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Partition a basis of a module $E$ into disjoint subsets $E^{\prime },E^{\prime \prime }$. Splitting the finite expression of each element gives $M=⟨E^{\prime }⟩+⟨E^{\prime \prime }⟩$. An element in both generated submodules would give, by subtraction, a relation among the disjoint parts of $E$; linear independence in a module forces every coefficient to vanish. The intersection is zero, proving the direct sum of modules decomposition. Each part is itself a basis of a module for the submodule it generates.

For a Noetherian ring $A$, a finitely generated module $M$ has a finite presentation of a module, and kernels of maps from finite free modules to $M$ are finitely generated. Its affine module sheaf is therefore a coherent sheaf. Conversely, if $\tilde{M}$ is coherent, choose a finite principal open subset cover on which its section modules are finitely generated. Lift their finitely many generators to numerators in $M$. The submodule generated by these numerators becomes all of $M$ on every chart, and hence is all of $M$: its quotient has zero localization at every point. This proves finite generation without assuming that global sections preserve arbitrary sheaf surjections.

A composition series of a module is a finite chain of submodules whose nonzero successive quotient modules are simple modules. Every finite-dimensional Lie algebra representation has such a chain, viewed as a module over the universal enveloping algebra.

A submodule $N$ of a free module with basis of a module $E$, generated by a countable set, is contained in the submodule generated by some countable subset $E^{\prime }\subseteq E$. Take $E^{\prime }$ to be the union of the finite coordinate supports in a free module of a countable generating family for $N$. This union is countable, and every generator belongs to $⟨E^{\prime }⟩$, which proves the inclusion. Combining this with basis partition gives a module direct sum places $N$ inside a countably generated direct summand.

At a primitive cube root of unity, the cubic homogeneous module has basis $x^3,x^{2}y,x{y}^2,y^3$. Its only simple modules occurring as submodules are the one-dimensional subspaces of $⟨x^3,y^3⟩$, on which $K=1$ and $E=F=0$. Its quotient by this two-dimensional submodule is simple, but it has no simple complement: any lift of the highest $K$ weight space contains $x^{2}y$, whose $E$ image is $x^3\ne 0$. Thus the cubic module is not a semisimple module.

A module is cyclic if one element $m$ generates it: every element is $rm$ for some scalar $r\in R$. The generating element need not give unique coefficients. The module homomorphism $R\to M$, $r\mapsto rm$, is surjective. Over a commutative ring, its kernel is the annihilator ${\mathrm{Ann}}_R(m)$, givingThus cyclic modules are exactly the quotient modules $R/I$ for ideals $I$ in $R$. For a left module over a noncommutative ring, use a left ideal and interpret the quotient as a module, with no claim that it is a quotient ring. See John A. Beachy's Abstract Algebra Online, section 10.1.

The module $\mathbb{Z}/6\mathbb{Z}$ is generated by $\overline{1}$, and also by $\overline{5}$, whose successive multiples are $\overline{0},\overline{5},\overline{4},\overline{3},\overline{2},\overline{1}$. The element $\overline{2}$ generates only $\{\overline{0},\overline{2},\overline{4}\}$, a submodule with three elements. A cyclic module can be infinite and free, as $\mathbb{Z}=\mathbb{Z}\cdot 1$ shows, or finite and torsion, as $\mathbb{Z}/6\mathbb{Z}$ shows. One generator counts directions of generation, not the number of elements and not freedom from relations.

A submodule $N$ of a module $M$ is a direct summand if some submodule $L$ gives $M=N\oplus L$. Equivalently, the inclusion has a left inverse R-module homomorphism, or $N$ is the image of an idempotent endomorphism of $M$.

The direct sum of a family of modules over the same ring consists of all tuples $(m_j)$ with $m_j\in M_j$ and only finitely many nonzero coordinates, with coordinatewise addition and scalar multiplication. For two submodules $N,P\subseteq M$, an internal direct sum $M=N\oplus P$ means $M=N+P$ and $N\cap P=\{0\}$. Equivalently, every element of $M$ is uniquely $n+p$: equality of two such expressions puts their difference in $N\cap P$. The projection $n+p\mapsto p$ is a module homomorphism with kernel $N$, image $P$, and square equal to itself.

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.

Partition the isomorphism classes of simple modules into two sets ${\mathcal{C}}_1,{\mathcal{C}}_2$, with ${\mathrm{Ext}}^1(S,T)={\mathrm{Ext}}^1(T,S)=0$ whenever $S,T$ lie in different sets. Every module $M$ of finite composition length then has a unique decomposition $M=U_1\oplus U_2$ into its largest submodules whose Jordan–Hölder factors belong to the respective sets.

For a simple $S\in {\mathcal{C}}_1$ and a finite-length $N$ with factors in ${\mathcal{C}}_2$, induction through the long exact sequence of the Ext functor gives ${\mathrm{Ext}}^1(S,N)=0$. Induct on the length of $M$: take $0\to N\to M\to S\to 0$, split $N=N_1\oplus N_2$, and suppose $S\in {\mathcal{C}}_1$. The quotient extension $0\to N_2\to M/N_1\to S\to 0$ splits. The inverse image of its $S$-summand complements $N_2$ in $M$ and has only factors in ${\mathcal{C}}_1$. Finally, Hom between modules with factors in different sets is zero, since any nonzero image would have a factor in both sets. Projecting any proposed submodule to the opposite summand therefore proves maximality and uniqueness.

If a submodule $N\subseteq M$ and the quotient module $M/N$ are finitely generated modules, then $M$ is a finitely generated module. Choose finitely many generators of $N$ and lift finitely many generators of $M/N$. For any $m\in M$, subtract an $R$-linear combination of the lifts to obtain an element of $N$, then express this using the generators of $N$. The union of the two finite lists generates $M$.

Suppose $A\subseteq B$ are unital rings and $B$ is a finitely generated module over $A$ on the right. If $A$ is a right Noetherian ring, then $B_A$ is a Noetherian module. Every right ideal of $B$ is an $A$-submodule, hence has finitely many generators over $A$; these also generate it over $B$. Thus $B$ is a right Noetherian ring. The same argument works on the left under the corresponding hypotheses. Finite-index group ring extensions give an important example, using coset representatives as a finite module basis.

A finite sum of simple modules inside one module is a semisimple module. Inductively, if $M$ is the sum already retained, then $M\cap S_j$ is a submodule of the simple module $S_j$, hence zero or $S_j$. In the first case append $S_j$ as a direct summand; in the second it adds nothing. This gives a direct sum of modules selected from the original family.

A free module is a module with a basis of a module $(e_i{)}_{i\in I}$: every element has exactly one expression $m={\sum }_{i\in I}{r}_i{e}_i$ with $r_i\in R$ and only finitely many $r_i$ nonzero. Equivalently, the coefficient map ${⨁}_{i\in I}R\to M$ is a module isomorphism. Existence means generation; uniqueness means linear independence in a module. For finite $I$ of size $n$, this says $M\cong R^n$, a finite free module. See Neil Strickland's Rings, Modules and Linear Algebra, section 11.

The integer lattice ${\mathbb{Z}}^2$ in the opening diagram is a free module over $\mathbb{Z}$ with basis of a module $e_1=(1,0)$, $e_2=(0,1)$. For example, $(3,2)=3e_1+2e_2$, and comparing the two coordinates makes the coefficients unique. A different basis of a module is $u=(1,0)$, $v=(1,1)$, because $(a,b)=(a-b)u+bv$; in particular $(3,2)=u+2v$.

By contrast, $w_1=(2,0)$ and $w_2=(0,1)$ are linearly independent in a module, but generate only the submodule $2\mathbb{Z}\times \mathbb{Z}$. They miss $(1,0)$ because $2a=1$ has no integer solution. They would form a basis over the rational numbers. Even generation need not let one extract a basis of a module: the elements $2,3$ generate $\mathbb{Z}$, since $1=3-2$, but neither singleton generates it, and $3\cdot 2-2\cdot 3=0$ is a nontrivial linear relation between them. These are concrete failures of the usual vector space argument that divides by any nonzero coefficient.

The cyclic module $\mathbb{Z}/6\mathbb{Z}$ has no basis of a module over $\mathbb{Z}$: every element has the nontrivial relation $6m=0$, so no nonempty subset is linearly independent in a module, while the empty subset cannot generate this nonzero module. Specifying the scalar ring is essential: over $R=\mathbb{Z}/6\mathbb{Z}$, the same additive object is a free module with the one-element basis of a module $\{\overline{1}\}$, since each coefficient is itself a unique class in $R$.

A nonzero module is indecomposable if it cannot be expressed as a direct sum of two nonzero submodules. The zero module is excluded. A module of finite composition length is indecomposable exactly when its endomorphism ring has no idempotents other than zero and one: an idempotent splits the module into its image and kernel.

The kernel of a module homomorphism $f:M\to N$ is the submodule of elements mapped to zero. Closure follows from $f(m+m^{\prime })=f(m)+f(m^{\prime })$ and $f(rm)=rf(m)$.

The left regular module of a unital associative algebra $A$ is its underlying additive group with $A$ acting by left multiplication. Its submodules are exactly the left ideals of $A$. The associative algebra is a semisimple algebra precisely when this module is a semisimple module.

For a finite-dimensional module, it suffices that every maximal proper submodule have a submodule complement in order that every submodule have one. Let $W$ be the sum of all simple modules contained in the given module. If $W$ is proper, choose a maximal submodule $M$ containing it. A complement $C$ is isomorphic to the simple quotient $V/M$ and hence is itself simple. This forces $C\subseteq W\subseteq M$, contradicting the complement. Thus $W=V$. Greedily adding simple summands not already in the sum gives a direct sum in finitely many steps. Finally, given any $U$, choose a maximal sum of these simple summands disjoint from $U$. A missing simple summand has zero intersection with their sum plus $U$, contradicting maximality unless their sum complements $U$. This proves the semisimple module criterion, including the zero module.

For a module $M$, ${\mu }_A(M)$ is the least cardinality of a generating set, equivalently the least size of a basis of a free module surjecting onto $M$. If $(A,\mathfrak{m})$ is a local ring and $M$ is finitely generated, then ${\mu }_A(M)={\dim }_{A/\mathfrak{m}}(M/\mathfrak{m}M)$. Any generating set spans this quotient, proving the lower bound. Lift a basis of the quotient and let $J$ be the submodule generated by those lifts. Then $N=M/J$ is finitely generated and satisfies $N=\mathfrak{m}N$. For generators $n_i$ write $n_i={\sum }_j{c}_{ij}{n}_j$ with $c_{ij}\in \mathfrak{m}$. The adjugate of $I-C$ shows that $\det (I-C)$ annihilates $N$; this determinant is one modulo $\mathfrak{m}$ and hence a unit. Thus $N=0$, giving the upper bound.

The module of integer sequences consists of all functions ${\mathbb{N}}_0\to \mathbb{Z}$, with pointwise addition and integer scalar multiplication. Its submodule of finitely supported sequences is the direct sum of modules ${⨁}_{j\ge 0}\mathbb{Z}$, a free module with coordinate basis of a module $(e_j)$. The full product is not a free module.

To prove this, suppose it had a basis of a module $E$. The countable basis support lemma places every finitely supported sequence in a submodule $C$ generated by a countable subset of $E$. The basis partition gives a module direct sum lemma supplies a projection $\pi$ onto its complementary direct summand, with kernel $C$. By countability of countably generated abelian groups, $C$ is countable. The Cantor diagonal argument makes the family of all sequences $w(j)=\pm 2^j$ uncountable, so choose one with $\pi (w)\ne 0$. Truncating $w$ before coordinate $i$ writes $w=f_i+2^i{g}_i$, where $f_i$ is finitely supported and $g_i$ is an integer sequence. Thus $\pi (w)=2^i\pi (g_i)$ for every $i$. The vanishing of infinitely divisible integer sequences forces $\pi (w)=0$, a contradiction.