A module has two kinds of structure at once: its elements add like an abelian group, and a ring acts on those elements like scalars. A map between modules should preserve both. If it preserves only addition, it may ignore the part of the module that remembers the ring; if it preserves only scalar multiplication, it may fail to be compatible with the underlying additive group. Module homomorphisms are the maps that preserve the full structure. This distinction appears even in familiar examples. The complex numbers can be regarded as a [vector space](/page/Vector%20Space) over $\mathbb{R}$ or as a module over $\mathbb{C}$. The same function may be compatible with one scalar system and incompatible with another. [example: Complex Conjugation as a Warning] Let $\mathbb{C}$ first be regarded as an $\mathbb{R}$-module, and define $f:\mathbb{C}\to\mathbb{C}$ by $f(z)=\overline{z}$. Write $z=x+iy$ and $w=u+iv$ with $x,y,u,v\in\mathbb{R}$. Then \begin{align*} f(z+w)=f((x+u)+i(y+v))=(x+u)-i(y+v) \end{align*} and \begin{align*} f(z)+f(w)=(x-iy)+(u-iv)=(x+u)-i(y+v). \end{align*} Therefore $f(z+w)=f(z)+f(w)$. For scalar multiplication by a real number $a\in\mathbb{R}$, \begin{align*} f(az)=f(ax+iay)=ax-iay. \end{align*} Also, \begin{align*} af(z)=a(x-iy)=ax-iay. \end{align*} Thus $f(az)=af(z)$ for every $a\in\mathbb{R}$ and $z\in\mathbb{C}$, so $f$ is an $\mathbb{R}$-module homomorphism. Now regard $\mathbb{C}$ as a $\mathbb{C}$-module over itself. Taking the complex scalar $i$ and the element $1\in\mathbb{C}$, we get \begin{align*} f(i\cdot 1)=f(i)=\overline{i}=-i. \end{align*} On the other hand, \begin{align*} i f(1)=i\overline{1}=i\cdot 1=i. \end{align*} Since $-i\neq i$, the scalar-compatibility identity $f(\lambda z)=\lambda f(z)$ fails for $\lambda=i$ and $z=1$. Thus the same additive map is compatible with the real scalar action but not with the complex scalar action, so the ring is part of the data in the phrase “module homomorphism.” [/example] The point is not merely terminological. Kernels, images, quotient modules, exact sequences, and presentations are all built from module homomorphisms. Once the right maps are chosen, many constructions from linear algebra and abelian group theory become instances of the same language. ## Definition Before defining the map, we recall the ambient objects. The parent concept is a [module](/page/Module): an abelian group equipped with a compatible action of a ring. The next definition is needed because a map between modules should preserve both the addition and the scalar action, not merely the underlying set. [definition: Module Homomorphism] Let $R$ be a ring, and let $M$ and $N$ be left $R$-modules. A function $f:M\to N$ is an $R$-module homomorphism if, for all $m_1,m_2\in M$ and all $r\in R$, \begin{align*} f(m_1+m_2)=f(m_1)+f(m_2) \end{align*} and \begin{align*} f(rm_1)=r f(m_1). \end{align*} [/definition] The first condition says that $f$ is a homomorphism of the underlying abelian groups. The second says that $f$ commutes with the action of $R$. In the special case where $R$ is a field, this is the usual notion of a [linear map](/page/Linear%20Map) between vector spaces. Many arguments use the phrase "$R$-linear map" instead of "$R$-module homomorphism." The next definition is useful because it records this standard synonym while keeping the ring in view. It is also needed so later references to linear maps over a ring point back to the same structure-preserving condition. [definition: $R$-Linear Map] Let $R$ be a ring, and let $M$ and $N$ be left $R$-modules. An $R$-linear map from $M$ to $N$ is an $R$-module homomorphism $f:M\to N$. [/definition] The zero map and identity map are the first tests of any proposed definition of structure-preserving map. They must be included, otherwise the class of modules would not have a usable algebra of maps. [example: Zero and Identity Homomorphisms] Let $M$ and $N$ be left $R$-modules. Define the zero map $0:M\to N$ by $0(m)=0_N$ for every $m\in M$. For $m_1,m_2\in M$, the value of the zero map is always $0_N$, so \begin{align*} 0(m_1+m_2)=0_N. \end{align*} Also, since $N$ is an abelian group under addition, $0_N$ is its additive identity, and hence \begin{align*} 0(m_1)+0(m_2)=0_N+0_N=0_N. \end{align*} Therefore $0(m_1+m_2)=0(m_1)+0(m_2)$. For $r\in R$ and $m\in M$, \begin{align*} 0(rm)=0_N. \end{align*} The scalar action on a module satisfies $r0_N=0_N$, because \begin{align*} r0_N=r(0_N+0_N)=r0_N+r0_N, \end{align*} and adding the additive inverse of $r0_N$ to both sides gives $0_N=r0_N$. Thus \begin{align*} 0(rm)=0_N=r0_N=r0(m). \end{align*} So the zero map preserves both addition and scalar multiplication, hence is an $R$-module homomorphism. The identity map $\operatorname{id}_M:M\to M$ is defined by $\operatorname{id}_M(m)=m$. For $m_1,m_2\in M$, \begin{align*} \operatorname{id}_M(m_1+m_2)=m_1+m_2. \end{align*} On the other hand, \begin{align*} \operatorname{id}_M(m_1)+\operatorname{id}_M(m_2)=m_1+m_2. \end{align*} Therefore $\operatorname{id}_M(m_1+m_2)=\operatorname{id}_M(m_1)+\operatorname{id}_M(m_2)$. For $r\in R$ and $m\in M$, \begin{align*} \operatorname{id}_M(rm)=rm=r\operatorname{id}_M(m). \end{align*} Thus $\operatorname{id}_M$ is an $R$-module homomorphism. The zero map gives the additive zero among homomorphisms $M\to N$, while the identity map is the neutral map for composition on $M$. [/example] Checking additivity and scalar compatibility separately can become repetitive. The practical question is whether the two defining identities can be replaced by one linear-combination test. Such a test must still recover ordinary addition and the action of each scalar, so it has to encode both requirements at once. [quotetheorem:8331] The criterion uses the standing convention from the parent [module](/page/Module) page that rings and modules are unital. With that convention, additivity is recovered from the displayed condition by taking coefficients $1_R$ and $1_R$. The criterion explains why module homomorphisms behave like linear maps. It also highlights a difference from arbitrary group homomorphisms: the coefficients $r_1$ and $r_2$ are visible in the test. For classification, the next problem is deciding when a homomorphism is a reversible change of coordinates. A bijection of underlying sets is not enough, because its inverse might fail to respect the scalar action. The definition therefore requires a bijective map that is already compatible with the full $R$-module structure. [definition: Module Isomorphism] Let $R$ be a ring, and let $M$ and $N$ be left $R$-modules. An $R$-module isomorphism from $M$ to $N$ is a bijective $R$-module homomorphism $f:M\to N$. [/definition] After this definition, we say that $M$ and $N$ are isomorphic as $R$-modules if there exists an $R$-module isomorphism $M\to N$. The phrase "as $R$-modules" matters: the same underlying abelian group may carry different scalar actions. The next theorem is needed to confirm that a bijective homomorphism really has a structure-preserving inverse. [quotetheorem:8332] This theorem justifies using bijective homomorphisms as isomorphisms. It says that the inverse map recovers module structure rather than merely recovering elements. ## Kernels, Images, and Quotients A homomorphism is useful because its failure to be injective and its failure to be surjective are themselves module-theoretic objects. For arbitrary functions, fibers and images are just subsets. For module homomorphisms, the kernel and image inherit module structure. The kernel records the elements that the map collapses to zero. The next definition is needed because this collapsed part controls when a homomorphism can be regarded as injective after passing to a quotient. [definition: Kernel of a Module Homomorphism] Let $R$ be a ring, let $M$ and $N$ be left $R$-modules, and let $f:M\to N$ be an $R$-module homomorphism. The kernel of $f$ is \begin{align*} \ker f=\{m\in M: f(m)=0_N\}. \end{align*} [/definition] The image records the part of the codomain that is actually reached. Surjectivity and factorisation depend on this reached submodule rather than on the ambient codomain alone, so the reachable elements need their own notation. [definition: Image of a Module Homomorphism] Let $R$ be a ring, let $M$ and $N$ be left $R$-modules, and let $f:M\to N$ be an $R$-module homomorphism. The image of $f$ is \begin{align*} \operatorname{im} f=\{n\in N: \text{there exists }m\in M\text{ such that }f(m)=n\}. \end{align*} [/definition] These two constructions are the first reason module homomorphisms are the right maps. Kernel and image would be far less useful if they were merely subsets; quotient and exactness arguments require them to remain inside the category of modules. [quotetheorem:8333] The theorem explains why familiar rank-nullity ideas have module-theoretic analogues even when dimension is unavailable. The kernel and image remain meaningful, though they may not be measured by a single integer. [example: Multiplication by an Integer] View $\mathbb{Z}$ as a module over itself, and fix $a\in\mathbb{Z}$. Define $\mu_a:\mathbb{Z}\to\mathbb{Z}$ by $\mu_a(n)=an$. For $m,n\in\mathbb{Z}$, \begin{align*} \mu_a(m+n)=a(m+n)=am+an=\mu_a(m)+\mu_a(n). \end{align*} For a scalar $r\in\mathbb{Z}$ and an element $n\in\mathbb{Z}$, \begin{align*} \mu_a(rn)=a(rn)=(ar)n=(ra)n=r(an)=r\mu_a(n), \end{align*} where $ar=ra$ because multiplication in $\mathbb{Z}$ is commutative. Thus $\mu_a$ preserves addition and the $\mathbb{Z}$-scalar action, so it is a $\mathbb{Z}$-module homomorphism. Its kernel is \begin{align*} \ker\mu_a=\{n\in\mathbb{Z}:an=0\}. \end{align*} If $a\neq 0$, then $an=0$ in the [integral domain](/page/Integral%20Domain) $\mathbb{Z}$ implies $n=0$, so \begin{align*} \ker\mu_a=\{0\}. \end{align*} If $a=0$, then for every $n\in\mathbb{Z}$, \begin{align*} \mu_0(n)=0n=0, \end{align*} so every integer lies in the kernel and \begin{align*} \ker\mu_0=\mathbb{Z}. \end{align*} The image is \begin{align*} \operatorname{im}\mu_a=\{\mu_a(n):n\in\mathbb{Z}\}=\{an:n\in\mathbb{Z}\}=a\mathbb{Z}. \end{align*} When $a=0$, this specializes to \begin{align*} \operatorname{im}\mu_0=0\mathbb{Z}=\{0\}. \end{align*} Thus multiplication by $a$ has image exactly the subgroup of multiples of $a$, while its kernel records whether multiplication by $a$ can kill a nonzero integer. [/example] Surjectivity has a different obstruction from injectivity. For a homomorphism $f:M\to N$, the elements reached by $f$ form $\operatorname{im} f$, but the unreached part of $N$ is not usually a well-defined complementary submodule. The useful construction is instead to collapse the reached submodule inside $N$ and record the quotient that remains. [definition: Cokernel of a Module Homomorphism] Let $R$ be a ring, let $M$ and $N$ be left $R$-modules, and let $f:M\to N$ be an $R$-module homomorphism. The cokernel of $f$ is the quotient module \begin{align*} \operatorname{coker} f=N/\operatorname{im} f. \end{align*} [/definition] The cokernel measures the part of $N$ not accounted for by $f$. It is the quotient-side companion to the kernel: kernels detect non-injectivity, while cokernels detect non-surjectivity. Together, the kernel and image answer the basic factorization question for an ordinary $R$-module homomorphism: what must be collapsed in the domain, and what submodule of the codomain is actually reached? [quotetheorem:862] This result is being used here only at the level of ordinary modules. It says that once the kernel has been collapsed, no further identifications remain: the quotient $M/\ker f$ is naturally the same module as the image of $f$ inside $N$. [example: Reduction Modulo $n$] Let $n\in\mathbb{N}$, and define $\pi_n:\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ by $\pi_n(a)=a+n\mathbb{Z}$. We first check that this is a $\mathbb{Z}$-module homomorphism. For $a,b\in\mathbb{Z}$, \begin{align*} \pi_n(a+b)=(a+b)+n\mathbb{Z}. \end{align*} Addition in the [quotient group](/theorems/790) is defined by adding representatives, so \begin{align*} (a+n\mathbb{Z})+(b+n\mathbb{Z})=(a+b)+n\mathbb{Z}. \end{align*} Therefore \begin{align*} \pi_n(a+b)=\pi_n(a)+\pi_n(b). \end{align*} For $r,a\in\mathbb{Z}$, scalar multiplication in the quotient module is given by $r(a+n\mathbb{Z})=ra+n\mathbb{Z}$, hence \begin{align*} \pi_n(ra)=ra+n\mathbb{Z}=r(a+n\mathbb{Z})=r\pi_n(a). \end{align*} Thus $\pi_n$ preserves addition and the $\mathbb{Z}$-scalar action. The map is surjective because every element of $\mathbb{Z}/n\mathbb{Z}$ has the form $a+n\mathbb{Z}$ for some $a\in\mathbb{Z}$, and \begin{align*} \pi_n(a)=a+n\mathbb{Z}. \end{align*} Its kernel is \begin{align*} \ker\pi_n=\{a\in\mathbb{Z}:a+n\mathbb{Z}=0+n\mathbb{Z}\}. \end{align*} The equality $a+n\mathbb{Z}=0+n\mathbb{Z}$ means exactly that $a-0\in n\mathbb{Z}$, so \begin{align*} \ker\pi_n=\{a\in\mathbb{Z}:a\in n\mathbb{Z}\}=n\mathbb{Z}. \end{align*} Also, \begin{align*} \operatorname{im}\pi_n=\mathbb{Z}/n\mathbb{Z} \end{align*} by surjectivity. By the *[First Isomorphism Theorem for Modules](/theorems/862)*, there is a $\mathbb{Z}$-module isomorphism \begin{align*} \overline{\pi}_n:\mathbb{Z}/\ker\pi_n\to\operatorname{im}\pi_n \end{align*} given by \begin{align*} \overline{\pi}_n(a+\ker\pi_n)=\pi_n(a). \end{align*} Substituting $\ker\pi_n=n\mathbb{Z}$ and $\operatorname{im}\pi_n=\mathbb{Z}/n\mathbb{Z}$ gives \begin{align*} \overline{\pi}_n:\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}. \end{align*} On an element $a+n\mathbb{Z}$, this isomorphism satisfies \begin{align*} \overline{\pi}_n(a+n\mathbb{Z})=a+n\mathbb{Z}. \end{align*} So in this case the induced isomorphism is the identity map on the quotient, and the example shows explicitly how quotienting by the kernel produces the image. [/example] ## Isomorphisms and Classification Classification problems ask when two modules are the same after the names of their elements are forgotten. A module homomorphism is the allowed change of coordinates; an isomorphism is a reversible change of coordinates. The endomorphisms of a module are its structure-preserving self-maps. The next definition is needed because self-maps can be added and composed, producing an algebraic object attached to the module. [definition: Module Endomorphism] Let $R$ be a ring, and let $M$ be a left $R$-module. An $R$-module endomorphism of $M$ is an $R$-module homomorphism $f:M\to M$. [/definition] When studying a single module, the useful coordinate changes are not arbitrary endomorphisms. A non-invertible self-map can collapse distinct elements or lose information, so it cannot identify two descriptions of the same module without changing the structure being studied. This leads to a separate notion for the self-maps that can serve as genuine changes of coordinates on $M$. The reversible self-maps are exactly the endomorphisms that preserve all module-theoretic information in both directions; these are the symmetries of the module. [definition: Module Automorphism] Let $R$ be a ring, and let $M$ be a left $R$-module. An $R$-module automorphism of $M$ is an $R$-module isomorphism $f:M\to M$. [/definition] For a free module, homomorphisms should be controlled by basis elements, as in linear algebra. The practical problem is to replace a map defined on every element by the smaller data of where the basis goes, then extend by $R$-linear combinations. [quotetheorem:8334] This theorem is the module version of the familiar fact that a linear map out of a finite-dimensional vector space is determined by the images of a basis. It also explains why matrices appear whenever free modules are chosen. [example: A Homomorphism from $R^2$] Let $R$ be a commutative ring, regard $R$ as a left $R$-module over itself, and choose $a,b\in R$. Define $f:R^2\to R$ by \begin{align*} f(x,y)=ax+by. \end{align*} For $(x_1,y_1),(x_2,y_2)\in R^2$, addition in $R^2$ is componentwise, so \begin{align*} f((x_1,y_1)+(x_2,y_2))=f(x_1+x_2,y_1+y_2)=a(x_1+x_2)+b(y_1+y_2). \end{align*} By distributivity in $R$, \begin{align*} a(x_1+x_2)+b(y_1+y_2)=ax_1+ax_2+by_1+by_2. \end{align*} Reordering the four summands in the additive abelian group of $R$ gives \begin{align*} ax_1+ax_2+by_1+by_2=(ax_1+by_1)+(ax_2+by_2)=f(x_1,y_1)+f(x_2,y_2). \end{align*} Thus $f$ preserves addition. For $r\in R$ and $(x,y)\in R^2$, scalar multiplication in $R^2$ is componentwise, so \begin{align*} f(r(x,y))=f(rx,ry)=a(rx)+b(ry). \end{align*} By associativity of multiplication in $R$, \begin{align*} a(rx)+b(ry)=(ar)x+(br)y. \end{align*} Since $R$ is commutative, $ar=ra$ and $br=rb$, hence \begin{align*} (ar)x+(br)y=(ra)x+(rb)y=r(ax)+r(by). \end{align*} By distributivity again, \begin{align*} r(ax)+r(by)=r(ax+by)=r f(x,y). \end{align*} Therefore $f$ is an $R$-module homomorphism. Conversely, let $g:R^2\to R$ be any $R$-module homomorphism. Set \begin{align*} a=g(1,0) \end{align*} and \begin{align*} b=g(0,1). \end{align*} For every $(x,y)\in R^2$, \begin{align*} (x,y)=x(1,0)+y(0,1). \end{align*} Using additivity and scalar compatibility of $g$, \begin{align*} g(x,y)=g(x(1,0)+y(0,1))=xg(1,0)+yg(0,1)=xa+yb. \end{align*} Since $R$ is commutative, $xa+yb=ax+by$, so \begin{align*} g(x,y)=ax+by. \end{align*} Thus every homomorphism $R^2\to R$ is obtained from the two values on $(1,0)$ and $(0,1)$. The kernel is therefore \begin{align*} \ker f=\{(x,y)\in R^2:f(x,y)=0\}=\{(x,y)\in R^2:ax+by=0\}. \end{align*} So the kernel is exactly the submodule of solutions to the linear equation $ax+by=0$ over $R$. [/example] Over a field, invertibility of square matrices controls isomorphisms of finite-dimensional vector spaces. Over a general ring, the same sentence must be treated with care because matrices may fail to have inverses even when their determinants behave unexpectedly. [example: Multiplication by $2$ on $\mathbb{Z}$] Define $\mu_2:\mathbb{Z}\to\mathbb{Z}$ by $\mu_2(n)=2n$, regarding $\mathbb{Z}$ as a module over itself. For $m,n\in\mathbb{Z}$, \begin{align*} \mu_2(m+n)=2(m+n). \end{align*} By distributivity in $\mathbb{Z}$, \begin{align*} 2(m+n)=2m+2n. \end{align*} Since $\mu_2(m)=2m$ and $\mu_2(n)=2n$, this gives \begin{align*} \mu_2(m+n)=\mu_2(m)+\mu_2(n). \end{align*} For a scalar $r\in\mathbb{Z}$ and an element $n\in\mathbb{Z}$, \begin{align*} \mu_2(rn)=2(rn). \end{align*} By associativity and commutativity of multiplication in $\mathbb{Z}$, \begin{align*} 2(rn)=(2r)n=(r2)n=r(2n). \end{align*} Therefore \begin{align*} \mu_2(rn)=r(2n)=r\mu_2(n). \end{align*} Thus $\mu_2$ preserves addition and the $\mathbb{Z}$-scalar action, so it is a $\mathbb{Z}$-module homomorphism. The map is injective because if $\mu_2(m)=\mu_2(n)$, then \begin{align*} 2m=2n. \end{align*} Subtracting $2n$ from both sides gives \begin{align*} 2m-2n=0. \end{align*} Factoring the left side gives \begin{align*} 2(m-n)=0. \end{align*} Since $\mathbb{Z}$ has no zero divisors and $2\neq 0$, we get $m-n=0$, hence $m=n$. The map is not surjective. If $1$ were in the image of $\mu_2$, then there would be some $n\in\mathbb{Z}$ such that \begin{align*} \mu_2(n)=1. \end{align*} This means \begin{align*} 2n=1. \end{align*} But $2n$ is even for every $n\in\mathbb{Z}$, while $1$ is odd, so no such integer $n$ exists. Hence $1\notin\operatorname{im}\mu_2$, and $\mu_2$ is not surjective. Therefore $\mu_2$ is an injective $\mathbb{Z}$-module endomorphism that is not a $\mathbb{Z}$-module isomorphism. This shows that module endomorphisms over rings can behave differently from endomorphisms of finite-dimensional vector spaces over fields, where injectivity forces surjectivity. [/example] The contrast is one of the main reasons module theory cannot be treated as linear algebra with different letters. Rings may have torsion, zero divisors, and non-invertible nonzero elements, and homomorphisms detect all of these phenomena. ## Hom Sets and Functorial Behaviour Once we have homomorphisms between two fixed modules, it is natural to collect all of them. This collection is not merely a set: it has addition, and under commutativity assumptions it often has a module structure of its own. The next definition is needed because the totality of maps from $M$ to $N$ is itself a central object of study, especially in commutative algebra and homological algebra. [definition: Hom Set of Modules] Let $R$ be a ring, and let $M$ and $N$ be left $R$-modules. The hom set from $M$ to $N$ is \begin{align*} \operatorname{Hom}_R(M,N)=\{f:M\to N: f\text{ is an }R\text{-module homomorphism}\}. \end{align*} [/definition] The notation emphasizes the ring because the same function may be linear over one ring and not over another. In computations, changing the subscript can change the answer. Once the allowable maps have been collected, the first structural question is whether two such maps can be added pointwise and still remain $R$-linear. [quotetheorem:8335] Pointwise addition gives Hom an abelian-group structure. A further question is whether Hom can carry the same kind of scalar action as the modules it connects. Scaling a map by an element of $R$ means scaling its values in $N$, but for left modules this can interfere with $R$-linearity unless the scalars commute in the needed way. [quotetheorem:8336] This extra structure is central in commutative algebra. It lets Hom be studied as a module, not only as a collection of maps. [example: Homomorphisms from a Cyclic Quotient] Let $n\in\mathbb{N}$, let $M=\mathbb{Z}/n\mathbb{Z}$, and let $A$ be an abelian group viewed as a $\mathbb{Z}$-module. If $f:M\to A$ is a $\mathbb{Z}$-module homomorphism and $a=f(1+n\mathbb{Z})$, then for every $k\in\mathbb{Z}$, \begin{align*} k+n\mathbb{Z}=k(1+n\mathbb{Z}). \end{align*} By scalar compatibility of $f$, \begin{align*} f(k+n\mathbb{Z})=f(k(1+n\mathbb{Z}))=k f(1+n\mathbb{Z})=ka. \end{align*} Thus $f$ is determined by the single element $a=f(1+n\mathbb{Z})$. This element must satisfy $na=0$. Indeed, \begin{align*} n(1+n\mathbb{Z})=n+n\mathbb{Z}=0+n\mathbb{Z}. \end{align*} Applying $f$ and using scalar compatibility gives \begin{align*} na=n f(1+n\mathbb{Z})=f(n(1+n\mathbb{Z}))=f(0+n\mathbb{Z}). \end{align*} Since every module homomorphism preserves the additive identity, $f(0+n\mathbb{Z})=0_A$, so $na=0_A$. Conversely, suppose $a\in A$ satisfies $na=0_A$. Define $g:\mathbb{Z}/n\mathbb{Z}\to A$ by \begin{align*} g(k+n\mathbb{Z})=ka. \end{align*} This is well-defined: if $k+n\mathbb{Z}=\ell+n\mathbb{Z}$, then $k-\ell\in n\mathbb{Z}$, so $k-\ell=qn$ for some $q\in\mathbb{Z}$. Hence \begin{align*} ka-\ell a=(k-\ell)a=(qn)a=q(na)=q0_A=0_A. \end{align*} Therefore $ka=\ell a$. The map $g$ is additive because, for $k,\ell\in\mathbb{Z}$, \begin{align*} g((k+n\mathbb{Z})+(\ell+n\mathbb{Z}))=g((k+\ell)+n\mathbb{Z})=(k+\ell)a=ka+\ell a=g(k+n\mathbb{Z})+g(\ell+n\mathbb{Z}). \end{align*} It is compatible with the $\mathbb{Z}$-scalar action because, for $r,k\in\mathbb{Z}$, \begin{align*} g(r(k+n\mathbb{Z}))=g(rk+n\mathbb{Z})=(rk)a=r(ka)=r g(k+n\mathbb{Z}). \end{align*} Thus $g$ is a $\mathbb{Z}$-module homomorphism with \begin{align*} g(1+n\mathbb{Z})=1a=a. \end{align*} So choosing a homomorphism $\mathbb{Z}/n\mathbb{Z}\to A$ is the same as choosing an element $a\in A$ with $na=0_A$. Therefore $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z},A)$ identifies with the $n$-torsion subgroup \begin{align*} A[n]=\{a\in A:na=0_A\}. \end{align*} [/example] The example computed one Hom set, but larger constructions use homomorphisms in chains. The key obstruction is whether two compatible maps remain compatible after composition. Multi-step arguments with kernels, images, and quotients would break down if composition could lose the $R$-action. [quotetheorem:8337] This result is the categorical backbone of the subject. It allows long chains of maps to be studied using kernels, images, quotients, and exactness. ## Exactness and Universal Constructions A single homomorphism has a kernel and an image. A sequence of homomorphisms becomes meaningful when the image of one map is compared with the kernel of the next. This comparison is the language of exactness. The next definition is needed because many algebraic constructions are controlled not by a single map, but by whether consecutive maps fit together without missing or extra elements. [definition: Exact Sequence of Modules] Let $R$ be a ring. A sequence of left $R$-modules and $R$-module homomorphisms \begin{align*} \cdots \longrightarrow M_{i-1} \xrightarrow{f_{i-1}} M_i \xrightarrow{f_i} M_{i+1} \longrightarrow \cdots \end{align*} is exact at $M_i$ if \begin{align*} \operatorname{im} f_{i-1}=\ker f_i. \end{align*} The sequence is exact if it is exact at every module where both adjacent maps are present. [/definition] Exactness says that everything killed by the next map came from the previous map. The most common situation has only three nonzero modules: one included in an ambient module and one obtained as the corresponding quotient, with zeros marking the endpoints. [definition: Short Exact Sequence of Modules] Let $R$ be a ring. A short exact sequence of left $R$-modules is an exact sequence of the form \begin{align*} 0 \longrightarrow A \xrightarrow{f} B \xrightarrow{g} C \longrightarrow 0. \end{align*} [/definition] The two zeros encode injectivity and surjectivity. Thus a short exact sequence says that $A$ sits inside $B$, that $C$ is the corresponding quotient, and that the maps make this identification precise. Concretely, the sequence is short exact exactly when $f$ is injective, $g$ is surjective, and \begin{align*} \operatorname{im} f=\ker g. \end{align*} Equivalently, after identifying $A$ with the submodule $f(A)\subset B$, the module $C$ is isomorphic to the quotient $B/f(A)$ by the map induced from $g$. Short exact sequences are everywhere in module theory because they package submodules and quotients into a single object. The homomorphisms carry the information about how the pieces fit together. [example: A Quotient Short Exact Sequence] Let $R$ be a ring, let $M$ be a left $R$-module, and let $K\subset M$ be a submodule. Define $\iota:K\to M$ by $\iota(k)=k$ and $\pi:M\to M/K$ by $\pi(m)=m+K$, giving the sequence \begin{align*} 0 \longrightarrow K \xrightarrow{\iota} M \xrightarrow{\pi} M/K \longrightarrow 0. \end{align*} First, $\iota$ is an $R$-module homomorphism. For $k_1,k_2\in K$, \begin{align*} \iota(k_1+k_2)=k_1+k_2=\iota(k_1)+\iota(k_2). \end{align*} For $r\in R$ and $k\in K$, \begin{align*} \iota(rk)=rk=r\iota(k). \end{align*} It is injective because if $\iota(k_1)=\iota(k_2)$, then $k_1=k_2$ in $M$, hence also in the subset $K$. Next, $\pi$ is an $R$-module homomorphism. For $m_1,m_2\in M$, addition in the quotient module is defined by adding representatives, so \begin{align*} \pi(m_1+m_2)=(m_1+m_2)+K=(m_1+K)+(m_2+K)=\pi(m_1)+\pi(m_2). \end{align*} For $r\in R$ and $m\in M$, scalar multiplication in the quotient module is defined by $r(m+K)=rm+K$, hence \begin{align*} \pi(rm)=rm+K=r(m+K)=r\pi(m). \end{align*} The map $\pi$ is surjective because every element of $M/K$ has the form $m+K$ for some $m\in M$, and then \begin{align*} \pi(m)=m+K. \end{align*} Now compute the exactness conditions. The image of the zero map $0\to K$ is $\{0_K\}$, and \begin{align*} \ker\iota=\{k\in K:\iota(k)=0_M\}=\{k\in K:k=0_M\}=\{0_K\}. \end{align*} Thus the sequence is exact at $K$. At $M$, \begin{align*} \operatorname{im}\iota=\{\iota(k):k\in K\}=\{k:k\in K\}=K. \end{align*} Also, \begin{align*} \ker\pi=\{m\in M:\pi(m)=0_{M/K}\}=\{m\in M:m+K=0_M+K\}. \end{align*} By the equality criterion for cosets, $m+K=0_M+K$ exactly when $m-0_M\in K$, so \begin{align*} \ker\pi=\{m\in M:m\in K\}=K. \end{align*} Therefore $\operatorname{im}\iota=\ker\pi$. Finally, since $\pi$ is surjective, \begin{align*} \operatorname{im}\pi=M/K. \end{align*} The kernel of the zero map $M/K\to 0$ is all of $M/K$, because every coset maps to the zero element of the zero module. Hence \begin{align*} \operatorname{im}\pi=\ker(M/K\to 0). \end{align*} The sequence is exact at every displayed module, so it is a short exact sequence. [/example] Universal constructions are often recognised by how homomorphisms into or out of them behave. For a quotient $M/K$, the central obstruction is well-definedness: a map on cosets can only come from a map on $M$ when all elements of $K$ are sent to zero. [quotetheorem:8338] This theorem is the engine behind many quotient arguments. Instead of defining maps on equivalence classes by hand each time, we check that a submodule is killed and then invoke the universal property. ## Representations as Module Homomorphisms ### Actions as Structure Modules also appear when an algebraic object acts on a vector space. In representation theory, the action is part of the structure, not extra decoration. A map that is only linear may preserve vector-space operations while ignoring how the algebraic object moves vectors. ### Intertwining Maps Preserving the action is the main requirement on maps between representations. Such maps are called intertwining maps because applying the action before the map gives the same result as applying the map before the action. ### Compatibility with Lie Algebra Actions For Lie algebras, the scalar field and the [Lie algebra](/page/Lie%20Algebra) action both matter. A map between representations must be linear over the base field and compatible with the action of every Lie algebra element; otherwise it is not a homomorphism of the represented structure. [definition: $\mathfrak{g}$-Module Homomorphism] Let $k$ be a field, let $\mathfrak{g}$ be a Lie algebra over $k$, and let $V$ and $W$ be $\mathfrak{g}$-modules. A $\mathfrak{g}$-module homomorphism from $V$ to $W$ is a $k$-linear map $T:V\to W$ such that, for all $x\in\mathfrak{g}$ and all $v\in V$, \begin{align*} T(xv)=xT(v). \end{align*} [/definition] This is the same structural idea in a different costume: the map must commute with the action. The displayed identity says that acting first and then mapping gives the same result as mapping first and then acting. [example: Intertwiners Between One-Dimensional Lie Algebra Modules] Let $k$ be a field, let $\mathfrak{g}=kx$ be a one-dimensional abelian Lie algebra, and let $V=kv$ and $W=kw$ be one-dimensional $\mathfrak{g}$-modules with \begin{align*} xv=\lambda v \end{align*} and \begin{align*} xw=\mu w \end{align*} for scalars $\lambda,\mu\in k$. A $k$-linear map $T:V\to W$ is determined by the image of the basis vector $v$, so write \begin{align*} T(v)=aw \end{align*} for a unique scalar $a\in k$. To check the intertwining condition on $v$, compute the two sides separately. Since $xv=\lambda v$ and $T$ is $k$-linear, \begin{align*} T(xv)=T(\lambda v)=\lambda T(v)=\lambda(aw)=(\lambda a)w. \end{align*} On the other hand, since $T(v)=aw$ and the action of $x$ on $W$ is $k$-linear, \begin{align*} xT(v)=x(aw)=a(xw)=a(\mu w)=(a\mu)w. \end{align*} Because $k$ is commutative, $a\mu=\mu a$, so the equality $T(xv)=xT(v)$ is equivalent to \begin{align*} (\lambda a)w=(\mu a)w. \end{align*} Since $w$ is a basis vector of the one-dimensional vector space $W$, equality of these scalar multiples is equivalent to \begin{align*} \lambda a=\mu a. \end{align*} Subtracting $\mu a$ from both sides gives \begin{align*} (\lambda-\mu)a=0. \end{align*} Thus $T$ is a $\mathfrak{g}$-module homomorphism precisely when $(\lambda-\mu)a=0$. If $\lambda\neq\mu$, then $\lambda-\mu\neq 0$, so $\lambda-\mu$ is invertible in the field $k$; multiplying $(\lambda-\mu)a=0$ by $(\lambda-\mu)^{-1}$ gives $a=0$, hence $T(v)=0$ and $T$ is the zero map. If $\lambda=\mu$, then $(\lambda-\mu)a=0a=0$ for every $a\in k$, so every $k$-linear map $V\to W$ commutes with the action of $x$. [/example] This example is a small model for a broad phenomenon: homomorphisms between representations preserve the restrictions imposed by the action. The obstruction is that a kernel or image formed only as a vector-space construction might fail to be stable under the Lie algebra action. When a representation is viewed as a module over the algebra encoding the action, the ordinary module result on kernels, images, and quotients applies to representation homomorphisms. In that setting, kernels and images are subrepresentations, and the quotient by the kernel identifies with the image. ## Beyond and Connected Topics Module homomorphisms are the maps behind most of algebra after groups and rings. The next natural step is [modules](/page/Module) themselves: submodules, quotient modules, direct sums, free modules, and finitely generated modules all use homomorphisms as their transition maps. In commutative algebra, the functor $\operatorname{Hom}_R(M,N)$ becomes a central object. It leads to projective modules, injective modules, dual modules, tensor-Hom adjunctions, and eventually derived functors such as $\operatorname{Ext}_R^n(M,N)$. The course-level continuation is Androma's [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). In homological algebra, exact sequences are the basic syntax. Chain complexes are sequences of module homomorphisms whose consecutive composites vanish, and resolutions replace complicated modules by better-behaved ones. A natural continuation is [Homological Algebra I: Complexes and Resolutions](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions). In representation theory, module homomorphisms become intertwiners. For Lie algebras, this leads to subrepresentations, irreducibility, Schur-type results, and decomposition questions. The relevant Androma continuation is [Lie Algebras I: Foundations](/page/Lie%20Algebras%20I%3A%20Foundations). At the undergraduate level, module homomorphisms unify familiar maps from abelian groups, vector spaces, ideals, and quotient rings. The appropriate broader reference is [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules), where modules appear as a common language for linear algebra over rings. ## References Androma, [Module](/page/Module). Androma, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Androma, [Homological Algebra I: Complexes and Resolutions](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions). Androma, [Lie Algebras I: Foundations](/page/Lie%20Algebras%20I%3A%20Foundations). Atiyah and Macdonald, *Introduction to Commutative Algebra* (1969). Dummit and Foote, *Abstract Algebra* (2004). Lang, *Algebra* (2002). Rotman, *An Introduction to Homological Algebra* (2009).