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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.

For a finite-dimensional algebra, the Auslander–Reiten quiver has one vertex for each isomorphism class of finite-dimensional indecomposable modules, and its arrows record irreducible module homomorphisms. A homomorphism is irreducible when it is neither a split injection nor a split surjection and every factorization forces the first map to split injectively or the second to split surjectively.

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.

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.

Evaluation at $e_i$ identifies a module homomorphism $h:kQ{e}_i\to X$ with $h(e_i)\in e_{i}X$. Given $x\in e_{i}X$, the inverse sends $p$ to $px$. The isomorphism is natural and shows that Hom functor evaluation on a vertex projective is exact.

For a nonzero associative algebra $A$ of finite dimension that is a simple ring, choose a minimal left ideal $I$. Right multiplication $I\to Ia$ is a module homomorphism, so $Ia$ is zero or a simple module. The sum ${\sum }_{a\in A}Ia$ is a nonzero two-sided ideal, hence $A$. Finite dimension reduces this to a finite sum. The finite sum of simple submodules lemma shows that the left regular module is a semisimple module.

For a module homomorphism $f:M\to N$, the map $m+\ker f\mapsto f(m)$ is a module isomorphism $M/\ker f\to \mathrm{Im}f$. The formula is well-defined because two representatives differ by an element of the kernel of a module homomorphism exactly when their images agree; the same observation proves injectivity, and surjectivity onto the image follows by its definition.

For a left module $M$ over a unital associative algebra $A$, right multiplication in $A$ gives a right module structure on ${\mathrm{Hom}}_A(M,A)$ by $(f\cdot a)(m)=f(m)a$. This remains a module homomorphism because $f(bm)a=bf(m)a$; associativity and the unit verify the action axioms. When $A=M_d(\mathbb{C})$ and $M={\mathbb{C}}^d$ is its simple module of columns, every such map is $m\mapsto mr$ for a unique row $r$, making the Hom space the simple row right module. This illustrates how left and right simple modules correspond for a semisimple algebra.

A homomorphism is a map between algebraic structures preserving their specified operations. For a single binary operation, this means $f(a\ast b)=f(a)\star f(b)$, where the source and target may use different operation symbols. A group homomorphism preserves the group operation; a ring homomorphism preserves addition and multiplication, with the identity convention specified; a module homomorphism preserves addition and scalar multiplication.

Reduction modulo $5$ is a ring homomorphism $\mathbb{Z}\to \mathbb{Z}/5\mathbb{Z}$: adding or multiplying first and then reducing gives the same residue as reducing first and then operating. For example, $12\cdot 14=168\equiv 3(\mathrm{mod}5)$, while the reduced operands give $2\cdot 4=8\equiv 3(\mathrm{mod}5)$. This lets a complicated integer calculation be projected into simpler finite arithmetic. It also explains why a polynomial equation over the integers imposes necessary modular arithmetic conditions on any integer solution.

A homomorphism can lose information: $0$ and $5$ have the same image under that reduction. An isomorphism loses none of the structure because it has an operation-preserving inverse. Preservation of the operation, rather than preservation of arbitrary labels or sizes, is the essential condition.

For $P$ cyclic of order $q=p^n$ in characteristic $p$, its group algebra is $k[t]/(t^q)$. The nilpotent Jordan normal form classifies its indecomposable finite-dimensional modules as $V_i$, $1\le i\le q$. A module homomorphism $V_i\to V_j$ is determined by an element annihilated by $t^i$, so its dimension is $\min (i,j)$.

For a module homomorphism $f:M\to N$ over a commutative ring $A$, $f$ is injective if and only if every $f_{\mathfrak{m}}$ is injective, with $\mathfrak{m}$ ranging over maximal ideals. In a localization of a module, $x/s$ maps to zero exactly when some denominator $t$ satisfies $tf(x)=0$. If $f$ is injective this gives $tx=0$, so the original fraction is zero. In general it identifies $\ker (f_{\mathfrak{m}})$ with $(\ker f{)}_{\mathfrak{m}}$. A nonzero element of $\ker f$ has a proper annihilator, contained in some maximal ideal; it remains nonzero in that localization. Thus vanishing of all localized kernels forces $\ker f=0$. No finiteness or Noetherian assumption is required.

For a free module $F$, a surjective module homomorphism $\varphi :P\to M$, and a module homomorphism $\psi :F\to M$, there is a module homomorphism $q:F\to P$ satisfying $\varphi q=\psi$. Choose a preimage under $\varphi$ of the image of each element of a basis of a module of $F$, and extend by linearity. Thus every free module is a projective module.

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 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 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.

A left module over the unital $K$-algebra $A$ is a $K$-vector space $V$ with a $K$-bilinear action $A\times V\to V$ satisfying $(ab)v=a(bv)$ and $1v=v$. For a vector $v$, the map $A\to V$, $a\mapsto av$, is an $A$-module homomorphism. Its kernel is $\mathrm{Ann}(v)$; explicitly this is a linear subspace and $b(av)=(ba)v=0$ for $a\in \mathrm{Ann}(v)$, so it is a left ideal.

If $V=Av$, this homomorphism is surjective, and the module isomorphism theorem gives $V\cong A/\mathrm{Ann}(v)$. Conversely the quotient by a left ideal is generated by $1+I$. Thus cyclic modules are exactly the quotients $A/I$ by left ideals.

For the first example set $e=\mathrm{diag}(1,1,0)$. Right multiplication by $e$ keeps the first two columns and zeroes the third. Every matrix with third column zero equals its product with $e$. Hence $V=Ae$ is cyclic, generated by $e$.

A composition series is a chain $0=V_0⊊V_1⊊\cdots ⊊V_r=V$ of submodules with every factor $V_i/V_{i-1}$ simple; its length is $r$. Suppose $0\subset S\subset V$ is one of length two, and put $T=V/S$.

For any nonzero proper submodule $U$, either $U\cap S=S$ or $U\cap S=0$, by simplicity of $S$. In the first case, simplicity of $V/S$ forces $U=S$. In the second, the quotient map embeds $U$ nontrivially into $T$, so $U\cong T$ and its image is all of $T$. Hence $V=S\oplus U$, and $V/U\cong S$. Thus every nonzero proper submodule is simple and has simple quotient; every composition series has length two.

The Jordan–Hölder theorem states that any two composition series of a finite-length module have equal length and the same multiset of simple factors, up to isomorphism and order. The preceding argument proves this for length two: either $U=S$, giving factors $S,T$, or $U\ne S$, giving factors $T,S$.

The map $A\to S$, $a\mapsto ax$, is a surjective module homomorphism with kernel $I={\mathrm{Ann}}_A(x)$. The first isomorphism theorem for modules gives $A/I\cong S$ as left modules; $I$ need only be a left ideal. Every simple module is cyclic, since any nonzero $x$ generates a nonzero submodule. As a quotient of the finite-dimensional regular module $A$, it is finite-dimensional. Extend $I$ to a composition series of $A$; its simple quotient $A/I$ is one of the factors. Jordan-Hölder shows that its isomorphism type belongs to the fixed finite list of factors of any one composition series of $A$. Hence there are finitely many simple-module isomorphism types.

For the final assertion let $W$ be the image of $\psi$. Its kernel is $\bigcap I_j$. We prove $W=\prod S_j$ by induction on $k$. It is true for $k=1$. Projection onto the first $k-1$ factors is onto by induction. Its kernel on $W$ is a submodule of $S_k$, so is zero or $S_k$. If zero, $W\cong {\prod }_{j<k}{S}_j$. But its projection onto $S_k$ is onto, making $S_k$ a composition factor of that product, contrary to Jordan-Hölder and the pairwise distinct types. Thus the kernel is all of $S_k$. Surjectivity on the first factors together with this kernel proves $W$ is the entire product. The first isomorphism theorem now givesas left $A$-modules, without asserting that the left ideals are two-sided.

For $a\in A$, $m\in M$,Additivity and complex linearity follow from those of $f$ and multiplication in $A$, so $f\cdot x$ is an module homomorphism. In particular the action stays inside $V$. Associativity givesBoth distributive laws hold pointwise. These are exactly the axioms for a right module, proving ${\mathrm{Hom}}_A(M,A)$ is a right $A$-module.

Since $A$ is a nonzero associative algebra of finite dimension, it has a nonzero left ideal $I$ of least positive dimension. Such a minimal left ideal is a simple module under left multiplication.

For $a\in A$, right multiplication defines the module homomorphismIts kernel is a submodule of $I$, so its image $Ia$ is either zero or a simple module isomorphic to $I$. Each $Ia$ is a left ideal, since $c(xa)=(cx)a$.

ConsiderIt is a left ideal. It is also stable under right multiplication, because $(Ia)b=I(ab)$, so it is a two-sided ideal. Moreover $I=I1\subseteq J$, giving $J\ne 0$. The assumption that $A$ is a simple ring therefore forces $J=A$.

Only finitely many summands are needed: express each member of a finite basis of $A$ as a finite sum of elements of the $Ia$, and collect the finitely many indices used. After discarding zero summands, $A$ is a finite sum of simple modules. Part (a)(ii) makes its left regular module a semisimple module. This is exactly the definition of a semisimple algebra, proving that a finite-dimensional simple algebra is semisimple. The finite dimension assumption is essential to this assertion about a simple ring.

A permutation sends $v_a$ to $v_{ga}$, $w_{ab}$ to $w_{ga,gb}$ and $x_{abc}$ to $x_{ga,gb,gc}$. Relabeling the summation indices givesLinearity therefore proves both maps are module homomorphisms. The map denoted $\varphi$ in the second displayed source formula is ${\varphi }_2$.