Let \(R\) be a ring. An \(R\)-module or module over \(R\) is an abelian group \((M, +)\) and a map \(\mu : R\times M \rightarrow M\) (scalar mulitplication) such that for all \(a,b \in R\) and \(x,y \in M\) we have

  1. \(a(x+y) = a x + a y\)

  2. \((a+b)x = a x + b x\)

  3. \((a b)x = a (b x)\)

  4. \(1 x = x\)

Alternatively, we may say that an \(R\)-module \(M\) is an abelian group with a ring homomophism \(R \rightarrow End(M)\) where \(End(M)\) is the ring of endomorphisms of \(M\).


  1. Any ideal \(I\triangleleft R\) is an \(R\)-module. (Scalar multiplication is ring multiplication.) In particular \(R\) is an \(R\)-module

2. If \(R\) is a field then \(R\)-modules are precisely the vector spaces over \(R\).

3. All abelian groups are \(\mathbb{Z}\)-modules.

4. Let \(R = K[x]\) for some field \(K\). Then an \(R\)-module is a \(K\)-vector space together with some linear transformation.

5. Let \(K\) be a field, \(G\) be a group and consider the group ring \(R = K[G]\). Then \(R\) modules are precisely \(K\)-representations of \(G\).

A mapping \(f:M\rightarrow N\) between \(R\)-modules \(M,N\) is an \(R\)-module homomorphism or \(R\)-linear if addition and scalar multiplication are preserved, that is for all \(x,y \in M\) and \(a \in R\) we have \(f(x+y) = f(x) + f(y), f(a x ) = a f(x)\). Alternatively we may say \(f\) is a homomorphism between abelian groups that respects the actions of the ring.

Let \(Hom_R (M,N)\) be the set of all \(R\)-module homomorphisms from \(M\) to \(N\). For all \(f,g \in Hom(M,N), a \in R\), define \(f+g\) and \(a f\) by \((f + g)(x) = f(x) + g(x), (a f) (x) = a f(x)\). Then it can be easily verified that \(Hom(M,N)\) is an \(R\)-module.

Let \(u:M'\rightarrow M\) and \(v:N\rightarrow N'\) be \(R\)-module homomorphisms. They induce \(R\)-module homomorphisms \(\bar(u) = f u : Hom(M,N)\rightarrow Hom(M',N)\) and \(\bar(v) = v f : Hom(M,N) \rightarrow Hom(M,N')\).

Example: If \(R\) is a field then \(R\)-module homomorphisms are linear transformations which may be written as matrices. The induced homomorphisms can be computed via matrix multiplications.

For any \(R\)-module \(M\) we have \(Hom_R(R,M) \cong M\). This can be seen by considering the map given by \(f \mapsto f(1)\) for all \(f \in Hom_R(R,M)\).

Ben Lynn 💡