closed linear subspace, and the quotient \[ q:Y\longrightarrow Y/L \tag{3.2} \] is a cokernel. A bounded \(t:Y\to T\) with \(tf=0\) vanishes on \(L\) by continuity. Its unique induced linear map \(\bar t([y])=t(y)\) is bounded: for every \(l\in L\), \[ \|\bar t([y])\| \le\|t\|\|y+l\|, \tag{3.3} \] and taking the infimum gives its bound by \(\|t\|\). Uniqueness follows from surjectivity of the quotient map. This proves every required universal property in \(\operatorname{Ban}\).

The complete additive comparison proof in the preceding lesson, §3, now applies. It identifies \[ \begin{gathered} \operatorname{Coim}f=X/\ker f,\\ \operatorname{Im}f=\overline{f(X)},\\ u_f([x])=f(x). \end{gathered} \tag{3.4} \] The image has its inherited Banach norm. The last map is bounded by the same infimum argument, and is the specified canonical comparison, rather than a chosen abstract identification.

4. A dense injection defeats the abelian axiom

Use \(H=\ell^2(\mathbb N,\mathbb C)\), with the square-sum norm. Its norm properties follow from finite sums: minimizing \(\sum_{n\le N}(|x_n|-t|y_n|)^2\) over real \(t\) proves finite Cauchy–Schwarz, with the zero \(y\) case immediate. Expanding \(\sum_{n\le N}|x_n+y_n|^2\) and applying that inequality gives the finite triangle inequality. Taking increasing limits proves both square summability of the sum and the triangle inequality in \(H\); positivity and homogeneity are coordinatewise.

Its completeness follows directly. A norm-Cauchy sequence has Cauchy coordinates, hence limits \(a_n\). Its norms are bounded by some \(M\). Passing to the limit in any finite sum gives \(\sum_{n\le N}|a_n|^2\le M^2\), so \(a\in H\). Once two sufficiently late vectors are within \(\varepsilon\), passing to the coordinate limit in every finite sum gives norm distance at most \(\varepsilon\) from \(a\). Thus the vectors converge to \(a\) in \(H\).

Define \[ D:H\to H,\qquad (Da)_n=a_n/n. \tag{4.1} \] It is linear, bounded with norm at most \(1\), and injective. Every finitely supported vector lies in its image, since multiplying its finitely many nonzero coordinates by their indices gives another element of \(H\). Such vectors are dense by convergence of the square-sum tails. But \(h=(1/n)_n\) is in \(H\) and not in the image: its only coordinatewise preimage would be \((1,1,\ldots)\), which is not square summable. The square summability of \(h\) follows, for example, by bounding the contribution of \(2^j\le n<2^{j+1}\) by \(2^{-j}\).

Therefore \(D(H)\) is dense and not closed. Section 3 gives \[ \begin{gathered} \ker D=0,\qquad \operatorname{coker}D=0,\\ \operatorname{Coim}D=H=\operatorname{Im}D,\\ u_D=D. \end{gathered} \tag{4.2} \] This comparison is not invertible, since \(h\) has no preimage. Thus \(\operatorname{Ban}\) is not abelian. It has all kernels and cokernels, and \(D\) is both monic and epic by the complete general zero-object tests in the preceding lesson, §1. In particular it is not balanced either.

5. The two quasi-abelian stability axioms

An additive category with kernels and cokernels is quasi-abelian when pullbacks of strict epimorphisms are strict epimorphisms and pushouts of strict monomorphisms are strict monomorphisms. This is the convention of Schneiders, Definition 1.1.3; strictness uses the same specified comparison as (3.4).

We first retain the exact bridge to universal arrows. By the full proof of Theorem 3.1 of Coimages, a strict epimorphism is a coequalizer of its kernel pair. In an additive category this is the cokernel of the difference of those projections; by the full biproduct calculation in the previous lesson, it is the cokernel of its ordinary kernel arrow. Conversely any cokernel is a coequalizer of the defining arrow and zero, hence is strict epic by that same theorem. Reversing its complete argument shows that strict monomorphisms are exactly kernels. The previous lesson's opposite and finite-diagram constructions check all hypotheses. We can therefore transport each strict mono or epi through the unique bounded isomorphism to a canonical closed-subspace inclusion or quotient from Section 3.

For pullbacks, take a closed \(K\subseteq X\), the quotient \(q:X\to X/K\), and any bounded \(g:Z\to X/K\). Their pullback is the closed subspace \[ \begin{gathered} E\subseteq X\oplus_1 Z, \\ E=\{(x,z):q(x)=g(z)\}. \end{gathered} \tag{5.1} \] Indeed it is the kernel of the bounded map \((x,z)\mapsto q(x)-g(z)\), with the universal property already proved in Section 3. Let \(p:E\to Z\) be projection. It is surjective, and its kernel is \(K\oplus0\). Two lifts of the same \(z\) differ by this kernel, so the induced map \[ \bar p:E/(K\oplus0)\longrightarrow Z \tag{5.2} \] is a linear bijection. It is bounded by \(1\). Its inverse is bounded as well. For each \(z\) and each \(\varepsilon>0\), the quotient norm allows a lift of \(g(z)\) with \[ \|x\|<\|g(z)\|+\varepsilon. \tag{5.3} \] Consequently the class of \((x,z)\), which is independent of that lift, has norm at most \((1+\|g\|)\|z\|+\varepsilon\). Taking the infimum over lifts and then letting \(\varepsilon\) decrease to zero gives \[ \|\bar p^{-1}(z)\| \le(1+\|g\|)\|z\|. \tag{5.4} \] Thus (5.2) is a bounded isomorphism. Section 3 identifies it with the coimage comparison of \(p\), so \(p\) is strict epic.

For pushouts, take a closed \(K\subseteq X\) and any bounded \(a:K\to Z\). The relation subspace \[ \begin{gathered} N=\{(k,-a(k)):k\in K\}, \\ N\subseteq X\oplus_1Z. \end{gathered} \tag{5.5} \] is closed. Convergence of a sequence in \(N\) makes its first coordinates converge to some \(k\in K\), and continuity of \(a\) determines the second-coordinate limit. Hence Section 3 and the retained signed pushout construction give the Banach pushout \[ \begin{gathered} P=(X\oplus_1Z)/N, \\ j(z)=[0,z]. \end{gathered} \tag{5.6} \] The map \(j\) is injective and bounded by \(1\). Put \(C=\max(1,\|a\|)\). For every \(k\in K\), \[ \begin{aligned} \|z\|&\le\|z+a(k)\|\\ &\quad+\|a\|\|k\|\\ &\le C\bigl(\|k\|\\ &\quad+\|z+a(k)\|\bigr). \end{aligned} \tag{5.7} \] Taking the infimum over representatives \((-k,z+a(k))\) of \(j(z)\) gives \(\|z\|\le C\|j(z)\|\). Thus \(j\) has bounded inverse on its range. That range is closed: a convergent sequence \(j(z_n)\) makes \(z_n\) Cauchy by this inequality, and its limit maps to the given limit in \(P\).

One can also exhibit its kernel property explicitly. The bounded map \[ \rho:P\to X/K,\qquad \rho([x,z])=[x] \tag{5.8} \] is well defined and contractive by the quotient infimum. Its kernel consists precisely of classes with \(x\in K\); such a class is \([0,z+a(x)]\). Thus its kernel is \(j(Z)\). A bounded map into \(P\) killed by \(\rho\) factors uniquely through \(j\), and the bound on \(j^{-1}\) makes that factor bounded. Therefore \(j\) is a kernel, hence strict monic.

Transporting these two computations through the bounded structural isomorphisms covers every strict epimorphism and strict monomorphism, not just the displayed models. Both defining axioms hold, so \(\operatorname{Ban}\) is quasi-abelian. Each stability proof includes its norm bound and its universal identification.

6. Four graded exercises with full solutions

Exercise 1 (introductory: a noncommutative finite test). Let \(R\) be the algebra of upper triangular \(2\times2\) matrices over \(\mathbb C\). Show that \(\operatorname{Mod}^f(R)\) is abelian. For the left-linear map \(f:R\to R\), \(f(r)=r e_{11}\), compute its kernel, image and coimage comparison. Check why the order of multiplication matters.

Solution. Every left ideal is a complex linear subspace of the three-dimensional vector space \(R\). A finite complex basis also generates it as a left \(R\)-module, since scalar matrices belong to \(R\). Hence \(R\) is left Noetherian, and (1.1) applies.

Writing \(r=\left(\begin{smallmatrix}a&b\\0&c\end{smallmatrix}\right)\), we get \(f(r)=a e_{11}\). Thus \(\ker f=R e_{22}\), consisting of matrices with \(a=0\), and \(\operatorname{im}f=R e_{11}=\mathbb C e_{11}\). The canonical comparison sends \([r]\in R/(R e_{22})\) to \(a e_{11}\). Its inverse sends \(a e_{11}\) to \([a e_{11}]\); their composites are identities because \(r-a e_{11}\in R e_{22}\). Both maps are left linear, so this is the specified isomorphism.

Right multiplication by \(e_{11}\) is left linear by associativity. Left multiplication by \(e_{11}\) would not be: with \(s=e_{12}\) and \(r=e_{22}\), one has \(e_{11}(sr)=e_{12}\) but \(s(e_{11}r)=0\). This explains the order in the exercise.

Exercise 2 (intermediate: diagonal strictness). Let \((w_n)\) be a bounded sequence of nonzero complex numbers. On \(H=\ell^2\), set \((D_wx)_n=w_nx_n\). Prove it is monic and epic, and is strict exactly when \(\inf_n|w_n|>0\). If the infimum is zero, exhibit a vector outside its range.

Solution. Boundedness follows from \(\|D_wx\|\le(\sup_n|w_n|)\|x\|\). Nonzero weights give injectivity, and every finitely supported vector lies in its image. Thus the image is dense, and the kernel and cokernel are zero by Section 3. The general tests make it monic and epic. Its canonical comparison is itself, as in (4.2).

If \(c=\inf_n|w_n|>0\), coordinatewise division by \(w_n\) sends every \(y\in H\) to an element of norm at most \(c^{-1}\|y\|\). It is a bounded inverse, so \(D_w\) is strict. Conversely, a bounded inverse with norm \(B\) would give \(1/|w_n|\le B\) by applying it to the unit vector \(e_n\), forcing a positive lower bound.

If the infimum is zero, choose distinct indices \(n_j\) with \(|w_{n_j}|<2^{-j}\). Distinctness can be maintained because any finite set of nonzero weights has positive minimum. Define \(y_{n_j}=w_{n_j}\), and set other coordinates zero. Its square sum is bounded by \(\sum_j4^{-j}\), so \(y\in H\). A preimage would have coordinate \(1\) at every \(n_j\), which is not square summable. Hence no preimage exists.

Exercise 3 (hard: complements and split squares). For a closed \(K\subseteq X\), with inclusion \(\iota:K\to X\), prove the quotient \(q:X\to X/K\) has a bounded linear section exactly when there is a bounded linear \(B:X\to K\) with \(B\iota=1_K\). In that case give a bounded section of every pullback in (5.1), and a bounded retraction of every pushout embedding in (5.6).

Solution. If \(s:X/K\to X\) is a bounded section, \(1_X-sq\) has range in \(K\), since \(q(1_X-sq)=0\). Its codomain restriction defines a bounded \(B:X\to K\) with \(\iota B=1_X-sq\) and \(B\iota=1_K\), using the subspace norm. Conversely, given \(B\), set \[ s([x])=(1_X-\iota B)x. \tag{6.1} \] It is independent of the representative because \(B\iota=1_K\). Taking the infimum of \(\|(1_X-\iota B)(x+\iota k)\|\le\|1_X-\iota B\|\|x+\iota k\|\) over \(k\in K\) proves boundedness. Since \(\iota Bx\in K\), \(qs=1\). This proves both directions.

For (5.1), the map \(z\mapsto(sg(z),z)\) lands in \(E\), is linear, and satisfies the bound \[ \|(sg(z),z)\|\le(1+\|s\|\|g\|)\|z\|. \] Projection to \(Z\) gives the identity, so it is the requested section. For (5.6), define \[ r([x,z])=z+a(Bx). \tag{6.2} \] The numerator map kills \((k,-a(k))\), since \(Bk=k\), so factors uniquely through the quotient. Its norm is bounded by \(\max(1,\|a\|\|B\|)\) against the sum norm, hence against the quotient norm after taking the infimum. Also \(rj=1_Z\). The formulas prove split stability under the additional complement hypothesis; no assertion that every closed subspace has such a projection is needed.

Exercise 4 (advanced: why strictness is essential). For \(D\) and \(h\) of Section 4, pull \(D:H\to H\) back along \(g:\mathbb C\to H\), \(g(\lambda)=\lambda h\). Compute the resulting map to \(\mathbb C\). Then push the monomorphism \(D:H\to H\) out along \(H\to0\). Explain why the two results do not contradict quasi-abelianity.

Solution. The pullback is the closed subspace of \(H\oplus_1\mathbb C\) where \(Dx=\lambda h\). If \(\lambda\ne0\), division by \(\lambda\) would put \(h\) in \(D(H)\), contradicting Section 4. Thus \(\lambda=0\), and injectivity of \(D\) gives \(x=0\). The pullback object is zero and its projection is \(0\to\mathbb C\), which is not epic: the identity and zero maps on \(\mathbb C\) agree after it but differ.

The signed pushout is the Banach cokernel of \(D\), namely \(H/\overline{D(H)}=0\). Its induced map from the target copy of \(H\) is \(H\to0\), which is not monic: the identity and zero maps into \(H\) agree after it but differ. The starting arrow \(D\) is both epic and monic, but is not strict. The stability axioms concern strict epi and strict mono arrows, so both counterexamples are compatible with the full proofs in Section 5. They show why replacing those hypotheses by ordinary epicity or monicity would be false.

7. References