# Chapter 130. Modules and Linear Algebra # Chapter 130. Modules and Linear Algebra ## 130.1 Introduction A module is the analogue of a vector space when the scalars come from a ring instead of a field. In a vector space, every nonzero scalar has a multiplicative inverse. In a module, this property may fail. The change is small in definition but large in consequence. Modules generalize vector spaces, and abelian groups are exactly modules over the ring of integers \(\mathbb{Z}\). They are central in commutative algebra, homological algebra, representation theory, algebraic geometry, and algebraic topology. The guiding comparison is: | Linear algebra | Module theory | |---|---| | Field \(F\) | Ring \(R\) | | Vector space over \(F\) | Module over \(R\) | | Scalar multiplication by field elements | Scalar multiplication by ring elements | | Every nonzero scalar invertible | Nonzero scalars may fail to be invertible | | Every vector space has a basis | A module may have no basis | | Dimension behaves uniformly | Rank may be subtler or unavailable | Thus module theory is linear algebra over rings, but with fewer guarantees. ## 130.2 Rings as Scalar Systems A ring \(R\) is a set with addition and multiplication. Addition makes \(R\) an abelian group, multiplication is associative, and multiplication distributes over addition. A field is a special kind of ring in which every nonzero element has a multiplicative inverse. Therefore every vector space is a module, but not every module is a vector space. Examples of rings include: | Ring | Description | |---|---| | \(\mathbb{Z}\) | Integers | | \(F[x]\) | Polynomials over a field | | \(M_n(F)\) | \(n \times n\) matrices over a field | | \(\mathbb{Z}/n\mathbb{Z}\) | Integers modulo \(n\) | | \(C^\infty(M)\) | Smooth functions on a manifold | | \(R/I\) | Quotient ring | A ring may be commutative or noncommutative. In a commutative ring, $$ ab = ba. $$ In a noncommutative ring, the order of multiplication matters. This distinction affects modules. Over a noncommutative ring, left modules and right modules must be distinguished. ## 130.3 Definition of a Module Let \(R\) be a ring. A left \(R\)-module is an abelian group \(M\) with scalar multiplication $$ R \times M \to M, \qquad (r,m)\mapsto rm, $$ such that for all \(r,s\in R\) and all \(m,n\in M\), $$ r(m+n)=rm+rn, $$ $$ (r+s)m=rm+sm, $$ $$ (rs)m=r(sm), $$ and $$ 1m=m $$ when the ring has a multiplicative identity \(1\). The first condition says scalar multiplication distributes over addition in the module. The second says it distributes over addition in the ring. The third says scalar multiplication is compatible with ring multiplication. The fourth says the identity scalar acts as the identity map. A right \(R\)-module is defined similarly, but scalars act on the right: $$ M \times R \to M, \qquad (m,r)\mapsto mr. $$ The compatibility condition becomes $$ (mr)s=m(rs). $$ For commutative rings, left and right modules are usually identified. ## 130.4 Modules Compared with Vector Spaces The definition of a module resembles the definition of a vector space. The essential difference is the scalar system. Over a field, one may divide by every nonzero scalar. This supports many familiar arguments. For example, if $$ av=0 $$ and $$ a\neq 0, $$ then multiplication by \(a^{-1}\) gives $$ v=0. $$ Over a ring, this inference can fail. For example, in a \(\mathbb{Z}\)-module, the equation $$ 3m=0 $$ may have nonzero solutions. In the abelian group \(\mathbb{Z}/3\mathbb{Z}\), $$ 3\overline{1}=0. $$ Such elements are called torsion elements. This single difference explains much of the complexity of module theory. ## 130.5 Basic Examples ### Example 1. Vector Spaces If \(F\) is a field, then an \(F\)-module is exactly a vector space over \(F\). Thus module theory contains ordinary linear algebra as a special case. ### Example 2. Abelian Groups Every abelian group \(A\) is a \(\mathbb{Z}\)-module. For \(n\in\mathbb{Z}\) and \(a\in A\), define $$ na = \underbrace{a+\cdots+a}_{n\text{ times}} $$ for positive \(n\), define $$ 0a=0, $$ and define negative multiples using additive inverses. Conversely, every \(\mathbb{Z}\)-module is an abelian group. Hence abelian group theory is module theory over \(\mathbb{Z}\). ### Example 3. Ideals If \(R\) is a ring and \(I\subseteq R\) is an ideal, then \(I\) is an \(R\)-module under ring addition and scalar multiplication from \(R\). This example is one reason modules are important in commutative algebra. Ideals can be studied as modules. ### Example 4. Quotient Rings If \(I\) is an ideal in a commutative ring \(R\), then the quotient $$ R/I $$ is an \(R\)-module by $$ r(a+I)=ra+I. $$ ### Example 5. Polynomial Modules Let \(F\) be a field. A module over \(F[x]\) is an \(F\)-vector space together with a linear operator. Multiplication by \(x\) acts as the operator. This viewpoint connects module theory with rational canonical form and Jordan canonical form. ## 130.6 Submodules A submodule is the module analogue of a subspace. Let \(M\) be an \(R\)-module. A subset \(N\subseteq M\) is a submodule if: 1. \(N\) is closed under addition, 2. \(N\) is closed under additive inverses, 3. \(N\) is closed under scalar multiplication by elements of \(R\). Equivalently, $$ m,n\in N,\ r\in R $$ imply $$ m+n\in N $$ and $$ rm\in N. $$ Examples include ideals of a ring, subgroups of an abelian group that are closed under integer multiplication, and invariant subspaces of a linear operator when viewed as modules over \(F[x]\). ## 130.7 Generated Submodules Let \(S\subseteq M\). The submodule generated by \(S\) is the smallest submodule of \(M\) containing \(S\). It is written $$ \langle S\rangle_R. $$ Explicitly, it consists of all finite sums $$ r_1s_1+\cdots+r_ks_k, $$ where $$ r_i\in R,\qquad s_i\in S. $$ If \(S=\{m_1,\ldots,m_k\}\), then $$ \langle m_1,\ldots,m_k\rangle_R = \{r_1m_1+\cdots+r_km_k:r_i\in R\}. $$ This is the analogue of span. ## 130.8 Finitely Generated Modules An \(R\)-module \(M\) is finitely generated if there exist elements $$ m_1,\ldots,m_k\in M $$ such that $$ M=\langle m_1,\ldots,m_k\rangle_R. $$ That is, every element of \(M\) can be written as $$ r_1m_1+\cdots+r_km_k. $$ Every finite-dimensional vector space is finitely generated as a module over its field. But finite generation is weaker than having a basis. A finitely generated module may have relations among its generators. For example, \(\mathbb{Z}/6\mathbb{Z}\) is generated by \(\overline{1}\) as a \(\mathbb{Z}\)-module, but $$ 6\overline{1}=0. $$ The generator satisfies a nontrivial relation. ## 130.9 Free Modules A free module is a module with a basis. An \(R\)-module \(M\) is free if there exists a set \(B\subseteq M\) such that every \(m\in M\) can be written uniquely as a finite sum $$ m=r_1b_1+\cdots+r_kb_k, $$ where \(r_i\in R\) and \(b_i\in B\). The standard free module of rank \(n\) is $$ R^n. $$ Its standard basis is $$ e_1,\ldots,e_n. $$ Every element of \(R^n\) has a unique expression $$ (r_1,\ldots,r_n) = r_1e_1+\cdots+r_ne_n. $$ Free modules behave most like vector spaces. However, unlike vector spaces, not every module is free. Some modules have no basis at all. ## 130.10 Torsion Let \(R\) be an integral domain. An element \(m\in M\) is a torsion element if there exists a nonzero \(r\in R\) such that $$ rm=0. $$ A module is torsion-free if its only torsion element is \(0\). For example, as a \(\mathbb{Z}\)-module, $$ \mathbb{Z}/n\mathbb{Z} $$ is torsion, since $$ n\overline{a}=0 $$ for every element \(\overline{a}\). The module \(\mathbb{Z}\) is torsion-free, since $$ na=0 $$ with \(n\neq0\) implies $$ a=0. $$ Torsion has no analogue in vector spaces over fields. It appears because nonzero scalars may fail to be invertible. ## 130.11 Quotient Modules Let \(N\) be a submodule of \(M\). The quotient module $$ M/N $$ is the set of cosets $$ m+N. $$ Addition is defined by $$ (m+N)+(m'+N)=(m+m')+N. $$ Scalar multiplication is defined by $$ r(m+N)=rm+N. $$ The quotient module identifies elements of \(M\) that differ by an element of \(N\). This construction generalizes quotient vector spaces and quotient groups. ## 130.12 Module Homomorphisms Let \(M\) and \(N\) be \(R\)-modules. A function $$ f:M\to N $$ is an \(R\)-module homomorphism if $$ f(m+m')=f(m)+f(m') $$ and $$ f(rm)=rf(m) $$ for all $$ m,m'\in M,\qquad r\in R. $$ A homomorphism preserves addition and scalar multiplication. Its kernel is $$ \ker f=\{m\in M:f(m)=0\}. $$ Its image is $$ \operatorname{im} f=\{f(m):m\in M\}. $$ Both are submodules. An isomorphism is a bijective homomorphism. Two modules are isomorphic when they have the same module structure, even if their elements are presented differently. ## 130.13 Exact Sequences A sequence of module homomorphisms $$ M_1 \xrightarrow{f} M_2 \xrightarrow{g} M_3 $$ is exact at \(M_2\) if $$ \operatorname{im} f = \ker g. $$ A longer sequence is exact if it is exact at every intermediate term. A short exact sequence has the form $$ 0\to A\xrightarrow{f} B\xrightarrow{g} C\to 0. $$ This says that \(f\) identifies \(A\) with a submodule of \(B\), and \(C\) is the corresponding quotient: $$ C\cong B/f(A). $$ Exact sequences are one of the main organizing tools of module theory. ## 130.14 Matrices over Rings Matrices over rings still define homomorphisms between free modules. If $$ A\in M_{m,n}(R), $$ then \(A\) defines a map $$ R^n\to R^m. $$ For a column vector \(x\in R^n\), $$ x\mapsto Ax. $$ When \(R\) is commutative, this behaves much like ordinary matrix multiplication. When \(R\) is noncommutative, order matters. Entries must be placed consistently, and one must distinguish left modules from right modules. Some familiar matrix facts fail over rings. For example, a matrix may have a left inverse without having a two-sided inverse. Row operations may also require care because a nonzero ring element may not be invertible. ## 130.15 Linear Systems over Rings A linear system over a ring still has the form $$ Ax=b. $$ But solving such a system is more delicate than over a field. Over a field, Gaussian elimination uses division by nonzero pivot elements. Over a ring, a nonzero pivot may have no inverse. Therefore ordinary elimination may fail. For example, over \(\mathbb{Z}\), the equation $$ 2x=1 $$ has no solution in \(\mathbb{Z}\), because \(1\) is not divisible by \(2\). The equation $$ 2x=4 $$ has the solution $$ x=2. $$ Thus solvability depends on divisibility in the ring. Over principal ideal domains, such as \(\mathbb{Z}\) or \(F[x]\), one can use Smith normal form to analyze systems of equations. ## 130.16 Smith Normal Form Let \(R\) be a principal ideal domain. For a matrix $$ A\in M_{m,n}(R), $$ there exist invertible matrices \(P\) and \(Q\) such that $$ PAQ = \begin{bmatrix} d_1 & 0 & \cdots & 0\\ 0 & d_2 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & d_r\\ 0 & 0 & \cdots & 0 \end{bmatrix}, $$ where $$ d_1\mid d_2\mid \cdots \mid d_r. $$ This is the Smith normal form. It generalizes diagonalization for matrices over rings. It is especially important over \(\mathbb{Z}\) and \(F[x]\). Over \(\mathbb{Z}\), it classifies finitely generated abelian groups. Over \(F[x]\), it is closely related to rational canonical form. ## 130.17 Modules over Principal Ideal Domains A principal ideal domain, or PID, is an integral domain in which every ideal is generated by a single element. Examples include $$ \mathbb{Z} $$ and $$ F[x] $$ when \(F\) is a field. Finitely generated modules over a PID have a strong classification theorem. If \(M\) is a finitely generated module over a PID \(R\), then $$ M \cong R^r \oplus R/(d_1) \oplus \cdots \oplus R/(d_k), $$ where $$ d_1\mid d_2\mid \cdots \mid d_k. $$ The free part $$ R^r $$ measures rank. The quotient terms measure torsion. This theorem generalizes the classification of finitely generated abelian groups. ## 130.18 Projective Modules A projective module is a direct summand of a free module. That means \(P\) is projective if there exists another module \(Q\) such that $$ P\oplus Q $$ is free. Every free module is projective, but not every projective module is free. Projective modules appear naturally in geometry. For example, sections of vector bundles over a suitable space correspond to projective modules over a ring of functions. This principle is one of the bridges between algebra and geometry. ## 130.19 Tensor Products Given modules \(M\) and \(N\), the tensor product $$ M\otimes_R N $$ is a module generated by formal symbols $$ m\otimes n $$ subject to bilinearity relations: $$ (m+m')\otimes n = m\otimes n + m'\otimes n, $$ $$ m\otimes(n+n')=m\otimes n + m\otimes n', $$ and $$ (rm)\otimes n = m\otimes(rn). $$ Tensor products allow bilinear maps to be treated as linear maps. They are fundamental in multilinear algebra, differential geometry, algebraic topology, and homological algebra. ## 130.20 Modules and Linear Operators Let \(V\) be a vector space over a field \(F\), and let $$ T:V\to V $$ be a linear operator. Then \(V\) can be made into an \(F[x]\)-module by defining $$ p(x)\cdot v = p(T)v. $$ In particular, $$ x\cdot v = T(v). $$ This construction converts the study of a linear operator into the study of a module over the polynomial ring \(F[x]\). The structure theorem for finitely generated modules over \(F[x]\) gives canonical forms for linear operators, including rational canonical form and, when the polynomial factors appropriately, Jordan canonical form. ## 130.21 Summary Modules extend linear algebra from fields to rings. The definitions remain close to vector spaces, but the theory becomes richer because ring elements may fail to be invertible. The main concepts are: | Concept | Meaning | |---|---| | Module | Vector-space-like object over a ring | | Submodule | Subspace analogue | | Generated submodule | Span analogue | | Free module | Module with a basis | | Torsion | Nonzero element killed by nonzero scalar | | Quotient module | Module of cosets | | Homomorphism | Structure-preserving map | | Exact sequence | Algebraic bookkeeping of kernels and images | | Smith normal form | Diagonal form over a PID | | Projective module | Direct summand of a free module | | Tensor product | Linearization of bilinear maps | Module theory preserves the language of linear algebra while removing the assumption that scalars form a field. It explains why linear algebra works so cleanly over fields and what remains when division is no longer available.