The preceding chapters described abelian categories using morphisms and universal properties. Here we construct their main concrete example: modules over a ring. Elements are available in this example, so we use them to verify those universal properties.
Conventions: rings and sets
Every ring here is associative and has an identity \(1\); it need not be commutative. Ring homomorphisms preserve \(1\), and every module is unital. The zero ring is allowed; its only unital module is the zero module. Indexing collections are sets. We work in ordinary set theory with the axiom of choice; its uses in lifting families of generators are stated explicitly. An abelian group is a group with commutative addition.
5.1. Left, right, and two-sided actions
Definition 5.1. Modules and their morphisms
A left \(R\)-module is an abelian group \(M\) with a function \(R\times M\to M\), written \(rm\), satisfying
\[ r(m+m')=rm+rm',\quad (r+r')m=rm+r'm,\quad (rr')m=r(r'm),\quad 1m=m. \]A right \(R\)-module has an action \(mr\), distributive in both variables, with \(m(rr')=(mr)r'\) and \(m1=m\). A left-module morphism \(f:M\to N\) is an additive function satisfying \(f(rm)=rf(m)\); the right-module version satisfies \(f(mr)=f(m)r\). Their categories are \(R\text{-}\mathrm{Mod}\) and \(\mathrm{Mod}\text{-}R\), respectively. Composition preserves these conditions, and identity functions satisfy them.
Alphabeta Math: left and right modules · module homomorphisms
Worked example for Definition 5.1
Counterexample: Left multiplication is not usually a map of left regular modules
The ring \(R\), with left multiplication, is its left regular module. Over a field, modules are vector spaces. A left \(\mathbb Z\)-module is exactly an abelian group: \(nm\) is repeated addition for \(n\geq0\), and its negative for \(n<0\). Indeed, distributivity and \(1m=m\) force this formula; conversely, the formula satisfies all four module laws. Thus module theory already includes integer arithmetic and linear algebra.
Definition 5.2. The opposite ring
The opposite ring \(R^{\mathrm{op}}\) has the same addition and identity as \(R\), with multiplication \(r\mathbin{*}r'=r'r\). A right \(R\)-module is therefore a left \(R^{\mathrm{op}}\)-module by \(r\cdot m=mr\).
Worked example for Definition 5.2
Counterexample: Left multiplication is not usually a map of left regular modules
Verification.
The opposite multiplication is associative because \((r*r')*r''=r''(r'r)=(r''r')r=r*(r'*r'')\). Its distributive laws and identity follow from those of \(R\). For the proposed left action, \((r*r')\cdot m=m(r'r)=(mr')r=r\cdot(r'\cdot m)\); the other module laws are unchanged. The construction reverses itself and preserves morphisms, so the two descriptions give the same category.
□Definition 5.3. Bimodules
An \(S\)-\(R\) bimodule \(B\) is an abelian group with a left \(S\)-module action and a right \(R\)-module action satisfying \((sb)r=s(br)\). A bimodule morphism \(u:B\to B'\) is additive and preserves both actions. Write \(S\text{-}\mathrm{Mod}\text{-}R\) for this category.
The ring \(R\) is an \(R\)-\(R\) bimodule by multiplication; the compatibility is precisely associativity. More generally, a unital ring map \(\phi:R\to S\) makes \(S\) an \(S\)-\(R\) bimodule by \(s\cdot r=s\phi(r)\). The order \(S\)-\(R\) matters: in Chapter 6, such a bimodule takes left \(R\)-modules to left \(S\)-modules.
5.2. Submodules, quotients, and the abelian axioms
Proposition 5.4. Quotients and their universal property
A submodule \(U\subseteq M\) is an additive subgroup closed under the \(R\)-action. The cosets \(M/U\) form a module with \((m+U)+(n+U)=m+n+U\) and \(r(m+U)=rm+U\). The projection \(q:M\to M/U\) is surjective, and every module morphism \(f:M\to N\) with \(f(U)=0\) factors uniquely through \(q\).
Proof.
The relation \(m\sim n\iff m-n\in U\) is reflexive, symmetric, and transitive because \(U\) is a subgroup. Its classes are exactly the cosets. Replacing \(m,n\) by \(m+u,n+v\) changes their sum by \(u+v\in U\), and replacing \(m\) by \(m+u\) changes \(rm\) by \(ru\in U\). Thus the operations are well defined. The group and module laws follow by applying the corresponding laws to representatives. In particular, \(0+U\) is the zero and \(-m+U\) is the additive inverse of \(m+U\).
If \(f(U)=0\), define \(\bar f(m+U)=f(m)\). Equal cosets have difference in \(U\), so this is well defined; the displayed operations show it is linear. It satisfies \(\bar f q=f\). Every coset is \(q(m)\), so this equation forces the formula and proves uniqueness.
□Proposition 5.5. Kernels, cokernels, and images
For \(f:M\to N\), the submodules \(\ker f=\{m:f(m)=0\}\) and \(\operatorname{im}f=\{f(m):m\in M\}\) give the categorical kernel \(\ker f\hookrightarrow M\) and cokernel \(N\twoheadrightarrow N/\operatorname{im}f\). The canonical map
\[M/\ker f\longrightarrow\operatorname{im}f,\qquad m+\ker f\longmapsto f(m)\]is an isomorphism. Moreover, monomorphisms are exactly injective module maps, and epimorphisms are exactly surjective module maps.
Alphabeta Math: first isomorphism theorem · kernels, images, and cokernels
Proof.
The two subsets are closed under subtraction and scalar multiplication by linearity of \(f\). If \(fg=0\), then \(g\) takes values in \(\ker f\); regarding \(g\) as a map into this submodule is its unique factorization through the inclusion. A map \(h:N\to L\) satisfies \(hf=0\) precisely when it kills \(\operatorname{im}f\), so Proposition 5.4 gives the cokernel property.
The displayed map is well defined since \(f\) kills its kernel. It is surjective by the definition of image and injective because \(f(m)=0\) means \(m\in\ker f\). A bijective linear map has a linear inverse: write elements as images and apply its linearity to their sums and scalar multiples. This proves the isomorphism.
Injective maps cancel on the left and surjective maps cancel on the right. Conversely, if \(f\) is not injective, choose \(0\ne m\in\ker f\). The map \(R\to M\), \(r\mapsto rm\), is nonzero but has zero composite with \(f\), so \(f\) is not monic. If \(f\) is not surjective, the quotient map \(N\to N/\operatorname{im}f\) is nonzero but becomes zero after \(f\), so \(f\) is not epic.
□Theorem 5.6. Module categories are abelian
\(R\text{-}\mathrm{Mod}\) is a locally small abelian category.
Alphabeta Math: Module categories are abelian
Proof.
Morphisms \(M\to N\) form a set, being a subset of all functions between two sets. They form an abelian group under pointwise addition: a sum or negative of linear maps is linear. Composition distributes over addition by additivity of the outer map and by pointwise evaluation. The zero module is a zero object. For two modules, \(M\oplus N=M\times N\) with componentwise operations has inclusions \(m\mapsto(m,0)\), \(n\mapsto(0,n)\) and coordinate projections. A map into it is uniquely determined by its two components; a map out is uniquely \(h(m,n)=h_M(m)+h_N(n)\). Thus it is a binary biproduct. Iterating this construction gives every finite biproduct: the maps from, or to, a tuple of \(n+1\) modules are specified uniquely by those for the first \(n\) modules and the last module. The empty case is the zero module. Hence the category is additive.
Kernels and cokernels exist by Proposition 5.5. The coimage is \(M/\ker f\). The image, defined categorically as the kernel of the cokernel, is exactly \(\operatorname{im}f\), since the quotient map kills exactly that submodule. Proposition 5.5 identifies the canonical coimage-to-image map as an isomorphism. This is the abelian criterion of Chapter 4.
□For example, multiplication by \(2\) on \(\mathbb Z\) is injective and has cokernel \(\mathbb Z/2\mathbb Z\). The image \(2\mathbb Z\) differs as a subset from its domain \(\mathbb Z\), but \(n\mapsto2n\) is the coimage-to-image isomorphism.
5.3. Arbitrary direct sums and free presentations
Proposition 5.7. Products and coproducts of modules
For a set-indexed family \((M_i)_{i\in I}\), the product is the set of all tuples \(\prod_i M_i\), with componentwise operations. The coproduct is the direct sum \(\bigoplus_i M_i\subseteq\prod_i M_i\), consisting of tuples with only finitely many nonzero coordinates. Their structural maps are coordinate projections and coordinate inclusions, respectively.
Proof.
Products satisfy the module laws coordinatewise. Finite support is preserved by sums, negatives, and scalar multiplication, so the direct sum is a submodule. Given maps \(g_i:X\to M_i\), the unique product map is \(x\mapsto(g_i(x))_i\); it is linear coordinatewise. Given maps \(f_i:M_i\to Y\), the unique coproduct map is \((m_i)_i\mapsto\sum_i f_i(m_i)\). The sum is finite, is linear, and has the required restrictions. Every finite-support tuple is a finite sum of coordinate inclusions, proving uniqueness. For the empty index set both constructions give the zero module.
□For infinitely many copies of \(\mathbb Z\), the tuple \((1,1,1,\ldots)\) belongs to the product but not the direct sum. This distinction will account for the finite support in the smallness hypothesis.
Theorem 5.8. Free modules and presentations
Write \(R^{(I)}=\bigoplus_{i\in I}R\), and let \(e_i\) be the tuple with \(1\) at \(i\) and \(0\) elsewhere. Any family \((m_i)_{i\in I}\) in \(M\) determines a unique linear map \(R^{(I)}\to M\) taking \(e_i\) to \(m_i\). Every module has an exact sequence
\[R^{(J)}\xrightarrow{d}R^{(I)}\xrightarrow{q}M\longrightarrow0.\]Exactness means \(\operatorname{im}d=\ker q\) and \(q\) is surjective; equivalently, \(q\) is a cokernel of \(d\). Such a sequence is a free presentation.
Alphabeta Math: free modules · universal property · free quotients
Proof.
A tuple \((r_i)\) equals the finite sum \(\sum_i r_i e_i\); thus the only possible map is \((r_i)\mapsto\sum_i r_i m_i\). This formula is linear, proving existence and uniqueness. Set \(I\) equal to the underlying set of \(M\), and send \(e_m\) to \(m\). The resulting \(q\) is surjective since \(q(e_m)=m\). Let \(K=\ker q\), let \(J\) be the underlying set of \(K\), and send \(e_k\in R^{(J)}\) to \(k\in R^{(I)}\). Its image is \(K\), so the sequence is exact. The quotient property gives its cokernel description. These choices use whole underlying sets, so no choice of a basis or generating subset is needed.
□For \(\mathbb Z/n\mathbb Z\), \(n\geq1\), a presentation is \(\mathbb Z\xrightarrow{n}\mathbb Z\twoheadrightarrow\mathbb Z/n\mathbb Z\). A presentation says which formal generators to use and which linear relations to impose.
5.4. Why the regular module remembers everything
Theorem 5.9. The regular module is a small projective generator
Evaluation at \(1\) gives a natural isomorphism of abelian groups \(\operatorname{Hom}_R(R,M)\cong M\). Consequently, \(R\) is projective, is a generator, and is small: the canonical map
\[\bigoplus_i\operatorname{Hom}_R(R,M_i)\longrightarrow\operatorname{Hom}_R\left(R,\bigoplus_i M_i\right)\]is an isomorphism. Here projective means every map from \(R\) through the target of an epimorphism lifts through that epimorphism; generator means \(\operatorname{Hom}_R(R,-)\) distinguishes distinct morphisms; and small means the displayed preservation of coproducts. Also \(\operatorname{End}_R(R)\cong R^{\mathrm{op}}\) as rings, with endomorphism multiplication given by composition.
Alphabeta Math: projectivity · the universal property of free modules · opposite rings
Worked example for Theorem 5.9
Counterexample: Left multiplication is not usually a map of left regular modules
Proof.
A linear \(h:R\to M\) satisfies \(h(r)=rh(1)\). Conversely, \(m\in M\) defines \(h_m(r)=rm\), which is linear. These constructions are inverse and additive; for \(f:M\to N\), \(fh_m=h_{f(m)}\), proving naturality.
If \(p:X\twoheadrightarrow Y\) and \(h:R\to Y\), choose \(x\in X\) with \(p(x)=h(1)\). Then \(h_x\) lifts \(h\). If \(f,g:M\to N\) differ, choose \(m\) with \(f(m)\ne g(m)\); then \(fh_m\ne gh_m\), proving the generator property. Under evaluation, the smallness map is exactly the identity between finite-support tuples, hence an isomorphism.
Finally, every endomorphism of the left regular module is \(\rho_a(r)=ra\) for its value \(a\) at \(1\). Addition corresponds to addition in \(R\), and \(\rho_a\circ\rho_b=\rho_{ba}\). The identity corresponds to \(1\). Thus \(a\mapsto\rho_a\) is a unital ring isomorphism from \(R^{\mathrm{op}}\).
□The next chapter constructs tensoring from explicit relations. Chapters 8 and 9 will use presentations and the regular module to recover an entire category and an entire functor from this small amount of data.
Sources and further reading
- Alphabeta Math: left and right modules, the universal property of free modules, and module categories are abelian. The constructions and proofs needed here are given above.
Worked examples
Each example checks the hypotheses and the calculation for the linked statements. The links in the statements return here.
Worked example 1. Matrix actions keep track of the sides
Applies to: Definition 5.1, Definition 5.2, Definition 5.3.
Let \(R=M_2(\mathbb R)\). Columns \(\mathbb R^2\) are left modules by matrix multiplication; rows are right modules. Distributivity and \((AB)v=A(Bv)\), \((wA)B=w(AB)\), and the identity matrix verify every module axiom. The map \(v\mapsto2v\) is a left-module morphism by \(2Av=A(2v)\). On rows the opposite-ring left action is \(A^{\mathrm{op}}\cdot w=wA\): applying \(B\) first and then \(A\) gives \(wBA\), precisely the reversed multiplication. For example \(E_{12}E_{21}=E_{11}\) but \(E_{21}E_{12}=E_{22}\), so the reversal matters. The matrices themselves form an \(R\)-\(R\) bimodule since \((AB)C=A(BC)\); multiplication by the central scalar 2 preserves both actions and is a bimodule map.
Worked example 2. A quotient over a polynomial ring
Applies to: Proposition 5.4, Proposition 5.5, Theorem 5.6.
Let \(R=\mathbb R[t]\), \(M=R\), and \(U=tR\). This is a submodule since multiples of \(t\) remain such after addition and multiplication. The map \(p(t)\mapsto p(0)\) induces \(R/tR\cong\mathbb R\), with \(R\)-action \(p(t)a=p(0)a\). A module map killing \(tR\) factors uniquely by \([p]\mapsto f(p)\). Multiplication by \(t\) on \(R\) has zero kernel and image \(tR\); its coimage-image comparison \(R\to tR\), \(p\mapsto tp\), has inverse \(tp\mapsto p\). This works because a polynomial killed by \(t\) is zero. The resulting cokernel is the displayed evaluation module. For arbitrary maps of \(R\)-modules, kernels and images are closed under the action, and the map \([m]\mapsto f(m)\) is always a module bijection. Together with the componentwise additive structure and direct sums, this verifies the abelian-category structure used in Theorem 5.6.
Further proof details
The general comparison used in this example has a direct proof. If \(m-m'\in\ker f\), then \(f(m)=f(m')\), so \([m]\mapsto f(m)\) is well defined. It preserves sums (and scalars in the module setting). It is onto \(\operatorname{im}f\) by the definition of image. If \(f(m)=0\), then \(m\in\ker f\), so its kernel is zero and it is injective. Its inverse sends \(f(m)\) to \([m]\); equality of two image representatives makes their difference a kernel element, proving the inverse well defined and additive, and linear when relevant. Kernels are closed under addition and scalars by applying \(f\); quotient operations are well defined because changing representatives changes a sum or scalar multiple by an element of the submodule. These formulas, together with the displayed direct sums, establish the asserted abelian-category comparison rather than merely naming an isomorphism theorem.
Worked example 3. Infinite tuples and a one-relation presentation
Applies to: Proposition 5.7, Theorem 5.8.
For countably many copies of \(R=\mathbb Z\), the product contains \((1,1,\ldots)\), whereas the direct sum does not. Projections factor any family of incoming maps by coordinate tuples. A family of outgoing maps from the summands extends by the finite sum \((n_i)\mapsto\sum_i n_i m_i\); uniqueness follows from the basis vectors. This also verifies the free-module property. As a finite presentation, \(\mathbb Z\xrightarrow{6}\mathbb Z\xrightarrow q\mathbb Z/6\to0\) is exact: \(q\) is onto and its kernel consists exactly of multiples of 6. The single generator \(\bar1\) and single relation \(6\bar1=0\) therefore recover this module as the specified cokernel.
Worked example 4. The regular module over a noncommutative ring
Applies to: Theorem 5.9.
Again let \(R=M_2(\mathbb R)\). A left-linear map \(h:R\to M\) is \(h(A)=Ah(I)\), so evaluation at \(I\) is a natural bijection. A preimage of \(h(I)\) lifts a map across a surjection; testing \(I\mapsto m\) detects distinct maps; finite support of \(h(I)\) proves smallness. Every endomorphism of \({}_RR\) is \(\rho_B(A)=AB\). Composition satisfies \(\rho_B\rho_C(A)=ACB=\rho_{CB}(A)\), so the endomorphism ring is \(R^{\mathrm{op}}\), not merely a set of matrices with an unspecified multiplication. This checks every part of the regular-generator assertion with noncommuting coefficients.
Counterexamples and limits of the hypotheses
Each construction below refutes the explicitly stated stronger claim. It does not refute the related theorem with its original hypotheses.
Counterexample 1. Left multiplication is not usually a map of left regular modules
Related statements: Definition 5.1, Definition 5.2, Theorem 5.9.
Claim refuted. For any ring element, left multiplication is automatically an endomorphism of the left regular module.
Proof.
Let \(R=M_2(\mathbb R)\), and write \(E_{ij}\) for the matrix unit with a single 1 in position \((i,j)\). Put \(a=E_{12}\), \(r=E_{21}\), and \(L_a(x)=ax\). This map is additive, but at \(x=I\), \(L_a(rx)=ar=E_{11}\), whereas \(rL_a(x)=ra=E_{22}\). The matrices differ, so left linearity fails. By contrast \(\rho_a(x)=xa\) is left linear because \(\rho_a(rx)=(rx)a=r(xa)\). Further \(\rho_a\rho_b(x)=(xb)a=x(ba)\), which explains the opposite multiplication in \(\operatorname{End}_R({}_RR)\cong R^{\mathrm{op}}\). The counterexample isolates the side convention, not a failure of associativity.
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