Draft lesson.

Finite subcoalgebras and profinite algebra reconstruction

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

A coalgebra operation produces a finite tensor from each vector. Its equations force that vector into a finite-dimensional subcoalgebra. This makes the coalgebra recoverable from finite pieces, and its continuous dual algebra recoverable from finite algebra quotients. We will prove those claims, identify every reconstruction map, and construct the universal completion of a discrete algebra. The examples explain why a right subcomodule can require a larger coalgebra, why quotient topology depends on closedness, and why completion can destroy an entire nonzero algebra.

Prerequisites: Completed tensors and continuous algebra objects, §§1–5; Pro representations and continuous duality, §§1–4; Ind representations and the locally finite coreflector, §§1–4; and Formal colimits and compact presentations, §§1–2.

Fix a discrete field \(k\) and a small discrete group \(G\). A \(G\)-coalgebra is a coalgebra with a \(G\)-action by coalgebra automorphisms. We consider those whose underlying representation is locally finite. A continuous \(G\)-algebra is an algebra object in \(\mathsf L(G)\), with its jointly continuous multiplication and action by algebra automorphisms. The continuous dual \(C\) and finite-topology full dual \(D\) give the inverse contravariant equivalences of the first prerequisite. Profinite-dimensional algebras here are inverse limits of finite-dimensional \(k\)-algebras, with their finite quotient topology. Zero coalgebras and zero unital algebras are allowed.

1. Finite subcomodules and coefficient coalgebras

A right comodule over a coalgebra \(H\) has \(\rho\colon M\to M\otimes H\) satisfying \((\rho\otimes1)\rho=(1\otimes\Delta)\rho\) and \((1\otimes\epsilon)\rho=1_M\). A subcomodule \(N\subset M\) satisfies \(\rho(N)\subset N\otimes H\). A subcoalgebra \(K\subset H\) has the stronger property \(\Delta(K)\subset K\otimes K\), with the restricted counit. A comodule map obeys \((f\otimes1)\rho_M=\rho_Nf\).

Lemma 1.1. Every finite subset of a comodule lies in a finite-dimensional subcomodule.

Proof. For \(m\in M\), write \(\rho(m)=\sum_{i=1}^r m_i\otimes h_i\) with the \(h_i\) linearly independent, by expanding in a basis of their finite span. Coassociativity gives

\sum_i\rho(m_i)\otimes h_i=\sum_i m_i\otimes\Delta(h_i). \tag{1.1}

Extend the coordinate functionals on \(\operatorname{span}(h_i)\) to \(H\). Applying the \(i\)-th functional to the last factor in (1.1) puts \(\rho(m_i)\) in \(\operatorname{span}(m_1,\ldots,m_r)\otimes H\). Thus that finite span is a subcomodule. The counit gives \(m=\sum_i m_i\epsilon(h_i)\), so it contains \(m\). For finitely many vectors, sum their finite subcomodules; the sum is still finite-dimensional and a subcomodule. \(\square\)

Let \(N\) be a finite-dimensional right comodule with basis \((n_i)_{i=1}^r\), and write

\rho(n_j)=\sum_i n_i\otimes h_{ij}. \tag{1.2}

Comparing the coefficients of \(n_i\) in the coassociativity equation, and then in the counit equation, gives

\Delta(h_{ij})=\sum_\ell h_{i\ell}\otimes h_{\ell j},\qquad \epsilon(h_{ij})=\delta_{ij}. \tag{1.3}

The span \(K(N)\) of these coefficients is therefore a finite-dimensional subcoalgebra. It is independent of the basis: it is the span of all \((\lambda\otimes1)\rho(n)\) for \(\lambda\in N^*\), \(n\in N\), which both contains and is contained in the displayed coefficient span.

Apply Lemma 1.1 to the regular right comodule \(H\), whose coaction is \(\Delta\). A finite-dimensional subspace \(E\subset H\) lies in a finite-dimensional right subcomodule \(N\). For its basis in (1.2), the other counit identity gives \(n_j=\sum_i\epsilon(n_i)h_{ij}\). Consequently \(E\subset N\subset K(N)\). We have proved the fundamental theorem of coalgebras: every finite-dimensional subspace lies in a finite-dimensional subcoalgebra, with the actual finite coefficient construction.

2. Least finite subcoalgebras and invariant pieces

Intersections of two subcoalgebras are subcoalgebras. For subspaces \(K,L\subset H\), choose complements giving

H=(K\cap L)\oplus K_0\oplus L_0\oplus T,\qquad K=(K\cap L)\oplus K_0,\quad L=(K\cap L)\oplus L_0. \tag{2.1}

The corresponding direct-sum tensor decomposition shows that \((K\otimes K)\cap(L\otimes L)=(K\cap L)\otimes(K\cap L)\). For \(h\in K\cap L\), its coproduct lies in this intersection. This proves the assertion with the restricted counit.

For finite-dimensional \(E\subset H\), choose among its finite-dimensional containing subcoalgebras one, \(K_E\), of least dimension; §1 supplies at least one. If any subcoalgebra \(L\) contains \(E\), then \(K_E\cap L\) also contains \(E\) and has dimension at most that minimum. It must equal \(K_E\). Thus \(K_E\subset L\), and \(K_E\) is the unique least subcoalgebra containing \(E\). This argument proves leastness even against infinite-dimensional containing subcoalgebras.

Now let \(H\) be a locally finite \(G\)-coalgebra. Every finite subspace is contained in a finite-dimensional invariant subspace \(E\), by summing the finite orbit spans of a basis. Each \(gK_E\) is a subcoalgebra containing \(gE=E\); leastness gives \(K_E\subset gK_E\), and their equal finite dimensions give equality. Hence \(K_E\) is invariant. The finite-dimensional invariant subcoalgebras form a small directed set: their sum is finite-dimensional and invariant, and its coproduct lies in the tensor of that sum with itself. Their union is exactly \(H\). We have obtained its canonical filtered presentation, with literal inclusions.

3. Filtered coalgebra colimits and finite compactness

Let \(\mathsf{Coalg}_{\mathrm{lf}}(G)\) denote this coalgebra category, and let \(\mathcal K\) be a small skeleton of its finite-dimensional objects. Such a skeleton exists: on each \(k^n\), coproduct and counit have finitely many scalar coefficients, and the action is a function \(G\to\operatorname{GL}_n(k)\); these data form a set. Coalgebra homomorphisms generally do not form vector spaces, so the Ind formulas below concern Hom sets.

For any small filtered diagram \((H_i)\), take the underlying colimit \(H\) in locally finite representations, with maps \(u_i\). Define its coproduct and counit by

\Delta_Hu_i=(u_i\otimes u_i)\Delta_i,\qquad \epsilon_Hu_i=\epsilon_i. \tag{3.1}

The right sides are compatible because all transitions are coalgebra maps, so the vector colimit gives unique linear maps. The coalgebra identities and equivariance hold on every \(u_i(h)\), which span the colimit, hence everywhere. Any coalgebra cocone gives its unique underlying colimit map, and (3.1) makes that map a coalgebra map. This proves the actual colimit universal property.

Theorem 3.1. The objects compact for filtered colimits in \(\mathsf{Coalg}_{\mathrm{lf}}(G)\) are precisely its finite-dimensional coalgebras.

Proof. For a coalgebra map \(f\colon K\to H=\varinjlim_iH_i\), finite-dimensional compactness in locally finite representations gives an equivariant linear lift \(f_i\colon K\to H_i\) with \(u_if_i=f\). This compactness holds for every small group, by the third prerequisite. The counit defect is already zero: \(\epsilon_i f_i=\epsilon_Hf=\epsilon_K\). Its coproduct defect is the linear map

d_i=\Delta_i f_i-(f_i\otimes f_i)\Delta_K\colon K\to H_i\otimes H_i. \tag{3.2}

Its image becomes zero in \(H\otimes H\). The canonical comparison \(\varinjlim_i(H_i\otimes H_i)\simeq H\otimes H\) is an isomorphism. Tensor preserves colimits by the first prerequisite; to replace its two-stage colimit by the diagonal, put every finite list of pairs of stage vectors into one common later stage. Any equality uses finitely many diagram relations, which become equal at another common stage; filteredness also equalizes any finite list of parallel arrows. These are exactly the inverse and equality tests for the comparison.

Choose a finite basis of \(K\). Each of its defect vectors in (3.2) therefore dies at some later stage, and one common later stage kills them all. If \(q\colon H_i\to H_j\) is that transition, the defect of \(qf_i\) is \((q\otimes q)d_i\), so it is zero. Counits are still preserved. Thus \(qf_i\) is an actual equivariant coalgebra lift. Conversely, two stage coalgebra maps with equal colimit maps become equal at a common later stage by underlying finite-dimensional compactness. Their transitions remain coalgebra maps. This proves both surjectivity and injectivity of the compact-Hom comparison, including general filtered categories. If an arbitrary coalgebra \(H\) is compact, apply its compactness to the canonical union of finite-dimensional invariant subcoalgebras in §2. Its identity factors through one literal finite-dimensional subcoalgebra, so its underlying vector space is contained in that finite-dimensional image. Thus \(H\) is finite-dimensional, proving the converse. \(\square\)

Section 2 and Theorem 3.1 give the equivalence

\operatorname{Ind}(\mathcal K)\simeq\mathsf{Coalg}_{\mathrm{lf}}(G),\qquad \operatorname{Hom}(\varinjlim_iK_i,\varinjlim_jL_j) \simeq\varprojlim_i\varinjlim_j\operatorname{Hom}_{G,\mathrm{coalg}}(K_i,L_j). \tag{3.3}

Here the map takes stage arrows to their colimit composites. Its inverse restricts a map to each \(K_i\), lifts it by Theorem 3.1, and uses the stage equality test to obtain compatible classes. Both composites are identities by testing on all structure maps. Identities and composition agree on those same maps. Every object is the canonical union in §2, so realization is essentially surjective as well as fully faithful, exactly the hypotheses proved in the fourth prerequisite.

4. Closed ideals and subcoalgebras

Write \(A=D(H)\) for a continuous \(G\)-algebra. For \(K\subset H\), let \(K^\perp\subset A\) be its annihilator; for \(J\subset A\), let \(J^\perp\subset H\) be its annihilator in the continuous dual.

A closed subspace \(J\subset A\) satisfies \(J=(J^\perp)^\perp\). Indeed, if \(a\notin J\), closedness gives a finite-topology neighbourhood \(a+E^\perp\) disjoint from \(J\), with \(E\subset H\) finite-dimensional. The restriction of \(a\) to \(E\) is outside the restriction image of \(J\), a linear subspace of \(E^*\). Finite bidual separation gives \(e\in E\) annihilating that image but with \(a(e)\ne0\). Thus \(e\in J^\perp\) detects \(a\). The opposite inclusion is immediate. Similarly \((K^\perp)^\perp=K\) for every algebraic subspace \(K\): extend a functional detecting any nonzero vector in \(H/K\).

Theorem 4.1. Annihilators give inverse inclusion-reversing bijections between invariant subcoalgebras of \(H\) and invariant closed two-sided ideals of \(A\). Under these bijections,

A/K^\perp\simeq D(K) \tag{4.1}

as continuous \(G\)-algebras, with the ordinary quotient topology. Finite-dimensional subcoalgebras correspond precisely to open finite-codimension ideals.

Proof. If \(K\) is a subcoalgebra and \(a\) vanishes on \(K\), then for \(b\in A\), both \((ab)(k)=(a\otimes b)\Delta k\) and \((ba)(k)=(b\otimes a)\Delta k\) vanish. Thus \(K^\perp\) is two-sided, and it is closed as an intersection of continuous evaluation kernels. Invariance follows from the contragredient action.

Conversely let \(J\) be closed and two-sided, and put \(K=J^\perp\), so \(J=K^\perp\). For \(k\in K\), all functionals \(j\otimes a\) and \(a\otimes j\), with \(j\in J,a\in A\), vanish on \(\Delta k\), because \(ja,aj\in J\). If \(q\colon H\to H/K\), the full dual of \(H/K\) is exactly \(K^\perp=J\). Scalar functionals in the two factors separate every nonzero algebraic tensor, by choosing bases of its finite coefficient spans. Therefore \((q\otimes1)\Delta k=0=(1\otimes q)\Delta k\). The two kernels are \(K\otimes H\) and \(H\otimes K\), whose intersection is \(K\otimes K\), by a complement to \(K\). Hence \(K\) is a subcoalgebra. The action makes it invariant, since \(J\) is invariant. Both double-annihilator identities proved above give the claimed inverse correspondences.

Restriction \(A\to D(K)\) is surjective by scalar-functional extension and has kernel \(K^\perp\). It is open: for finite \(E\subset H\), its image of \(E^\perp\) is the annihilator of \(E\cap K\) in \(D(K)\). The complete extension proof for this identity is in Pro representations and continuous duality, §4. Thus the quotient topology is exactly the finite topology. Restriction preserves multiplication and unit by the subcoalgebra equations, proving (4.1) on every vector and operation.

Finally, finite \(K\) makes \(K^\perp\) open with codimension \(\dim K\). Conversely, if \(K^\perp\) is open, it contains \(E^\perp\) for some finite \(E\); double annihilators give \(K\subset E\), proving finite dimension. Every open ideal is closed, because its complement is a union of open cosets. This proves the last assertion. \(\square\)

5. All finite algebra quotients and all Pro maps

Every finite subspace of \(H\) lies in a finite-dimensional invariant subcoalgebra \(K\), by §2. Its annihilator is consequently an invariant open two-sided ideal inside the original finite-topology neighbourhood. Such ideals form a basis. Restriction therefore gives the canonical reconstruction

A\xrightarrow{\sim}\varprojlim_K K^* \simeq\varprojlim_J A/J, \tag{5.1}

where \(J\) runs over invariant open finite-codimension two-sided ideals. A compatible family of restrictions glues to a functional on the union \(H\); the reverse map restricts that functional. They are inverse, continuous in both directions, equivariant and algebra maps, because multiplication and unit are tested on every \(K\). The transitions are the actual surjective quotient maps.

Let \(\mathcal F\) be a small skeleton of finite-dimensional unital \(G\)-algebras. Finite duality gives \(\mathcal F\simeq\mathcal K^{\mathrm{op}}\). On an arbitrary small cofiltered diagram \((A_i)\), form the filtered coalgebra colimit \(H_A=\varinjlim_i A_i^*\) from §3. Continuous duality realizes it as

\mathcal R(A)=D(H_A)\simeq\varprojlim_i A_i, \tag{5.2}

with the actual vector limit topology, by the Pro realization prerequisite. Its projections are algebra maps, being duals of the coalgebra structure maps, so its multiplication and unit are precisely the coordinatewise limit operations. This proves existence and the stated topology without any epimorphic-transition assumption.

The equivalence \(\operatorname{Pro}(\mathcal F)\simeq\mathsf{Alg}_{\mathrm{cont}}(G)\) includes all the Hom bijections

\begin{aligned} \operatorname{Hom}_{G,\mathrm{cont\ alg}}(\mathcal R(A),\mathcal R(B)) &\simeq\operatorname{Hom}_{G,\mathrm{coalg}}(H_B,H_A)\\ &\simeq\varprojlim_j\varinjlim_i \operatorname{Hom}_{G,\mathrm{alg}}(A_i,B_j). \end{aligned} \tag{5.3}

For the forward map, dualize each target projection of a continuous algebra map. Theorem 3.1 lifts its map \(B_j^*\to H_A\) to a coalgebra map into some \(A_i^*\). Finite duality gives the algebra map \(A_i\to B_j\), with the prescribed projection composite. Stage equalities give independence and outer compatibility. Conversely such compatible classes give continuous algebra maps to every \(B_j\) by composing with the source projections, hence the unique map into their limit. Later representatives give the same composites. Testing every target projection proves both inverse identities, naturality, and preservation of identity and composition. No extension of an algebra map from a projection image to an entire stage is assumed.

The inverse object construction is the canonical quotient system (5.1). Its comparison to an original Pro diagram uses the maps \(A/\ker p_i\to A_i\); the other comparison uses (5.3) to lift the quotient maps \(A\to A/J\) to stage classes. Both realize the identity of \(A\), so full faithfulness makes their two composites the Pro identities. The same uniqueness proves naturality. Thus this is an equivalence with explicit inverse comparisons on arbitrary presentations.

6. The finite dual and universal completion

Let \(R\) be any discrete unital \(k\)-algebra with \(G\)-action by automorphisms; its underlying representation need not be locally finite. Let \(\mathcal I_G(R)\) consist of its invariant two-sided ideals of finite codimension, including \(R\) itself. Intersections keep finite codimension, because \(R/(I\cap J)\) embeds in \(R/I\times R/J\). Define the equivariant finite dual and completion by

R_G^\circ=\bigcup_{I\in\mathcal I_G(R)}(R/I)^*\subset R^*,\qquad \widehat R_G=D(R_G^\circ)\simeq\varprojlim_I R/I. \tag{6.1}

The union uses pullback along quotient maps. Every term is a finite-dimensional \(G\)-coalgebra, and their directed union has the colimit structure of §3. Explicitly \(\Delta\lambda\) is the finite tensor representing \((r,s)\mapsto\lambda(rs)\) in a quotient through which \(\lambda\) factors, and \(\epsilon(\lambda)=\lambda(1)\). Choosing a different quotient gives the same tensor after passing to \(I\cap J\), so these operations are well-defined. The limit comparison restricts a functional to each \((R/I)^*\), and its inverse evaluates compatible quotient vectors on the union; finite quotient biduality makes them inverse with the actual limit topology.

The canonical map \(\iota_R\colon R\to\widehat R_G\) is evaluation: \(\iota_R(r)(\lambda)=\lambda(r)\). The displayed coproduct and counit make it a unital algebra map, and the contragredient action makes it equivariant. Its kernel and density are exactly

\ker\iota_R=\bigcap_{I\in\mathcal I_G(R)}I,\qquad \overline{\iota_R(R)}=\widehat R_G. \tag{6.2}

For the kernel, the functionals on each finite quotient separate its vectors. For density, any finite observation in \(R_G^\circ\) lies in one \((R/I)^*\), by taking the intersection of finitely many defining ideals. The map \(R\to R/I\) is onto, so evaluation realizes every possible observation there. This proves density, including when the completion is zero. Injectivity is precisely separation by invariant finite-dimensional algebra quotients.

Theorem 6.1. For every continuous \(G\)-algebra \(A\), restriction along \(\iota_R\) gives inverse natural bijections

\operatorname{Hom}_{G,\mathrm{cont\ alg}}(\widehat R_G,A) \simeq\operatorname{Hom}_{G,\mathrm{alg}}(R,U A), \tag{6.3}

where \(U\) forgets the topology.

Proof. Given \(f\colon R\to U A\), use the finite ideal basis of §5. For each basic \(J\subset A\), the inverse image \(I=f^{-1}J\) is an invariant two-sided ideal of finite codimension, since \(R/I\) embeds in the finite-dimensional \(A/J\). The induced algebra map \(R/I\to A/J\), composed with the projection of \(\widehat R_G\), gives a continuous map to that quotient. These maps agree on the dense image of \(R\); after refining ideals, their closed equalizers give compatibility. Completeness (5.1) produces the continuous equivariant algebra map \(\widehat f\colon\widehat R_G\to A\). It agrees with \(f\) on every finite quotient, hence everywhere on \(R\). Any two extensions have a closed equalizer containing that dense image, so are equal. Conversely restriction of a continuous algebra map supplies \(f\); the same uniqueness recovers the original map. This proves both inverse identities. Composition in either variable commutes with restriction and these unique extensions, proving naturality. \(\square\)

This is an adjunction \(\widehat{(-)}_G\dashv U\). For an algebra map \(h\colon R\to S\), its completion is the unique extension of \(\iota_Sh\); equality on dense \(R\) proves the identity and composition laws. The unit is \(\iota_R\), and the counit \(e_A\colon\widehat{U A}_G\to A\) extends the identity of \(U A\). The first triangle \(e_{\widehat R_G}\widehat{\iota_R}_G=1_{\widehat R_G}\) holds because its restriction to \(R\) is \(\iota_R\), followed by uniqueness of extension. The second, \(U(e_A)\iota_{U A}=1_{U A}\), is exactly the counit's defining extension identity. Thus all adjunction maps and both triangles are explicit.

7. Four worked exercises

Exercise 7.1 (advanced: a right subcomodule can need a larger coalgebra). For \(n\geq2\), let \(H\) have basis \(h_{ij}\), coproduct \(\Delta h_{ij}=\sum_\ell h_{i\ell}\otimes h_{\ell j}\), and counit \(\epsilon(h_{ij})=\delta_{ij}\), with trivial group action. Verify the coalgebra laws. Compute its subcoalgebras, and the coefficient coalgebra of the right subcomodule \(N_i=\operatorname{span}(h_{i1},\ldots,h_{in})\).

Solution. Both iterated coproducts of \(h_{ij}\) are \(\sum_{a,b}h_{ia}\otimes h_{ab}\otimes h_{bj}\), and either counit leaves \(h_{ij}\), proving the laws on every basis vector. For its dual basis \(E_{ij}\), multiplication is \(E_{ij}E_{pq}=\delta_{jp}E_{iq}\), and the unit is \(\sum_iE_{ii}\). Thus the dual algebra is \(\operatorname{Mat}_n(k)\). Any nonzero two-sided ideal contains a matrix \(X\) with some nonzero \(X_{ij}\). Multiplying gives \(E_{pi}XE_{jq}=X_{ij}E_{pq}\), for every \(p,q\). Division by the nonzero scalar gives every matrix unit, so the ideal is the whole algebra. Theorem 4.1 in finite dimension now gives only the subcoalgebras \(0,H\).

The formula for \(\Delta h_{ij}\) keeps its first tensor factor in \(N_i\), so \(N_i\) is a right subcomodule of dimension \(n\). In its displayed basis, its coefficients are all \(h_{\ell j}\), so its coefficient coalgebra is the whole \(n^2\)-dimensional \(H\). Since \(0<N_i<H\), it is not a subcoalgebra. Each nonzero vector of \(H\) likewise has \(H\) as its least containing subcoalgebra. This calculates the enlargement in the fundamental theorem rather than identifying a right subcomodule with a subcoalgebra.

Exercise 7.2 (advanced: scaling makes the full completion adic). Take \(R=\mathbb Q[u]\), \(G=\mathbb Z\), and let its generator act by \(u\mapsto u/2\). Compute all invariant finite-codimension ideals, \(R_G^\circ\), \(\widehat R_G\), and the dual action. Then classify every ideal of the completed algebra and the corresponding subcoalgebras.

Solution. Every nonzero ideal of \(\mathbb Q[u]\) is generated by one monic polynomial \(f\): take a nonzero monic element of least degree and divide any other ideal element by it; the lower-degree remainder must vanish. If its degree is \(d\), invariance gives \(f(u/2)=2^{-d}f(u)\), since generators differ by a nonzero constant and leading coefficients determine that constant. Comparing coefficient \(u^j\), for \(j<d\), gives \(a_j2^{-j}=2^{-d}a_j\), hence \(a_j=0\). Thus \(f=u^d\). Conversely every \((u^d)\) is invariant. The zero ideal has infinite codimension; \(d=0\) gives the whole ideal and zero quotient. These are exactly all invariant finite-codimension ideals.

The finite dual therefore has basis the coefficient functionals \(\epsilon_n\), \(n\geq0\), with \(\Delta\epsilon_n=\sum_{i+j=n}\epsilon_i\otimes\epsilon_j\) and counit \(\delta_{n0}\), as in Completed tensors and continuous algebra objects, Exercise 7.2. The contragredient action is \(g\epsilon_n=2^n\epsilon_n\). Formula (6.1) gives \(\widehat R_G=\mathbb Q[[u]]\), with \(ga(u)=a(u/2)\), by the inverse maps assigning a compatible family its series coefficients and a series all its truncations. The unit embeds each polynomial as its coefficient series; it is injective because \(\bigcap_d(u^d)=0\), and dense because every finite truncation has a polynomial representative.

A series with nonzero constant coefficient has an inverse: solve \(b_0a_0=1\) and then \(b_na_0=-\sum_{i<n}b_i a_{n-i}\), recursively. If an ideal \(J\) is nonzero, choose a member of minimum nonzero order \(d\). It is \(u^d\) times a unit, so \(u^d\in J\); every member has order at least \(d\), so \(J=(u^d)\). Consequently all ideals are \(0\) and \((u^d)\), including \((u^0)=A\). They are closed, since each \((u^d)\) imposes the vanishing of its first \(d\) coefficients, and all are invariant. Their annihilators are respectively the whole coalgebra, \(\operatorname{span}(\epsilon_0,\ldots,\epsilon_{d-1})\) for \(d\geq1\), and zero for \(d=0\). These give every subcoalgebra by Theorem 4.1, also with the group omitted. The equivariance requirement was essential in the polynomial ideal classification; the result computes all the required quotients, rather than silently choosing only a preferred tower.

Exercise 7.3 (advanced: dense ideals and every closed product ideal). With trivial group action, take \(A=k^{\mathbb N}\) with coordinate multiplication and product topology. Classify all closed ideals. Show that \(J=k^{(\mathbb N)}\) is a dense proper ideal and compute the topology and continuous dual of its algebraic quotient.

Solution. The dual coalgebra has coordinate basis \(e_n\), with \(\Delta e_n=e_n\otimes e_n\), as proved in the coordinate-algebra exercise of Completed tensors and continuous algebra objects. For a closed ideal \(L\), let \(S\) consist of coordinates at which some member is nonzero. Multiplying such a member by a coordinate idempotent and the inverse nonzero scalar puts that coordinate idempotent in \(L\). Hence \(L\) contains every finitely supported function on \(S\). Their closure is all functions supported on \(S\), because every finite coordinate condition has such a representative. Closedness gives those functions in \(L\), and the definition of \(S\) forces every member of \(L\) to vanish outside \(S\). Thus

L=\{a:a_n=0\text{ for }n\notin S\},\qquad A/L\simeq k^{\mathbb N\setminus S}. \tag{7.1}

The quotient map is coordinate restriction, surjective and open, with the indicated product topology. Its dual subcoalgebra is the span of \(e_n\) for \(n\notin S\). Open finite-codimension ideals occur exactly when that complement is finite: then the finitely many coordinates give a basic open kernel; if it is infinite, no finite coordinate kernel can be contained in \(L\).

The finitely supported functions \(J\) form a two-sided ideal, since multiplication cannot enlarge their finite supports. They are dense by matching any finite coordinate list, and proper because the constant function \(1\) is absent. Thus the algebraic quotient is nonzero. Its quotient topology is indiscrete: a nonempty quotient-open set pulls back to a nonempty open set stable under translation by \(J\). That pullback contains a basic coset fixing finitely many coordinates. Adding finitely supported functions changes those coordinates arbitrarily and reaches every element of \(A\); the pullback is therefore all of \(A\). Every continuous scalar functional vanishing on \(J\) vanishes on its closure \(A\), so the quotient's continuous dual is zero. Its separated continuous quotient is \(A/\overline J=0\). This identifies exactly the role of closedness in Theorem 4.1.

Exercise 7.4 (advanced: a nonzero algebra with zero profinite completion). For a characteristic-zero field, take \(R=k\langle x,y\rangle/(xy-yx-1)\), with trivial group. Prove that \(R\ne0\) but \(R^\circ=0\) and \(\widehat R=0\). Explain why characteristic zero matters by giving a finite-dimensional representation in characteristic \(p\).

Solution. On \(k[t]\), let \(x\) act by differentiation and \(y\) by multiplication by \(t\). The monomial calculation \(\frac{d}{dt}(tf)-t\frac{d}{dt}f=f\) gives the defining relation, and the identity operator is nonzero. This unital representation proves \(R\ne0\).

If a finite-dimensional unital representation existed on a nonzero space of dimension \(d\), its operators would satisfy \(XY-YX=1\). But \(\operatorname{tr}(XY)=\sum_{i,j}X_{ij}Y_{ji}=\operatorname{tr}(YX)\), while \(\operatorname{tr}(1)=d\ne0\) in characteristic zero, a contradiction. A nonzero finite-dimensional algebra quotient would give such a representation by left multiplication on itself, so none exists. The only finite-codimension ideal is therefore \(R\), giving zero finite dual and zero completion. Formula (6.2) has kernel all of \(R\). There is also no unital map to a nonzero continuous algebra: a finite ideal neighbourhood excluding its unit would give a nonzero finite-dimensional quotient of its image, contradicting the same representation argument.

In characteristic \(p\), differentiation descends to \(k[t]/(t^p)\), because \(\frac{d}{dt}(t^p f)=t^p f'\). Multiplication by \(t\) also descends, and the same identity \(XY-YX=1\) holds on that nonzero \(p\)-dimensional space. The image is a nonzero finite-dimensional algebra, so its kernel is a proper finite-codimension ideal. Thus the zero-completion argument specifically uses characteristic zero; its finite-dimensional trace obstruction disappears when the dimension is \(p\).