Module (mathematics, $M$)

A left $R$-module consists of an abelian group $(M,+)$ and scalar multiplication $R\times M\to M$, written $(r,m)\mapsto rm$, where $R$ is a ring with identity element $1$. For every $r,s\in R$ and $m,n\in M$, requireThe module elements can be added to one another and multiplied by scalars; a product of two module elements is not part of the definition. These conventions agree with ^[ref]. For a noncommutative ring, the side matters: a right module instead has $(mr)s=m(rs)$. Over a commutative ring, the two conventions agree.

Over a field $k$, these are exactly the vector space axioms. A nonzero $a\in k$ has a multiplicative inverse, so $am=0$ implies $m=a^{-1}(am)=0$. Over the integers, $2$ has no multiplicative inverse: $2x=1$ has no solution in $\mathbb{Z}$. In the module $\mathbb{Z}/6\mathbb{Z}$, division can also lose uniqueness: $2\overline{1}=2\overline{4}=\overline{2}$. A ring element which is a unit still acts invertibly on every module; the difficulty lies with scalars which are not units.

Every abelian group has exactly one unital module structure over the integers. For $n>0$, define $nm$ by adding $m$ to itself $n$ times, put $0m=0$, and put $(-n)m=-(nm)$. The module axioms force this action, so no additional choice is involved. Moreover, every Abelian group homomorphism respects it and is a module homomorphism. See David McKinnon's Modules, the integer-scalar example.

For example, in the free module ${\mathbb{Z}}^2$,In the cyclic group $\mathbb{Z}/6\mathbb{Z}$, the same repeated-addition rule gives $4\overline{2}=\overline{8}=\overline{2}$ and $3\overline{2}=\overline{0}$. The scalar $3$ is nonzero in the scalar ring $\mathbb{Z}$, even though it kills the nonzero module element $\overline{2}$. Thus the scalar ring and the module must be kept distinct.

A linear operator $T:V\to V$ on a vector space over a field $k$ makes $V$ a module over the polynomial ring $k[t]$: let $t$ act as $T$, let constants act as the original scalars, and define $p(t)\cdot v=p(T)v$. The identity $(pq)(T)=p(T)q(T)$ for polynomials $p,q$ verifies the multiplication axiom. The submodules are precisely the invariant subspaces of $T$. This construction is described in Keith Conrad's Introductory Notes on Modules, section 6.

For a computed example, take $k=\mathbb{Q}$, $V=k^2$, andThen $T{e}_2=e_1$, $T{e}_1=0$, and $T^2=0$. ConsequentlyEvery $(a,b)$ equals $(b+at)\cdot e_2$. The module homomorphism $k[t]\to V$, $p\mapsto p(T)e_2$, is therefore surjective, and its kernel is the ideal $(t^2)$. It induces a module isomorphism $V\cong k[t]/(t^2)$. This is a cyclic module and a torsion module over $k[t]$, since $t^{2}V=0$. Over $k$ it is a free module with basis $(e_1,e_2)$ and has no nonzero torsion elements of a module. Over $\mathbb{Q}$ its nonzero elements also have infinite additive order of a group element. Changing scalars changes the questions one asks about the same addition law.

A module is finitely presented if it has finitely many generators and finitely many defining relations, equivalently an exact sequence of the displayed form with m and n finite. Over a commutative ring a relation matrix records the presentation, and its minors define the Fitting ideals.

// Target: knot-theory.bigb

For a module over a commutative ring presented by a matrix with n generators, its zeroth Fitting ideal is generated by all n-by-n minors of the relation matrix. It is independent of the chosen finite presentation. Over a unique factorization domain, the greatest common divisor of these minors defines the module order up to a unit. This does not require a Smith normal form over the ring. The square Seifert-matrix presentation of the Alexander module makes the order its determinant.

// Target: knot-theory.bigb

A locally free module is one whose associated sheaf is locally a free module. In the finite constant-rank case, it becomes free of rank $r$ on a principal open subset cover of $\mathrm{Spec}A$. Equivalently it is a finite projective module of constant rank $r$. A surjection onto it splits, so the kernel is also finite projective. Its associated sheaf is a locally free sheaf and defines an algebraic vector bundle.

// Target: algebraic-geometry.bigb

A module is finitely presented when it admits the displayed exact sequence for finite integers $r,s$. Thus it has finitely many generators and finitely many relations. On a Noetherian ring, every finitely generated module has a finite presentation.

Changing the coefficient ring of a module along a ring homomorphism can mean restriction of scalars, extension of scalars, or coextension of scalars. The underlying module may be retained with a smaller ring action, or replaced by a tensor or Hom construction with the new ring.

For a ring homomorphism $R\to A$ and a left $R$-module $M$, the left $A$-action on ${\mathrm{Hom}}_R(A,M)$ is $(a\cdot h)(b)=h(ba)$. This construction is right adjoint to restriction of scalars.

For a ring homomorphism $R\to A$, tensoring an $R$-module with the $(A,R)$-bimodule $A$ gives an $A$-module. This functor is left adjoint to restriction of scalars.

For a ring homomorphism $\theta :R\to A$, an $A$-module becomes an $R$-module by the displayed action. Its underlying abelian group stays the same.

A module isomorphism is a bijection of modules over the same ring preserving addition and scalar multiplication. Its inverse preserves these operations too. When a linear operator $T$ gives a vector space an $F[t]$-module structure by $t\cdot v=T(v)$, a module isomorphism is exactly an invertible linear map intertwining the two operators. This translates classification of modules into matrix similarity and rational canonical form.

A torsion module over an integral domain $R$ is a module in which every element is a torsion element of a module. Each element may require a different nonzero annihilating scalar. The module $\mathbb{Z}/6\mathbb{Z}$ is torsion over $\mathbb{Z}$, and the polynomial module from a linear operator $\mathbb{Q}[t]/(t^2)$ is torsion over $\mathbb{Q}[t]$.

If the torsion module is a finitely generated module, one common nonzero scalar kills the whole module: multiply annihilating scalars for a finite generating set. Without finite generation this can fail. The quotient module $\mathbb{Q}/\mathbb{Z}$ is torsion, since the class of $a/b$ is killed by $b$, but no nonzero integer $n$ kills it all: the class of $1/(2|n|)$ survives multiplication by $n$. See Keith Conrad's Modules over a PID, exercise 6.5(c).

For a module $M$ over an integral domain $R$, its torsion submodule isThis is a submodule. If $rm=0$ and $sn=0$ with $r,s\ne 0$, then $rs(m+n)=0$ and $rs\ne 0$; also $r(am)=a(rm)=0$ for every $a\in R$. The integral domain hypothesis supplies both the nonzero product and commutativity. These are the conventions of ^[ref].

A mixed example is the direct sum $M={\mathbb{Z}}^2\oplus \mathbb{Z}/6\mathbb{Z}$. If a nonzero integer kills $(a,b,\overline{c})$, then $a=b=0$. Conversely, $6$ kills every $(0,0,\overline{c})$. ThusThe torsion submodule isolates the finite-order part while the quotient module keeps the independent integer directions. In general $M/T(M)$ is a torsion-free module: if $r(m+T(M))=0$ with $r\ne 0$, some $s\ne 0$ kills $rm$, so $sr\ne 0$ kills $m$.

The hypothesis matters. In the ring $R=\mathbb{Z}/6\mathbb{Z}$ viewed as a module over itself, the elements $\overline{2}$ and $\overline{3}$ are killed by nonzero scalars $\overline{3}$ and $\overline{2}$, but their sum $\overline{5}$ is a unit and is killed by no nonzero scalar. Thus using arbitrary nonzero scalars in a ring with zero divisors need not give a submodule. A common broader convention uses regular elements of a ring as annihilators; the integral domain convention here avoids that ambiguity.

For a module $M$ over an integral domain $R$, a torsion element is an $m\in M$ with $rm=0$ for some $0\ne r\in R$. This includes zero. Equivalently, its annihilator ${\mathrm{Ann}}_R(m)=\{r\in R:rm=0\}$ is a nonzero ideal. The torsion elements of a module form the torsion submodule.

For $M=\mathbb{Z}/6\mathbb{Z}$, computed annihilators areFor these three elements, the additive order of a group element is respectively $6$, $3$, and $2$: these are the smallest positive integers killing them. Torsion over $\mathbb{Z}$ agrees exactly with finite additive order of a group element. Over a general integral domain, torsion concerns the chosen scalars instead; the polynomial module from a linear operator gives a torsion example whose nonzero elements have infinite additive order of a group element.

An $R$-module homomorphism is an additive map $f:M\to N$ satisfying $f(rm)=rf(m)$.

A module endomorphism is a module homomorphism from a module to itself. These maps form the endomorphism ring ${\mathrm{End}}_R(M)$ under pointwise addition and composition. An invertible module endomorphism is a module automorphism.

A module automorphism is an invertible module endomorphism. Equivalently, it is a bijective module homomorphism from a module to itself; its inverse is automatically a module homomorphism. They form a group under composition.

A module retraction onto a submodule $U\stackrel{i}{\to }X$ is a module homomorphism $r:X\to U$ with $ri=I_U$. It gives $X=i(U)\oplus \ker r$: write $x=i(r(x))+(x-i(r(x)))$. Thus a nonzero proper submodule admitting a retraction contradicts indecomposability. Extending a projection from a larger submodule can produce such a retraction through the long exact sequence of Ext groups.

The endomorphism ring ${\mathrm{End}}_R(M)$ consists of the $R$-module homomorphisms $M\to M$, with pointwise addition and composition as multiplication.

An endomorphism ring is local when its nonunits form a proper two-sided ideal; commutativity is not required. The Fitting lemma gives this property for an indecomposable finite-length module. If two nonunits had an invertible sum, multiplication by its inverse would give $I=f+g$ with noninvertible $f,g$. Fitting makes $f$ nilpotent, so $I-f=g$ would be invertible, a contradiction. Products with a nonunit are nonunits by finite-length injectivity/surjectivity. The quotient by the nonunit ideal is a division ring.

For a finite-dimensional module over an algebraically closed field, write a Krull-Schmidt decomposition $X={⨁}_a{M}_a^{\oplus m_a}$ with distinct indecomposable types. Each endomorphism ring of $M_a$ is a local endomorphism ring with residue division algebra $k$, by the finite-dimensional division algebra over an algebraically closed field result. Modulo the Jacobson radical of $\mathrm{End}(X)$, the blocks of one type become $M_{m_a}(k)$ and all maps between different types vanish. A composite through a different indecomposable type cannot be invertible, since that would make one type a direct summand of the other. This yields the displayed product and the Levi decomposition of a quiver automorphism group.

A nonzero module is a brick when every endomorphism is either zero or an isomorphism. Equivalently, its endomorphism ring is a division ring.

An indecomposable finite-dimensional quiver representation that is not a brick module contains a brick with nonzero self-extensions. The Ringel form then gives $q_Q(\dim B)=1-\dim {\mathrm{Ext}}_Q^1(B,B)\le 0$, impossible for a positive definite Tits form of a quiver.

The Fitting lemma supplies a nonzero nilpotent endomorphism $f$ of a non-brick indecomposable $X$. Minimize its nonzero rank; nilpotence and minimality give $f^2=0$. Set $I=\mathrm{im}f\subset K=\ker f={⨁}_j{K}_j$. Choose a nonzero component $u:I\to K_j$. The square-zero endomorphism $X\stackrel{f}{\to }I\stackrel{u}{\to }{K}_j\hookrightarrow X$ has rank at least $\dim I$, so $u$ is injective.

If ${\mathrm{Ext}}^1(I,K_j)=0$, the projection $K\to K_j$ extends to $X\to K_j$ by the long exact sequence of Ext groups, giving a module retraction and contradicting indecomposability. Thus this extension group is nonzero. Since path algebras are hereditary, $u$ induces a surjection ${\mathrm{Ext}}^1(K_j,K_j)\twoheadrightarrow {\mathrm{Ext}}^1(I,K_j)$. Hence $K_j$ is a proper indecomposable submodule with self-extensions. Iterate until a brick module is reached; dimensions strictly decrease.

This minimal-rank argument is given in section 2 of William Crawley-Boevey's quiver lectures.

A submodule $N$ of a module $M$ is a subset closed under addition, additive inverses, and scalar multiplication by every element of $R$.

A submodule $N\subseteq M$ is an essential submodule if it intersects every nonzero submodule of $M$ nontrivially. A nonzero module is a uniform module precisely when every nonzero submodule is an essential submodule.

An essential right ideal is an essential submodule of the right regular module of a ring. Every right ideal containing an essential right ideal is essential.

If $c$ is a regular element of a ring in a right Noetherian ring and a nonzero right ideal $B$ had $B\cap cR=0$, cancellation would make $B+cB+c^{2}B+\cdots$ an infinite direct sum of nonzero right ideals. Its finite partial sums contradict the ascending chain condition. Thus $cR$ is an essential right ideal.

An extension $N\subseteq M$ is essential when $N$ is an essential submodule of $M$. Essentiality is transitive. It is an intersection condition, not a statement that the quotient is small in dimension or finitely generated.

A primary decomposition expresses an ideal or submodule as an intersection of primary components.

Every proper ideal in a Noetherian ring is a finite intersection of primary ideals. This existence theorem does not give uniqueness of the individual components: embedded primary components may vary.

A proper submodule $N⊊M$ is primary when $rm\in N$ and $m\notin N$ imply $r^{k}M\subseteq N$ for some positive integer $k$.