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.