Character duals and one-sided injectivity

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

A character records an element of an abelian group by its value in \(\mathbb Q/\mathbb Z\). Enough characters distinguish every nonzero element. They also distinguish homomorphisms, even when the group is infinite. For a module, the same construction reverses the side on which the ring acts. Keeping that side visible explains why the character dual of a projective left module is an injective right module.

We assume ordinary modules over a unital ring, short exact sequences, and the lifting definition of projectivity. The abelian framework is developed in Trace coreflections and balanced nonabelian categories. Basic references are Stacks, Injective abelian groups, Stacks, Injective modules, and the ordinary module discussion in Stacks, Injective modules over algebras. The stronger functorial embedding theorem is already proved in Injective modules and bounded below derived functors, Lemmas 2.1–2.2. We use its ordinary module result here.

All groups and index sets lie in a fixed universe. We use ordinary choice for families of extensions. Rings need not be commutative; modules are unital. All Hom groups below are algebraic. Put

1. Evaluation preserves the module side

Two established group facts will be used throughout. An abelian group \(J\) is injective precisely when it is divisible, meaning \(nJ=J\) for every positive integer \(n\); this is Stacks, Lemma 15.55.1. Given \(q+\mathbb Z\in E\), the class \(q/n+\mathbb Z\) multiplies by \(n\) to \(q+\mathbb Z\). Thus \(E\) is injective. Consequently character duality is exact, and the evaluation homomorphism into the double character dual is injective; these are Stacks, Lemmas 15.56.6–7. The element-separating character in the latter statement extends from the cyclic subgroup generated by that element. These group results apply to the underlying groups of modules.

If \(M\) is a left \(R\)-module, make \(M^\vee\) a right \(R\)-module by

Indeed,

If \(N\) is a right \(R\)-module, its dual is a left module with

Here \((r(s\psi))(n)=\psi(nrs)=((rs)\psi)(n)\). Both actions are additive in each variable, and the identity of \(R\) acts as the identity. Thus the second dual returns to the original side.

A left module map \(f:M\to N\) induces the right module map

For example,

Identity maps and composites are preserved contravariantly:

Exactness on the underlying groups gives exactness with these actions. In particular, a short exact sequence of left modules gives a short exact sequence of right modules in the reverse direction.

Define evaluation by

Proposition 1.1. Evaluation is a natural monomorphism of left \(R\)-modules. Character duality is faithful: for left modules \(M,N\), the map

is injective. Both statements hold with left and right exchanged.

Proof. Evaluation is additive. Its left linearity follows from the two side conventions:

The group evaluation theorem recalled above makes it monic. For \(f:M\to N\), evaluation at \(\psi\in N^\vee\) gives

Hence \(f^{\vee\vee}e_M=e_Nf\). If \(f^\vee=g^\vee\), then \(f^{\vee\vee}=g^{\vee\vee}\), so naturality gives \(e_Nf=e_Ng\). Cancel the monomorphism \(e_N\) to get \(f=g\). Replacing \(R\) by \(R^{\mathrm{op}}\) proves the right module statements. \(\square\)

This proves the Hom assertion for arbitrary abelian groups by taking \(R=\mathbb Z\). It needs neither finite generation nor an isomorphism \(M\cong M^{\vee\vee}\). Exercise 4 will show that evaluation can fail to be surjective even for \(M=\mathbb Z\).

2. Dualizing a projective splitting

An injective right module \(J\) has the following extension property: every right linear map \(U\to J\) extends across every monomorphism \(U\hookrightarrow V\). We use the ordinary noncommutative result that the dual of the left regular module,

is injective as a right module. Its action is \((\xi a)(r)=\xi(ar)\). This is the coinduced-module result in Stacks, Section 22.17, with ground ring \(\mathbb Z\). The dual of the right regular module is instead injective on the left, with action \((a\xi)(r)=\xi(ra)\). The distinction matters when \(ar\ne ra\).

Lemma 2.1. A product of injective right modules is injective. A direct summand of an injective right module is injective.

Proof. For a map \(u:U\to\prod_{i\in I}J_i\) and a monomorphism \(U\hookrightarrow V\), extend each coordinate \(u_i:U\to J_i\) to a map \(v_i:V\to J_i\). Choice selects such an extension for every \(i\). The product map \((v_i)_i\) extends \(u\). For an empty index set the product is the zero module, whose extension property is immediate.

If \(s:J\to K\) and \(p:K\to J\) satisfy \(ps=1_J\), extend \(su:U\to K\) to \(v:V\to K\). Then \(pv\) extends \(u\). This proves the summand assertion. \(\square\)

Theorem 2.2. If \(P\) is a projective left \(R\)-module, then \(P^\vee\) is an injective right \(R\)-module.

Proof. Take a free left module \(F=R^{(I)}\) and a surjection \(q:F\to P\). For example, one may use a basis indexed by the elements of \(P\). The lifting definition of projectivity applied to \(1_P\) provides a left linear section \(s:P\to F\) with \(qs=1_P\). Dualizing gives right linear maps

Restriction to the summands of \(F\) identifies

as right modules. The inverse sends \((\xi_i)_i\) to the character \((r_i)_i\mapsto\sum_i\xi_i(r_i)\); this sum is finite on every element of \(F\). Both maps respect (1.1). Every factor in (2.3) is injective by the retained regular-module result. Lemma 2.1 makes \(F^\vee\) injective and then makes its summand \(P^\vee\) injective. \(\square\)

The theorem uses a splitting of a possibly infinite free module. No finite-rank or finite-presentation condition is required. Applying it to \(R^{\mathrm{op}}\) gives the corresponding assertion for projective right modules. Exercise 2 computes both sides over a noncommutative triangular ring and exhibits a nonprojective module whose dual is not injective.

3. The two injective embeddings have the same coordinates

We recall the ordinary module consequence of Injective modules and bounded below derived functors, Lemmas 2.1–2.2: every left \(R\)-module admits a natural monomorphism into an injective left module. Write its coinduced module as

This supplies enough injectives over every unital ring. We now compare its coordinates with a construction that uses the two duals and a free module. The comparison will also explain which side the free module occupies.

Let \(F(N)\) be the free right \(R\)-module on the underlying set of a right module \(N\), with basis symbols \([n]\). Its canonical surjection is

It is functorial: a right linear \(v:N\to N'\) gives \(F(v)([n])=[v(n)]\). The ordinary character-dual construction in Stacks, Section 22.17 uses this free module. Write it as

Theorem 2.2, on the opposite side, makes \(J(M)\) injective. The map \(q_{M^\vee}\) is surjective, so exact character duality makes its dual injective. Proposition 1.1 therefore makes \(j_M\) injective. Thus (3.3) agrees with the required existence statement. The stronger functorial theorem (3.1) remains the underlying module foundation.

For \(\chi\in M^\vee\), define

Proposition 3.1. The correspondence \(\chi\mapsto S_\chi\) is a bijection from \(M^\vee\) to \(\operatorname{Hom}_R(M,D)\). There is a natural isomorphism of left modules

such that \(T_Mj_M=\iota_M\). Its component indexed by \(S_\chi\) is

Proof. The formula (3.4) is left linear, since

For a left linear \(h:M\to D\), define \(\chi_h(m)=h(m)(1)\). Then

Conversely \(S_\chi(m)(1)=\chi(m)\). These identities prove bijectivity, including naturality under precomposition by a left module map.

Every element of \(F(M^\vee)\) is a finite sum of basis terms \([\chi]r\). Thus (3.6) gives a bijection: its inverse sends \((\xi_{S_\chi})_\chi\) to

For \(a\in R\), the component of \(T_M(a\eta)\) at \(S_\chi\), evaluated at \(r\), is \(\eta([\chi]ra)\). This equals \((a(T_M\eta)_{S_\chi})(r)\) by the left action on \(D\). Hence \(T_M\) is left linear.

Finally,

This proves \(T_Mj_M=\iota_M\). For a left map \(u:M\to N\), use \(J(u)=F(u^\vee)^\vee\). Formula (3.6) shows that \(J(u)\) reindexes the coordinate at \(\psi\in N^\vee\) by \(\psi u\in M^\vee\). Under (3.4) this is exactly the coordinate rule \(h\mapsto hu\) for \(I(u)\). Therefore (3.5) is natural. \(\square\)

The coordinate comparison does not say that the functors \(I\) and \(J\) are additive. A functorial choice of injective embeddings need not be an additive functor, as Exercise 3 makes explicit.

4. Exercises with complete solutions

Exercise 1 — Introductory: cyclic characters and two different arrows. For a positive integer \(n\), let \(C_n=\mathbb Z/n\mathbb Z\). Identify \(C_n^\vee\) and \(C_n^{\vee\vee}\) explicitly, and compute evaluation. If \(n\mid m\), compute the duals of the reduction \(\rho:C_m\to C_n\) and the injection \(\lambda:C_n\to C_m\), \(\lambda(1)=m/n\). Extend the evaluation calculation to a finite direct sum of cyclic groups.

Solution. An additive homomorphism from \(C_n\) to \(E\) is determined by the image of \(1\), which is killed by \(n\). The elements of \(E\) killed by \(n\) are precisely \(j/n+\mathbb Z\), for \(j\in\mathbb Z/n\mathbb Z\): if \(nq\) is an integer, \(q\) has this form modulo \(\mathbb Z\). Thus

In these coordinates evaluation sends \(a\) to the character \(j\mapsto aj/n+\mathbb Z\), so it is the identity of \(C_n\) after the second identification (4.1). The case \(n=1\) gives the zero group and fits the same calculation.

For \(n\mid m\), precomposing \(\chi_j:C_n\to E\) by \(\rho\) sends \(1\in C_m\) to \(j/n=(m/n)j/m\) modulo \(\mathbb Z\). Thus \(\rho^\vee\), in the cyclic coordinates, is multiplication by \(m/n\) from \(C_n\) to \(C_m\). Precomposing \(\chi_t:C_m\to E\) by \(\lambda\) sends \(1\in C_n\) to \((m/n)t/m=t/n\). Thus \(\lambda^\vee\) is reduction from \(C_m\) to \(C_n\). The two arrows exchange under duality.

For \(A=\bigoplus_{i=1}^d C_{n_i}\), restriction to each summand identifies \(A^\vee\) with \(\prod_{i=1}^d C_{n_i}\). The inverse adds the finitely many coordinate characters. Because the index set is finite, this product is the same group as the direct sum. The resulting pairing is

Evaluation is the identity on every coordinate, hence an isomorphism. This conclusion applies to these finite groups; Proposition 1.1 alone does not assert it for infinite groups. \(\square\)

Exercise 2 — Intermediate: a triangular ring exposes the side reversal. Let \(k=\mathbb F_p\), and let \(R\) be the ring of upper triangular \(2\times2\) matrices over \(k\). Compute the right character duals of the left projectives \(Re_{11}\) and \(Re_{22}\). Let \(S_2\) be the one-dimensional left module on which a matrix acts by its lower right entry. Prove that \(S_2\) is not projective and that \(S_2^\vee\) is not injective, using explicit nonsplit maps.

Solution. Write

For any \(k\)-vector space \(V\), every character \(V\to E\) takes values in \(E[p]\). The map \(k\to E[p]\), \(j\mapsto j/p+\mathbb Z\), identifies that group with \(k\). Additive maps between groups killed by \(p\) are \(k\)-linear. Therefore the character duals here are exactly the \(k\)-linear duals, with the right action (1.1).

The left ideal \(P_1=Re_{11}\) has basis \(e_{11}\); its action is multiplication by \(a\). The left ideal \(P_2=Re_{22}\) has ordered basis \(e_{12},e_{22}\), and its action on the coordinates \((u,v)\) is

Both are projective, since \(R=P_1\oplus P_2\) as left modules and the left regular module is free. Their duals have right actions

The second formula follows by evaluating the row functional \((s,t)\) on (4.4). Theorem 2.2 makes both right modules in (4.5) injective.

Let \(S_1=P_1\). There is a short exact sequence of left modules

where \(i(u)=(u,0)\) and \(q(u,v)=v\). If \(q\) had a left linear section, the image of \(1\in S_2\) would be \((d,1)\) for some \(d\in k\). But \(e_{12}\) kills \(S_2\), while

No section exists. Projectivity of \(S_2\) would provide one by lifting its identity through \(q\), so \(S_2\) is not projective.

Dualizing (4.6) gives the monomorphism

The right action on \(S_2^\vee\) is multiplication by \(c\). A retraction of (4.8), being \(k\)-linear and equal to the identity on its image, would have the form \(\tau(s,t)=ds+t\). Right linearity with respect to \(e_{12}\) would require

The left side is \(\tau(0,s)=s\), a contradiction at \(s=1\). Hence (4.8) has no right linear retraction. If \(S_2^\vee\) were injective, its identity would extend across (4.8) and provide exactly such a retraction. It is therefore not injective. \(\square\)

Exercise 3 — Advanced: natural embeddings do not give an additive functor. Use the functor \(J\) of (3.3). Give its action on a left module map \(u:M\to N\) in the coordinates of (3.6), and verify its composition law and the naturality of \(j\). For \(R=\mathbb Z\), show that \(J\) is not additive by computing its action on the zero endomorphism of \(M=\mathbb Z\). Explain why this does not contradict the natural isomorphism \(J\cong I\).

Solution. With coordinates indexed by \(\chi\in M^\vee\), write an element of \(J(M)\) as \((\xi_\chi)_\chi\), each \(\xi_\chi\in D\). The map \(F(u^\vee)\) sends \([\psi]r\), for \(\psi\in N^\vee\), to \([\psi u]r\). Dualizing therefore gives

The expression is left linear because every coordinate is selected without changing its left \(R\)-module action. For \(v:N\to L\), the component of \(J(v)J(u)\xi\) at \(\omega\in L^\vee\) is \(\xi_{(\omega v)u}=\xi_{\omega(vu)}\), which is the component of \(J(vu)\xi\). The identity map selects each original coordinate. Thus (4.10) is a functorial action.

The coordinate of \(j_M(m)\) at \(\chi\), evaluated at \(r\), is \(\chi(rm)\). Its coordinate after \(J(u)\) at \(\psi\) is \((\psi u)(rm)=\psi(ru(m))\), since \(u\) is left linear. This equals the corresponding coordinate of \(j_N(u(m))\), proving naturality directly.

Now take \(R=M=\mathbb Z\). A homomorphism \(\mathbb Z\to E\) is determined by its value at \(1\), so both \(M^\vee\) and \(D\) identify with \(E\). In (4.10), every \(\psi\) composed with the zero endomorphism is the zero character. Hence

Choose the input whose zero-character coordinate is \(1/2+\mathbb Z\) and whose other coordinates are zero. Its image is the nonzero constant family. Thus \(J(0)\) is a nonzero group homomorphism. An additive functor sends a zero morphism to a zero morphism, so \(J\) is not additive. Also \(J(0\text{ module})=D\cong E\ne0\), giving the same obstruction on objects.

Proposition 3.1 is an isomorphism of ordinary functors. It conjugates the maps \(J(u)\) and \(I(u)\) through the module isomorphisms \(T_M,T_N\); it makes no additivity assertion. In fact it transfers this nonadditivity to \(I\). Every \(j_M\) and every \(J(u)\) is individually linear, which is compatible with nonadditivity of the assignment on Hom groups. \(\square\)

Exercise 4 — Expert: the algebraic bidual of the integers. For a prime \(p\), define

where \(E[p^n]=\{x\in E:p^nx=0\}\), and the transition maps in the inverse limit are reduction. Prove an isomorphism of rings \(\operatorname{End}_{\mathbb Z}(E)\cong\prod_p\mathbb Z_p\). Under the group identification \(\mathbb Z^{\vee\vee}\cong\operatorname{End}_{\mathbb Z}(E)\), compute evaluation and exhibit an endomorphism outside its image. Deduce that faithful character duality is not full on all abelian groups.

Solution. Every element of \(E\) has finite order. If its order is \(d=\prod_{p\mid d}p^{a_p}\), the Chinese remainder theorem gives integers \(b_p\) such that, for every prime divisor \(q\ne p\) of \(d\),

Thus \(x=\sum_{p\mid d}b_px\), and \(b_px\in E_p\), since \(p^{a_p}b_p\) is divisible by \(d\). This proves that the \(E_p\) span \(E\). To prove that the sum is direct, suppose a finite sum \(\sum_p x_p=0\) has \(x_p\in E_p\). Choose exponents \(a_p\) killing the respective elements and coefficients as in (4.13). Multiplying the sum by \(b_p\) kills every other term and fixes \(x_p\). Hence every \(x_p=0\). We have proved

Every homomorphism \(E\to E\) preserves each \(E_p\), because it preserves the equation \(p^nx=0\). Conversely, any family of endomorphisms of the \(E_p\) defines an endomorphism of (4.14): each input has only finitely many nonzero components. Addition and composition are coordinatewise. Therefore

as rings. In particular, a map from \(E_p\) to \(E_q\) is zero for \(p\ne q\): its values are killed by a power of both primes and are therefore zero by Bézout's identity.

The subgroup \(E[p^n]\) is cyclic, generated by \(u_n=1/p^n+\mathbb Z\). An endomorphism of \(E_p\) restricts on it to multiplication by a unique residue \(c_n\in\mathbb Z/p^n\mathbb Z\). Since \(pu_{n+1}=u_n\), these residues satisfy

Conversely, compatible residues define an endomorphism on the increasing union \(E_p=\bigcup_n E[p^n]\). On \(E[p^n]\) use multiplication by \(c_n\). Compatibility makes the definitions agree on overlaps. This defines an additive map on the union; sums of two elements can be computed in a single sufficiently large cyclic subgroup. Addition of endomorphisms adds residues, and composition multiplies them. Thus \(\operatorname{End}(E_p)\cong\mathbb Z_p\) as rings. Substituting in (4.15) proves the requested ring isomorphism.

Evaluation at \(1\) identifies \(\mathbb Z^\vee\) with \(E\). An integer \(t\) is sent by double evaluation to \(x\mapsto tx\). Accordingly,

This map is injective also by the direct computation: an integer divisible by every positive integer is zero.

Let \(\epsilon:E\to E\) be the projection onto \(E_2\) in (4.14). It is nonzero because it fixes \(1/2+\mathbb Z\), and it is not the identity because it kills \(1/3+\mathbb Z\). If \(\epsilon\) were multiplication by an integer \(t\), its idempotence would make multiplication by \(t^2-t\) zero on \(E\). Injectivity of the integer evaluation map implies \(t^2-t=0\), so \(t=0\) or \(t=1\). Both possibilities contradict the two values just computed. Thus evaluation is not surjective.

Every group endomorphism of \(\mathbb Z\) is multiplication by an integer. Its character dual is multiplication by that same integer on \(E\). The endomorphism \(\epsilon\) therefore belongs to \(\operatorname{Hom}_{\mathbb Z}(\mathbb Z^\vee,\mathbb Z^\vee)\) but is not the dual of any map \(\mathbb Z\to\mathbb Z\). This proves failure of fullness, alongside the faithfulness proved in Proposition 1.1. \(\square\)

References

• The Stacks Project Authors, The Stacks Project, Injective abelian groups, Lemma 15.55.1; Injective modules, Lemmas 15.56.6–9; and the ordinary module body of Injective modules over algebras.
• Injective modules and bounded below derived functors, Section 2, Lemmas 2.1–2.2, in Derived categories and sheaf operations, for the complete ordinary Baer and functorial embedding proofs.