Finite orbit spans and actions on formal unions

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There are two ways to combine an algebra action with a filtered union. One can take a filtered system of actual modules, so every displayed stage is stable under every algebra element. Or one can first form a formal union of vector spaces and let operators move between its finite stages. These constructions have different images. The distinction is measured by a general question: does every vector belong to a finite-dimensional stable subspace?

We answer that question for any unital algebra, without finite-generation assumptions. The regular polynomial module is the first test: its formal union admits multiplication by the variable, but no nonzero finite-dimensional subspace is stable under that multiplication. Weyl modules then give a second test where a trace calculation rules out every finite-dimensional stable subspace in characteristic zero. Good filtrations describe their actual stage maps; positive characteristic and rank zero show the precise limits of the obstruction.

We use the formal Hom convention and constant embedding from Ind-objects through their elements, and the realization adjunction in Formal colimits and compact presentations, Proposition 1.1. The action and underlying-colimit proofs are in Ring actions and tensor–Hom, Sections 2–3; the additive structure is in Ind-abelian categories, Section 2. The termwise extension is Ind-objects, Theorem 3.1. The field, algebras, vector spaces and filtered index sets belong to a fixed universe, with a larger universe for their categories. An internal action is a unital algebra homomorphism into the endomorphism algebra.

1. A formal union remembers finite images

Let \(k\) be any field and let \(M\) be a vector space. Write \(\mathcal V=k\text{-}\mathsf{Mod}\) and \(\iota:\mathcal V\to\operatorname{Ind}(\mathcal V)\) for the constant embedding. The finite-dimensional subspaces of \(M\), ordered by inclusion, form a filtered set: \(V+V'\) is a common upper bound. Define \[ F(M)=\underset{V\subset M,\ \dim V<\infty}{\text{formal colim}}\,\iota V. \tag{1.1} \] The inclusions give a natural map \(j_M:F(M)\to\iota M\). Ordinary realization sends \(F(M)\) to \(M\), because every vector belongs to a one-dimensional subspace.

Lemma 1.1. The map \(j_M\) is monic. For every vector space \(T\), including an infinite-dimensional one, it identifies maps \(\iota T\to F(M)\) with the linear maps \(T\to M\) that have finite-dimensional image.

Proof. The formal Hom formula gives \[ \begin{gathered} \operatorname{Hom}(\iota T,F(M))\\ =\underset{V}{\operatorname{colim}}\operatorname{Hom}_k(T,V). \end{gathered} \tag{1.2} \] A map on the right has image in its finite stage. Conversely a finite-image map factors through its image subspace. Two stage maps induce the same map to \(M\) exactly when they agree after inclusion into \(V+V'\); those inclusions are injective. This proves the claimed identification and its injectivity.

For an arbitrary ind-object \(A=\langle T_a\rangle\), maps \(A\to F(M)\) form the inverse limit over \(a\) of the groups in (1.2). Maps \(A\to\iota M\) form the corresponding inverse limit of \(\operatorname{Hom}_k(T_a,M)\). The preceding injections give an injection between these limits: equality can be checked on each coordinate. Thus postcomposition with \(j_M\) is injective for every \(A\), which is monicity. \(\square\)

In particular \(F(M)\ne0\) when \(M\ne0\). If \(v\ne0\), the stage map \(\iota(kv)\to F(M)\) has nonzero composite into \(\iota M\).

The same construction is functorial in linear maps \(u:M\to N\): send a finite subspace \(V\) into the finite subspace \(u(V)\). The resulting \(F(u)\) satisfies \[ j_NF(u)=(\iota u)j_M. \tag{1.3} \] Functoriality follows either on stages or by canceling the monic \(j_N\).

2. When the finite pieces can be modules

Let \(R\) be any unital \(k\)-algebra and \(M\) an ordinary left module. First give \(F(M)\) its action by allowing each algebra element to move a finite subspace to another finite subspace.

More generally, any ordinary module over a unital \(k\)-algebra \(R\) gives an internal action on \(F(M)\). For \(r\in R\), a finite subspace \(V\) maps into the finite subspace \(rV\), so Section 1 gives an endomorphism \(F(\xi_M(r))\). Additivity follows by composing with \(j_M\) and canceling it; composition and the unit follow from functoriality. Thus these operators give a unital algebra map \(R\to\operatorname{End}(F(M))\). This observation does not require a common degree bound for all elements of \(R\).

For a unital \(k\)-algebra \(R\), let \(R\text{-}\mathsf{Mod}\) be its ordinary left-module category, and let \((\operatorname{Ind}(\mathcal V))_R\) denote the internal-action category. There is a natural functor \[ \begin{gathered} \Phi_R:\operatorname{Ind}(R\text{-}\mathsf{Mod})\\ \longrightarrow (\operatorname{Ind}(\mathcal V))_R. \end{gathered} \tag{2.1} \] It sends a filtered system \((V_a)\) of actual modules to the formal filtered colimit of the underlying vector spaces, with the action induced from their actions. The existing internal-colimit construction cited at the beginning gives this colimit with equivariant structure maps. A morphism represented by later-stage module maps induces the corresponding formal equivariant map; later-stage equality and composition are respected. This specifies \(\Phi_R\) on arrows as well as objects.

Lemma 2.1. If the internal module \(F(M)\) is isomorphic to \(\Phi_R\langle V_a\rangle\), every structural map from \(\iota V_a\) to \(F(M)\) has, after composition with \(j_M\), a finite-dimensional \(R\)-stable image in \(M\).

Proof. The structural map and \(j_M\) are equivariant. Full faithfulness of the constant embedding identifies their composite with an actual linear map \(u_a:V_a\to M\). Its equivariance is exactly \(R\)-linearity. Lemma 1.1 says that its image is finite-dimensional, even if \(V_a\) itself is infinite-dimensional. Linearity over \(R\) makes that image an \(R\)-submodule. \(\square\)

Theorem 2.2 (local finiteness criterion). The internal module \(F(M)\) is in the essential image of \(\Phi_R\) if and only if every element of \(M\) belongs to a finite-dimensional \(R\)-submodule. No finite generation of \(R\) or \(M\) is assumed. The zero module is included.

Proof. Suppose first that \(M\) is locally finite. Its finite-dimensional \(R\)-submodules form a filtered set, since sums of two such submodules are again finite-dimensional submodules. Given any finite vector subspace \(V\), choose a finite basis \(v_1,\ldots,v_s\), and for each \(v_i\) choose a finite-dimensional submodule \(L_i\) containing it. Their finite sum contains \(V\). Thus the finite-dimensional submodules are cofinal among all finite vector subspaces. Their ordinary inclusions form an ind-system of \(R\)-modules, and its image under \(\Phi_R\) has underlying object \(F(M)\). The induced action agrees after composition with \(j_M\); monicity identifies the two actions. This proves sufficiency, including \(M=0\).

Conversely, take a presentation \(F(M)\simeq\Phi_R\langle V_a\rangle\). Lemma 2.1 gives maps \(u_a:V_a\to M\) with finite-dimensional submodule images. Applying ordinary realization to the underlying formal isomorphism gives \[ \underset{a}{\operatorname{colim}}\,V_a\ \simeq\ M, \tag{2.2} \] with structural maps exactly \(u_a\). Every vector in a filtered vector-space colimit is represented at some stage. To verify this directly, construct that colimit as the quotient of the direct sum of the stages by the transition relations. A vector is represented by a finite sum of vectors from finitely many stages. Filteredness moves these stages to one common stage, and the relations identify the sum with one vector there. Thus every element of \(M\) lies in some \(u_a(V_a)\), proving local finiteness. The maps \(u_a\) need not be injective and the \(V_a\) need not be finite-dimensional; only their images have been shown finite-dimensional. This completes both directions.

The forward direction bounds the images of the structural maps; it does not bound the dimensions of their source modules or assume that those maps are injective. The converse constructs actual stable stages, cofinal among all finite vector subspaces. This is what changes an action on the completed object into a system of actions on its displayed pieces.

Corollary 2.3. Suppose \(M\ne0\) and \(M\) has no nonzero finite-dimensional \(R\)-submodule. Then \(F(M)\), with its internal action, does not belong to the essential image of \(\Phi_R\).

Proof. Assume an isomorphism \(\Phi_R\langle V_a\rangle\simeq F(M)\). By Lemma 2.1 every \(u_a\) is zero. Monicity of \(j_M\) makes every structural map \(\iota V_a\to F(M)\) zero. Those maps form a colimit cocone: composing the original colimit cocone with the isomorphism preserves its universal property. The identity and zero endomorphisms of \(F(M)\) therefore have the same composites with every structural map. Uniqueness in the colimit property gives \(1_{F(M)}=0\). An object whose identity is zero is a zero object in an additive category: every arrow from or to it equals its composite with this zero identity. This contradicts the nonzero test in Section 1. \(\square\)

The finite-image statement is a property of the formal target. It does not claim that an arbitrary map from an infinite-dimensional space into an ordinary filtered union factors through a finite stage.

3. The polynomial shift as a first test

Take \(R=k[t]\) and \(M=R\). Multiplication by \(t\) takes the degree-\(m\) space to the degree-\(m+1\) space. Thus it is a well-defined internal endomorphism of \(F(M)\); it is not an endomorphism of any nonzero displayed degree space. The local-finiteness criterion detects this distinction over every field.

Proposition 3.1. For the regular polynomial action,

The last isomorphism respects addition, scalar multiplication, composition and the unit. The nonzero internal module \(F(M)\) is outside the image of \(\Phi_R\).

Proof. No nonzero finite-dimensional subspace of \(k[t]\) is stable under multiplication by \(t\). Indeed, if such a subspace contains a nonzero polynomial \(p\), it contains every \(t^jp\). These polynomials are linearly independent: a finite relation would give \(p\sum_j a_jt^j=0\), and the polynomial ring is a domain. Thus an equivariant map \(\iota M\to F(M)\) has zero composite into \(\iota M\) by Lemma 1.1, and is zero by monicity. The first Hom group is zero, and Corollary 2.3 proves the last assertion.

By realization, maps \(F(M)\to\iota M\) are actual linear maps \(M\to M\). The formal equivariance equations are equivalent to the ordinary ones: the realization bijection is natural with respect to the generator actions, so it identifies both sides of each equation. Such maps are the \(R\)-linear endomorphisms of \(R\). Each is multiplication by \(p=u(1)\), and every \(p\in R\) works. Thus this Hom group is \(R\).

For an internal endomorphism of \(F(M)\), compose with \(j_M\). The preceding computation identifies this composite with multiplication by a unique \(p\). Conversely multiplication by \(p\) maps each finite subspace into a finite subspace and induces \(F(u_p)\). It is equivariant because the ordinary multiplication operators commute, and its composite is the specified map. Monicity of \(j_M\) makes the lift unique. For polynomials \(p,q\), addition gives \(F(u_p)+F(u_q)=F(u_{p+q})\), while composition gives \(F(u_p)F(u_q)=F(u_{pq})\). Indeed, after composing with the monic \(j_M\), these equalities are the ordinary linear identities \(u_p+u_q=u_{p+q}\) and \(u_pu_q=u_{pq}\). Also \(F(u_1)=1\), and scalar multiplication corresponds to multiplication of the polynomial by that scalar. Thus the identification preserves addition, scalar multiplication, composition and the unit, proving that the endomorphism algebra is \(k[t]\). In particular the zero first Hom group does not prevent a nonzero formal object or a nonzero map in the reverse direction.

The map \(j_M\) in the middle line is the operation associated with the polynomial \(1\). Its existence does not provide a map in the other direction. Realization forgets the finite-image restriction, whereas a constant source mapping to the formal union retains it.

4. Weyl operators and the stages they cross

For any exhaustive increasing filtration \((M_m)_{m\in\mathbb Z}\) by finite-dimensional subspaces, the analogous formal object is canonically \(F(M)\). Indeed, a basis of any finite-dimensional subspace lies in finitely many stages; its vectors all lie in one common later stage. Thus the filtration stages are cofinal among the finite-dimensional subspaces. Equivalently, inclusions of each finite subspace into a sufficiently large stage define inverse maps between the two formal systems. Changing a chosen stage changes nothing after passage to a common later stage. Both composites are the identity on each stage.

Let \(n\ge1\). The Weyl algebra \(W_n(k)\) is the unital associative \(k\)-algebra generated by \(x_1,\ldots,x_n,d_1,\ldots,d_n\), with central scalars and relations \[ \begin{gathered} {[x_i,x_j]}=0,\\ {[d_i,d_j]}=0,\\ {[d_i,x_j]}=\delta_{ij}1. \end{gathered} \tag{4.1} \] Here \([a,b]=ab-ba\). The trace application in Section 5 imposes characteristic zero; the stage construction here works over every field.

Suppose \(M\) is a left \(W_n(k)\)-module with an exhaustive increasing filtration by finite-dimensional spaces, such that \[ \begin{gathered} x_iM_m\subset M_{m+1},\\ d_iM_m\subset M_{m+1}. \end{gathered} \tag{4.2} \] Put \(F=\langle M_m\rangle\). Each generator gives compatible stage maps \(M_m\to M_{m+1}\), hence an endomorphism of \(F\). Scalar multiplication gives the central \(k\)-action. The unit is the identity.

Proposition 4.1. These operators give an internal \(W_n(k)\)-action on \(F\), and the canonical \(j:F\to\iota M\) is equivariant. The resulting internal module is canonically independent of the filtration.

Proof. A composition of two generators on stage \(M_m\) is represented in \(M_{m+2}\). Their commutators in that space satisfy (4.1), because the original module does. The identity on \(M_m\), included into \(M_{m+2}\), represents the identity on \(F\). Consequently all the defining relations hold as formal endomorphisms. A word in the generators acts by the corresponding composite. Extending linearly gives an algebra homomorphism from the free associative algebra; the relations just checked show that it factors through \(W_n(k)\). It preserves the unit and central scalars.

For each generator the composite with \(j\) is its actual action on \(M_m\) followed by inclusion in \(M\). Hence \(j\) is equivariant. Given a second filtration, the cofinal comparison just proved supplies a canonical underlying isomorphism \(c:F\to F'\) satisfying \(j'c=j\). For any generator \(r\), both \(j'c\xi_F(r)\) and \(j'\xi_{F'}(r)c\) equal \((\iota\xi_M(r))j\). Monicity of \(j'\) gives equivariance of \(c\); equivariance for generators and scalars implies equivariance for all words and linear combinations. The inverse is equivariant by the same argument. These isomorphisms satisfy the cocycle identity, since their composites into \(\iota M\) coincide and the target comparison is monic. \(\square\)

A good filtration for the order-one filtration on \(W_n(k)\) additionally has \(M_m=0\) for sufficiently negative \(m\), and, for sufficiently large \(m\), \[ \begin{aligned} M_{m+1}&=M_m\\ &\quad+\sum_{i=1}^n(x_iM_m+d_iM_m). \end{aligned} \tag{4.3} \] Thus Proposition 4.1 applies to every finitely generated Weyl module with the given good filtration. Independence uses only finite-dimensionality, exhaustiveness and (4.2). No assertion that an individual \(M_m\) is a Weyl submodule enters the argument.

Take \(M=k[z_1,\ldots,z_n]\). Let \(x_i\) multiply by \(z_i\), and let \(d_i\) be partial differentiation with respect to \(z_i\). Multiplications commute. Partial derivatives commute, as checked on each monomial. The product rule gives, on a polynomial \(f\), \[ d_i(z_jf)-z_jd_i(f)=\delta_{ij}f. \tag{4.4} \] Thus the operators define a Weyl action. Multiplication of \(1\) by monomials spans \(M\), so \(1\) generates it as a Weyl module.

For \(m\ge0\), let \(M_m\) consist of polynomials of total degree at most \(m\); put \(M_m=0\) for \(m<0\). Each stage has the finite monomial basis with total exponent at most \(m\), and their union is \(M\). Multiplication raises the bound by one; differentiation lowers it by one and hence also satisfies (4.2). For every \(m\ge0\), each degree-\(m+1\) monomial is \(z_i\) times a degree-\(m\) monomial for some \(i\). Together with \(M_m\), these span \(M_{m+1}\). The derivative terms already lie in \(M_m\), proving (4.3).

In characteristic zero, this gives a concrete nonzero example of Corollary 5.2. Over every field, the action is recorded by maps between finite degree spaces, although no nonzero finite degree space is stable under multiplication by every variable. In fact this polynomial module fails the local-finiteness criterion over every field: a nonzero stable subspace containing \(f\) must contain \(z_1^j f\) for all \(j\geq0\). These are linearly independent, since a finite dependence would give \(f\sum_j a_jz_1^j=0\) in the polynomial domain. Thus no nonzero finite-dimensional stable subspace exists, independently of the trace argument. Ordinary realization recovers the original polynomial module; retaining its finite stages changes the possible formal maps into it.

5. What characteristic zero excludes

Lemma 5.1. If \(k\) has characteristic zero and \(n\ge1\), every finite-dimensional \(W_n(k)\)-module is zero.

Proof. For two \(q\times q\) matrices \(A,B\) over \(k\), \[ \begin{aligned} \operatorname{tr}(AB)&=\sum_{i,j}a_{ij}b_{ji},\\ \operatorname{tr}(BA)&=\sum_{i,j}b_{ij}a_{ji} =\sum_{i,j}a_{ij}b_{ji}. \end{aligned} \tag{5.1} \] The last equality exchanges the two finite indices and uses commutativity in \(k\). Thus a commutator has trace zero. On a finite-dimensional Weyl module \(U\), the first pair of generators satisfies \([d_1,x_1]=1_U\). Taking traces gives \[ 0=\operatorname{tr}([d_1,x_1]) =\dim_k(U)\,1_k. \tag{5.2} \] In characteristic zero a positive integer times \(1_k\) cannot vanish. Hence \(\dim_k U=0\). \(\square\)

Corollary 5.2. Let \(k\) have characteristic zero and \(n\ge1\). For any nonzero finitely generated \(W_n(k)\)-module \(M\) endowed with a good filtration, the formal object \(\langle M_m\rangle\) is a filtration-independent internal Weyl module outside the essential image of \(\Phi_{W_n(k)}\).

Proof. Proposition 4.1 gives the action and independence, and Section 4 identifies the object with \(F(M)\). Every finite-dimensional Weyl-stable subspace is itself a finite-dimensional Weyl module, so Lemma 5.1 rules it out unless it is zero. Corollary 2.3 applies. \(\square\)

The positive-rank hypothesis matters. With the empty-generator convention \(W_0(k)=k\), a finitely generated module is finite-dimensional. An exhaustive increasing filtration then reaches all of \(M\) at some stage, by applying exhaustiveness to a finite basis. Its formal object is the constant \(\iota M\), which is the image of the constant module under \(\Phi_k\). The nonzero obstruction therefore concerns \(n\ge1\).

6. Graded exercises with full solutions

Exercise 1 (developing): uniform comparison of good filtrations. Let \((M_m)\) and \((N_m)\) be two good filtrations on the same finitely generated Weyl module, over any field. Prove that there are integers \(c,d\) with \(M_m\subset N_{m+c}\) and \(N_m\subset M_{m+d}\) for every integer \(m\). Describe the canonical formal isomorphism supplied by these bounds.

Solution. Choose \(b\) with \(M_m=0\) for \(m<b\), and choose \(a\ge b\) such that the recurrence (4.3) holds for all \(m\ge a\). Finite-dimensionality and exhaustiveness give \(\ell\) with \(M_a\subset N_\ell\). Induction using the recurrence for \(M\) and the generator inclusions for \(N\) yields \[ M_{a+r}\subset N_{\ell+r}\qquad(r\ge0). \tag{6.1} \] Set \(c=\ell-b\). For \(b\le m\le a\), we have \(M_m\subset M_a\subset N_\ell\subset N_{m+c}\). For \(m\ge a\), (6.1) gives \(M_m\subset N_{m+\ell-a}\subset N_{m+c}\), because \(a\ge b\). For \(m<b\), the desired inclusion holds since \(M_m=0\). Interchanging the filtrations gives \(d\).

These inclusions are compatible stage maps into reindexed later stages. Their formal composites are the inclusions of the original stages into later stages, which represent the identity. If a displayed shift is negative, it still defines an order-preserving cofinal reindexing of \(\mathbb Z\); or one may enlarge \(c,d\) to nonnegative values without losing the inclusions. The underlying isomorphism is the cofinal comparison from Section 4 and is equivariant by Proposition 4.1. No bounded dimension of \(M\) was assumed.

Exercise 2 (intermediate: stable torsion pieces). Let \(R=k[t]\) and \(M=\bigoplus_{n\geq1}R/(t^n)\). Decide whether \(F(M)\) comes from an ind-system of modules. Compute the equivariant Hom groups from \(\iota R\) to \(F(M)\) and from \(F(M)\) to \(\iota R\), and contrast them with Proposition 3.1.

Solution. A vector of the direct sum has finitely many nonzero coordinates, all killed by one sufficiently large power of \(t\). Its cyclic submodule is spanned by finitely many successive powers of that vector, so is finite-dimensional. Theorem 2.2 therefore puts \(F(M)\) in the image of \(\Phi_R\).

An \(R\)-linear map \(R\to M\) is determined by the image \(m\) of \(1\). Its image is the finite-dimensional cyclic submodule \(Rm\), so Lemma 1.1 gives a lift to \(F(M)\); monicity of \(j_M\) makes that lift unique and equivariant. Every \(m\) works. Hence \(\operatorname{Hom}_R(\iota R,F(M))\simeq M\), as an additive group and \(k\)-vector space.

By the same natural realization bijection used in Proposition 3.1, maps \(F(M)\to\iota R\) correspond to actual \(R\)-linear maps \(M\to R\). Every \(m\) is killed by some \(t^a\), so its image is killed by \(t^a\) in the domain \(k[t]\), and is zero. Thus this second Hom group is zero. The asymmetry has reversed: stable finite orbit spans permit maps from the constant regular module into the formal union, while torsion prevents maps back into the torsion-free regular module.

Exercise 3 (intermediate): characteristic \(p\) changes the conclusion. Let \(k\) have characteristic \(p>0\). On \(U=k[z]/(z^p)\), let \(x\) be multiplication by \(z\) and let \(d\) be differentiation. Prove that this defines a nonzero \(p\)-dimensional \(W_1(k)\)-module. Give it a good filtration, and show that its formal object belongs to the image of \(\Phi_{W_1(k)}\). Explain exactly why the trace obstruction does not apply.

Solution. Differentiation preserves \((z^p)\): the derivative of \(z^ph\) is \(pz^{p-1}h+z^ph'=z^ph'\). It therefore descends to \(U\). The product rule descends too, giving \([d,x]=1\). At the top basis vector, this can be checked without the product rule: \[ \begin{aligned} (dx-xd)(z^{p-1})&=-(p-1)z^{p-1}\\ &=z^{p-1}. \end{aligned} \tag{6.2} \] The classes \(1,z,\ldots,z^{p-1}\) form a basis, so the module is nonzero and has dimension \(p\). It is generated by \(1\). Take the images of the total-degree filtration on \(k[z]\), with zero negative stages. They are finite, exhaustive, and satisfy the generator inclusions. Once \(m\ge p-1\), both stages in (4.3) equal \(U\), and \(U+xU+dU=U\); hence the filtration is good. It is eventually constant, so its formal object is \(\iota U\), the image under \(\Phi\) of the constant Weyl module.

The trace computation still gives \(0=(\dim U)1_k=p1_k\). This equality holds in characteristic \(p\) and no longer forces the integer dimension to be zero. The finite-dimensional submodule excluded in characteristic zero is here the whole module.

Exercise 4 (advanced: torsion and a growing nilpotent module). Prove that a \(k[t]\)-module is locally finite exactly when every vector is annihilated by a nonzero polynomial. Apply the criterion to \(P=k[t,t^{-1}]/k[t]\). Give \(F(P)\) as an actual filtered system of modules, with its transition maps, without a finite-generation assumption.

Solution. If the cyclic span of \(m,tm,t^2m,\ldots\) is finite-dimensional, a finite linear dependence gives a nonzero polynomial \(f\) with \(f(t)m=0\). Conversely, if \(f(t)m=0\) and \(f\) has degree \(d\geq1\), its invertible leading coefficient expresses \(t^dm\) in the span of \(m,\ldots,t^{d-1}m\). Repeated multiplication reduces every higher power to that span. If \(f\) is a nonzero constant, \(m=0\). Thus every torsion vector has a finite-dimensional cyclic submodule. These implications prove the equivalence, over any field.

In \(P\), every vector is a finite linear combination of the classes of \(t^{-1},t^{-2},\ldots\), hence is killed by a power of \(t\). Put \(L_n=\operatorname{span}_k(t^{-1},\ldots,t^{-n})\) for \(n\geq1\). It is stable under \(t\), since \(t\,t^{-1}=0\) in the quotient and the other negative powers shift down one position. The map \(R/(t^n)\to L_n\), \(1\mapsto t^{-n}\), is a module isomorphism: its powers give the displayed basis and its kernel is exactly \((t^n)\).

The inclusions \(L_n\subset L_{n+1}\) correspond to the explicit module maps \(R/(t^n)\to R/(t^{n+1})\), \(1\mapsto t\). They are injective, since the monomial bases give their images and kernels. The union of the \(L_n\) is \(P\), and every finite vector subspace is contained in one \(L_n\) by taking the largest negative exponent in a finite basis. Hence these stable stages are cofinal among all finite subspaces. Their image under \(\Phi_R\) is \(F(P)\), with the same action by the equivariant cofinal comparison from Theorem 2.2. The module \(P\) itself has no finite generating set: a finite set lies in one \(L_n\), and its generated module stays in that finite stage. This illustrates both the positive classification and its lack of a finite-generation hypothesis.

7. References

The general local-finiteness criterion applies to arbitrary algebras; the Weyl application uses its finite-dimensional trace obstruction. The polynomial shift, torsion examples and positive-characteristic module distinguish the hypotheses.

The Stacks Project, Remark 4.22.4 (05PW) gives the ind-object Hom convention. The complete internal prerequisite proofs are linked in the introduction; the remark is used for the convention. Linked Stacks material retains its GNU Free Documentation License 1.2. No third-party text is incorporated here. No differential-operator sheaf or characteristic-variety theorem is needed for these algebraic conclusions.