Natural scalars on modules and finite abelian groups

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

Multiplication by an integer gives an endomorphism of every abelian group and commutes with every group homomorphism. A natural scalar is any operation with these two properties. Which operations exist depends on the category. On all modules over a ring they are exactly its central elements. On finite abelian groups they form a much larger ring: one compatible sequence of residues for each prime.

We assume modules, natural transformations and elementary integer arithmetic. The ring-action lesson proves that the natural endomorphisms of the identity of an additive category form a commutative ring. The projector lesson explains how orthogonal projectors give finite direct-sum coordinates. We will construct the necessary projectors explicitly, so no classification theorem for finite abelian groups is needed.

We fix a universe containing the ring and the underlying sets in question. Natural transformations on a large category can first be taken in a larger universe. The two computations below identify their rings with sets in the original universe. Basic open references are Mathlib's category-center construction, its module filtered-colimit construction, and Stacks, Grothendieck's AB conditions.

1. The regular module determines every component

Let \(R\) be an associative unital ring, and let \(R\text{-}\mathsf{Mod}\) denote its left modules. Multiplication in an endomorphism ring means composition: \(uv=u\circ v\). Write \[ \mathfrak Z(\mathcal C)=\operatorname{End}(\operatorname{id}_{\mathcal C}). \] An element \(\eta\) has components \(\eta_X:X\to X\) satisfying \[ \eta_Y f=f\eta_X\qquad(f:X\to Y). \tag{1.1} \] Addition is pointwise. The commutativity proved in the ring-action lesson uses (1.1) with \(f\) another component; it requires compatibility with all arrows, including these endomorphisms.

The left regular module \(R\) is a useful probe. For each \(x\in M\), the map \[ f_x:R\longrightarrow M,\qquad f_x(r)=rx \tag{1.2} \] is left linear. Conversely every left-linear map out of \(R\) has this form, with \(x=f_x(1)\). Thus evaluation at \(1\) identifies \(\operatorname{Hom}_R(R,M)\) with the underlying additive group of \(M\), naturally in \(M\). In particular these maps detect every module homomorphism.

Theorem 1.1. There is a canonical ring isomorphism \[ \begin{gathered} Z(R)\xrightarrow{\ \sim\ } \mathfrak Z(R\text{-}\mathsf{Mod}),\\ z\longmapsto(x\mapsto zx), \end{gathered} \tag{1.3} \] where \(Z(R)=\{z\in R:zr=rz\text{ for all }r\in R\}\).

Proof. Given \(\eta\), put \(a=\eta_R(1)\). Left linearity says \[ \eta_R(r)=ra. \tag{1.4} \] This is right multiplication by \(a\). For any \(b\in R\), right multiplication \(\rho_b(r)=rb\) is a left-linear endomorphism of \(R\). Naturality at \(\rho_b\), evaluated at \(1\), gives \[ ba=\eta_R(b)=\rho_b(a)=ab. \tag{1.5} \] Consequently \(a\) is central. Naturality at (1.2) then gives, for every module and every element, \[ \eta_M(x)=f_x(\eta_R(1))=ax. \tag{1.6} \] The component on the regular module therefore determines every component.

Conversely, for central \(a\), the rule \(x\mapsto ax\) is additive and left linear: \(a(rx)=r(ax)\). For a left-linear \(f:M\to N\), we have \(f(ax)=af(x)\). The family is natural. Its value on \(1\in R\) is \(a\), so the constructions are inverse.

Pointwise addition corresponds to \(a+b\), composition to \(ab\), and the identity family to \(1\). This proves the ring assertion. The proof also covers the zero ring, for which every unital module is zero. \(\square\)

For example, the natural scalars on all abelian groups are the integers, by taking \(R=\mathbb Z\). A noncentral element of a noncommutative ring does not define a scalar on all its left modules: on the regular module its left multiplication fails left linearity.

The same regular-module probe appears in the Grothendieck structure of module categories. Here is the structural verification over an arbitrary ring. Kernels and quotient cokernels have the induced module structures; the map from a module modulo a kernel to the image is an isomorphism. Thus the category is abelian. Small direct sums exist, and the colimit of a small diagram is the quotient of the direct sum of its objects by the relations identifying an element with its image under each diagram arrow. The probe (1.2) is a generator.

For a filtered diagram, every colimit element has a representative \([j,x]\) at one stage. Two representatives are equal exactly when their images agree at a common later stage. This filtered equality rule follows from formal colimits; parallel arrows can be equalized after a further transition. Sums use a common stage, and scalar multiplication is \[ r[j,x]=[j,rx]. \tag{1.7} \] These operations are independent of the representatives because every transition is linear. They give the colimit its module structure and its universal property for linear maps, as in Mathlib's construction cited above.

For a filtered diagram of short exact sequences, a representative in the first term that maps to zero becomes zero in the middle term at a later stage; injectivity there makes it zero in the first term. A middle representative mapping to zero in the last colimit becomes a kernel element at a later stage, where exactness supplies its preimage. A representative in the last term lifts at its own stage. These three statements prove exactness of the colimit sequence. Hence \(R\text{-}\mathsf{Mod}\) is Grothendieck in the fixed universe. No commutativity of \(R\) enters this argument.

2. Cyclic groups give compatible residues

Let \(\mathsf{FinAb}\) be the category of finite abelian groups and all group homomorphisms. For a prime \(p\) and \(n\geq1\), set \[ C_{p,n}=\mathbb Z/p^n\mathbb Z. \] Every endomorphism of this cyclic group is multiplication by a unique residue. Thus a natural scalar \(\eta\) gives \[ a_{p,n}=\eta_{C_{p,n}}(\overline1) \in\mathbb Z/p^n\mathbb Z. \tag{2.1} \] The quotient map \(C_{p,n+1}\to C_{p,n}\) is a morphism in \(\mathsf{FinAb}\). Applying naturality to \(\overline1\) shows \[ a_{p,n+1}\equiv a_{p,n}\pmod {p^n}. \tag{2.2} \]

We define the ring of \(p\)-adic integers algebraically. Its elements are sequences of residues \(a_n\in\mathbb Z/p^n\mathbb Z\), for \(n\geq1\), satisfying the compatibility below: \[ \begin{gathered} \mathbb Z_p=\varprojlim_{n\geq1}\mathbb Z/p^n\mathbb Z,\\ a_{n+1}\equiv a_n\pmod {p^n}. \end{gathered} \tag{2.3} \] Its addition, multiplication and unit are coordinatewise; reduction is a ring homomorphism, so these operations preserve compatibility. No metric or convergence theorem is part of this definition.

For an integer \(m\), its residues give an element of \(\mathbb Z_p\). This map \(\mathbb Z\to\mathbb Z_p\) is injective: an integer divisible by every \(p^n\) is zero, because \(p^n\) eventually exceeds its absolute value. Naturality has produced one such compatible sequence for each prime. To recover the operation on every finite group, we need its primary coordinates.

3. Primary coordinates from integer projectors

For an abelian group \(A\), write \[ A_p=\{x\in A:\exists k\geq1,\ p^k x=0\}. \tag{3.1} \] This is a subgroup: a common larger exponent kills the sum or difference of two of its elements. Every group homomorphism sends \(A_p\) into the corresponding subgroup of its target.

Proposition 3.1. Every finite abelian group has the canonical decomposition \[ A=\bigoplus_p A_p, \tag{3.2} \] with only finitely many nonzero summands. Its component projections are multiplication by suitable integers on \(A\).

Proof. Each element of a finite group has finite order, by repetition among its nonnegative multiples. A product of the finitely many element orders is a positive integer \(N\) annihilating \(A\). If \(N=1\), then \(A=0\), and the decomposition is empty. Otherwise factor \[ N=\prod_{p\in S}p^{n_p},\qquad n_p\geq1. \tag{3.3} \] For each \(p\in S\), put \(d_p=N/p^{n_p}\). Since \(d_p\) and \(p^{n_p}\) are coprime, Bézout's identity gives an integer \(t_p\) with \[ t_p d_p\equiv1\pmod {p^{n_p}}. \] Set \(c_p=t_p d_p\). It is \(1\) modulo \(p^{n_p}\), and \(0\) modulo every \(q^{n_q}\) for \(q\ne p\). Coprimality of these prime powers implies \[ \begin{gathered} \sum_{p\in S}c_p\equiv1\pmod N,\\ c_p^2\equiv c_p\pmod N,\\ c_pc_q\equiv0\pmod N\quad(p\ne q). \end{gathered} \tag{3.4} \] Indeed each congruence holds modulo every factor in (3.3), and an integer divisible by all these coprime factors is divisible by their product.

Multiplication by \(c_p\) is therefore an idempotent \(e_p\) on \(A\); the \(e_p\) are orthogonal and sum to the identity. Its image is \[ e_p A=\ker(p^{n_p}:A\to A). \tag{3.5} \] For the forward inclusion, \(p^{n_p}c_p\) is a multiple of \(N\). For the reverse inclusion, \(c_p-1\) is a multiple of \(p^{n_p}\), so \(e_p x=x\) when \(p^{n_p}x=0\).

Every \(x\) is the sum of its \(e_p x\). If \(\sum_p x_p=0\) with \(x_p\in e_p A\), applying \(e_q\) gives \(x_q=0\). This proves the direct-sum isomorphism and its inverse coordinates.

It remains to identify these images intrinsically with (3.1). Suppose \(p^k x=0\). Decompose \(x\) into the just-constructed coordinates. For \(q\ne p\), its \(q\)-coordinate is killed by both \(p^k\) and \(q^{n_q}\). Bézout's identity makes that coordinate zero. If \(p\notin S\), all coordinates vanish; otherwise only \(e_p x\) remains. Conversely (3.5) consists of elements of \(A_p\). Hence the summands and projections do not depend on \(N\) or the chosen \(t_p\). \(\square\)

For example, on \(\mathbb Z/12\mathbb Z\), multiplication by \(9\) projects to its \(2\)-primary subgroup, and multiplication by \(4\) to its \(3\)-primary subgroup. These integers are orthogonal idempotents modulo \(12\), and \(9+4=1\) there. The projection onto the \(2\)-primary summand is preserved by every homomorphism of finite abelian groups, even though different groups can require different integer representatives for it.

4. Reconstructing every natural scalar

Theorem 4.1. The cyclic residues (2.1) give a ring isomorphism \[ \mathfrak Z(\mathsf{FinAb}) \xrightarrow{\ \sim\ } \prod_{p\text{ prime}}\mathbb Z_p. \tag{4.1} \]

Proof. Start with a family \(\alpha=(\alpha_p)_p\), where \(\alpha_p=(a_{p,n})_{n\geq1}\) is compatible. On \(A_p\) for finite \(A\), choose \(n\) such that \(p^n A_p=0\), and let \(\alpha\) act by multiplication by any integer representing \(a_{p,n}\). Changing that integer changes it by a multiple of \(p^n\), which acts by zero. Choosing a larger exponent gives the same operation by compatibility. Two arbitrary choices can be compared at their maximum, so the operation is independent of the exponent.

Use (3.2) to define \[ \eta^\alpha_A\left(\sum_p x_p\right) =\sum_p a_{p,n_p}x_p. \tag{4.2} \] Here residues act by integer representatives, and the sums have finite support. The decomposition is unique, so this is a well-defined group endomorphism.

For a homomorphism \(f:A\to B\), we have \(f(A_p)\subseteq B_p\). Choose a common exponent killing both of these finite primary subgroups. On this summand, \[ f(a_{p,n}x)=a_{p,n}f(x). \tag{4.3} \] Adding the finitely many primary components proves naturality. On \(C_{p,n}\), the operation is exactly multiplication by \(a_{p,n}\), so its residues recover \(\alpha\).

Conversely, let \(\eta\) be natural and let \(\alpha\) be its cyclic residues. For \(x\in A_p\) with \(p^n x=0\), the map \[ \begin{gathered} u_x:C_{p,n}\longrightarrow A,\\ u_x(\overline m)=mx \end{gathered} \tag{4.4} \] is well defined. Naturality gives \[ \eta_A(x)=u_x(\eta_{C_{p,n}}(\overline1)) =a_{p,n}x. \tag{4.5} \] By additivity and (3.2), this determines \(\eta_A\) on every element and is exactly (4.2). The two constructions are inverse.

On each primary component, adding operations adds the residues, and composing operations multiplies them. The identity gives residue \(1\) at every prime and exponent. Thus the bijection preserves the entire ring structure. \(\square\)

The integer operation \(x\mapsto mx\) corresponds to the diagonal residue family of \(m\). Every finite group sees only finitely many primes and finite residue levels. The ring (4.1) also permits a separate compatible choice at each prime, without requiring one integer to realize all choices simultaneously.

For each prime \(p\), take \(\alpha_p=1\) and all other coordinates zero. Its natural operation is the projection onto \(A_p\). On each finite group the nonzero primary projectors form a finite orthogonal identity sum. There is no assertion of an infinite sum in the endomorphism ring: (4.2) is an elementwise finite sum.

5. Exercises with complete solutions

Exercise 1 (introductory: a noncommutative ring). Let \(k\) be a field and \(R=M_2(k)\). Compute the natural scalars on all left \(R\)-modules. Use the matrix units to determine the center, and explain why the action of \(E_{11}\) on the regular module is not a component of a natural scalar.

Solution. Write a central matrix as \[ B=\begin{pmatrix}a&b\\c&d\end{pmatrix}. \] The equality \(BE_{11}=E_{11}B\) forces \(b=c=0\). The equality with \(E_{12}\) then forces \(a=d\). Conversely every \(aI_2\) commutes with all matrices. Hence \(Z(R)=kI_2\), and Theorem 1.1 identifies all natural scalars with \(k\): each acts on every module by \(x\mapsto(aI_2)x\).

Left multiplication by \(E_{11}\) on \(R\) fails left linearity. In fact \[ E_{11}(E_{12}I_2)=E_{12},\qquad E_{12}(E_{11}I_2)=0. \] Thus it is not even an \(R\)-module endomorphism of the left regular module. Right multiplication by \(E_{11}\) is left linear, but fails naturality: it does not commute with right multiplication by \(E_{12}\). Formula (1.5) detects precisely this failure. Both tests distinguish an individual matrix action from an operation compatible with every module map.

Exercise 2 (intermediate: a bounded exponent). Fix a positive integer \(N\). Let \(\mathsf{FinAb}_N\) be the full category of finite abelian groups annihilated by \(N\). Determine its natural scalar ring, including \(N=1\). For \(N=12\), compute the primary projectors and the operation of residue \(7\) on \(\mathbb Z/4\oplus\mathbb Z/3\).

Solution. The group \(C_N=\mathbb Z/N\mathbb Z\) belongs to the category. A natural operation on it is multiplication by a unique \(a\in\mathbb Z/N\mathbb Z\). Every \(x\in A\) gives a homomorphism \(C_N\to A\) taking \(\overline1\) to \(x\). Naturality forces the operation on \(A\) to be \(x\mapsto ax\). Conversely this rule is well defined because \(NA=0\), and every group homomorphism commutes with it. Addition and composition correspond to the two operations on residues. Therefore \[ \mathfrak Z(\mathsf{FinAb}_N)\simeq\mathbb Z/N\mathbb Z. \tag{5.1} \] For \(N=1\), every object is zero, so there is just one natural transformation and its ring is the zero ring \(\mathbb Z/1\mathbb Z\).

For \(12=4\cdot3\), the residues \(c_2=9\) and \(c_3=4\) have the required values: \[ \begin{array}{c|cc} &\bmod4&\bmod3\\\hline 9&1&0\\ 4&0&1 \end{array} \] Their sum is \(13\equiv1\pmod {12}\), their product is \(36\equiv0\), and their squares are themselves modulo \(12\). Thus on the indicated direct sum they are respectively \((x,y)\mapsto(x,0)\) and \((x,y)\mapsto(0,y)\). Since \(7\equiv3\pmod4\) and \(7\equiv1\pmod3\), its operation is \[ (x,y)\longmapsto(3x,y). \tag{5.2} \]

Exercise 3 (advanced: finite probes cannot force one integer). Consider the natural operation \(e_2\) projecting each finite abelian group onto its \(2\)-primary summand. Prove that it is a nonzero idempotent and cannot be multiplication by one integer on all finite abelian groups. Prove nevertheless that, for any finite list of finite abelian groups, a single integer gives its operation on every group in that list.

Solution. The family corresponding to \(\alpha_2=1\) and \(\alpha_p=0\) for \(p\ne2\) is \(e_2\), by Theorem 4.1. Its square has the same residues, so it is idempotent. On \(\mathbb Z/2\) it is the identity and is nonzero.

If one integer \(m\) realized it everywhere, then the groups \(\mathbb Z/2^n\) would give \(m\equiv1\pmod {2^n}\) for every \(n\). Hence \(m=1\). The groups \(\mathbb Z/3^n\) would give \(m\equiv0\pmod {3^n}\) for every \(n\), hence \(m=0\). This is impossible.

For a finite list, choose a common positive annihilator \(N\), for instance a product of the annihilators used in Proposition 3.1. Write \(N=2^a b\) with \(b\) odd. If \(a>0\) and \(b>1\), Bézout's identity for \(b\) and \(2^a\) supplies an integer \(t\) with \[ \begin{gathered} t\equiv1\pmod {2^a},\\ t\equiv0\pmod b. \end{gathered} \tag{5.3} \] For example take \(t=sb\) where \(sb\equiv1\pmod {2^a}\). Its action is identity on every \(2\)-primary component and zero on every odd-primary component in the list. If \(a=0\), use \(t=0\); if \(b=1\) and \(a>0\), use \(t=1\). When \(N=1\) all groups are zero and any integer works. Thus every finite list has an integer representative, although all finite groups together do not.

Exercise 4 (advanced: torsion groups and a Prüfer group). Let \(\mathsf{TorsAb}\) be the category of all torsion abelian groups: every element is killed by a positive integer. Prove that restriction gives a ring isomorphism \[ \mathfrak Z(\mathsf{TorsAb}) \xrightarrow{\ \sim\ } \mathfrak Z(\mathsf{FinAb}). \tag{5.4} \] For \(P_p=\mathbb Z[1/p]/\mathbb Z\), compute \(\operatorname{End}(P_p)\) and identify how a natural scalar acts on it.

Solution. A subgroup generated by finitely many torsion elements is finite. For two generators killed by \(m,n\), it is an image of the finite group \(\mathbb Z/m\oplus\mathbb Z/n\); the same argument works for any finite number. An element of a torsion group lies in a finite cyclic subgroup and therefore has a finite, unique primary decomposition. The decompositions agree on inclusions of finite subgroups, since the primary subgroups are defined intrinsically by (3.1).

Given \(\alpha\in\prod_p\mathbb Z_p\), act on each element by its finite primary decomposition and compatible residues, as in (4.2). To check additivity for \(x,y\), put them in their finite generated subgroup and use the already-proved finite-group action there. Every homomorphism preserves primary elements, and choosing an exponent killing an element and its image proves (4.3). Thus these operations form a natural scalar on all torsion groups.

Any natural scalar on torsion groups is determined by its finite cyclic components: if \(mx=0\), naturality at \(\mathbb Z/m\to A\), \(\overline1\mapsto x\), determines its value at \(x\). Its restriction is therefore injective, and the construction from \(\alpha\) proves surjectivity. Since restriction preserves addition, composition and identity, it is the ring isomorphism (5.4).

For the Prüfer group, let \[ u_n=p^{-n}+\mathbb Z\in P_p. \] It has order \(p^n\), satisfies \(p u_{n+1}=u_n\), and its cyclic subgroups exhaust \(P_p\). Moreover \[ P_p[p^n]=\langle u_n\rangle\simeq C_{p,n}. \tag{5.5} \] Indeed, choose a representative \(q\in\mathbb Z[1/p]\) of a class killed by \(p^n\). Then \(p^n q\) is an integer, and the class is \((p^n q)u_n\).

Every endomorphism \(h\) preserves (5.5), giving unique residues \(a_n\) with \(h(u_n)=a_nu_n\). Applying \(h\) to \(p u_{n+1}=u_n\) gives \(a_{n+1}\equiv a_n\pmod {p^n}\). Conversely compatible residues give compatible endomorphisms of the increasing cyclic subgroups: the inclusion sends \(u_n\) to \(p u_{n+1}\), and this equality is respected. They therefore define one homomorphism on the union. Addition and composition are coordinatewise addition and multiplication of residues, proving \[ \operatorname{End}(P_p)\simeq\mathbb Z_p. \tag{5.6} \] Under (5.4) and (4.1), a natural scalar \((\alpha_q)_q\) acts on \(P_p\) by exactly \(\alpha_p\); every element is \(p\)-primary. Its values at all \(u_n\) specify the endomorphism completely.

References